(a) Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) 401.15
(b)How much work (in J) does she do if her center of mass rises 0.260 m?
(c) What is her useful power output (in W) if she does 30 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)
Answers
the force the woman exerts to do a push-up at constant speed is 333 N.
the work the woman does is 152 J.
her useful power output is 76 W.
(a) To calculate the force the woman exerts to do a push-up, we need to use torque methods. The woman is doing a push-up at constant speed, which means that the net torque on her body is zero. The only torque acting on her body is due to her weight W, which acts at the center of mass of her body. The distance between her center of mass and her hands is 0.76 m, and the angle between her body and the horizontal is 45 degrees.
The torque due to her weight about her hands is given by:
τ = r x W = (0.76 m) x (cos 45°)(W)
where r is the distance between her hands and her center of mass and cos 45° is the component of the distance perpendicular to the weight vector. Since the woman is at constant speed, the torque she exerts about her hands must be equal and opposite to the torque due to her weight. Therefore:
τ = (0.76 m)(cos 45°)(W) = (1/2)(W)(0.76 m)
Solving for W, we get:
W = 2(τ/0.76 m) = 2[(0.5)(mg)(0.76 m)/(0.76 m cos 45°)] = 333 N
Therefore, the force the woman exerts to do a push-up at constant speed is 333 N.
(b) The work the woman does is equal to the change in her potential energy as her center of mass rises. The woman's mass is not given, so we will assume a value of 60 kg. The gravitational potential energy of the woman is given by:
U = mgh
where m is the mass of the woman, g is the acceleration due to gravity (9.81 m/s^2), and h is the height her center of mass rises (0.26 m). Therefore:
U = (60 kg)(9.81 m/s^2)(0.26 m) = 152 J
Therefore, the work the woman does is 152 J.
(c) The useful power output of the woman is the work she does per unit time, taking into account the work done in lowering her body. Each push-up involves two phases: lifting her body and lowering her body. When she lowers her body, the work done is negative, as the force she exerts is in the opposite direction to the displacement. The work done in lowering her body is equal to the work done in lifting her body, so the total work done in one push-up is zero.
The woman does 30 push-ups in 1 minute, which means she does one push-up every 2 seconds. Therefore, the useful power output of the woman is:
P = (152 J)/(2 s) = 76 W
Therefore, her useful power output is 76 W.
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Related Questions
Ten identical balls made of steel, each 27g are immersed in a measuring cylinder containing 20cm3 of water. The water level rises to a reading of 50cm3. What is the density of the steel?
Answers
Ten steel balls, each weighing 27g, are placed in a measuring cylinder filled with 20cm3 of water. The measurement of the water level increases to 50cm3.Then the density of the steel is 9 g/cm³.
Since the steel balls are fully immersed in water, the amount of water they displaced is equal to the volume of balls.
The volume of water before the balls were added is 20 cm³.
The water level rises to a reading of 50cm³ which means it is the total volume of water and the steel balls together. Therefore, the volume of the steel balls is:
50 cm³ - 20 cm³ = 30 cm³
It is given that there are ten identical balls, so the volume of each ball is:
30 cm³ ÷ 10 = 3 cm³
Each ball has a mass of 27 g.
So, the density of steel is calculated by following formula:
Density = Mass ÷ Volume
Density = 27 g ÷ 3 cm³
Density = 9 g/cm³
Therefore, the density of the steel balls is 9 g/cm³.
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A particular star is d = 56.1 light-years (ly) away, with a power output of P = 3.00 x 1026 W. Note that one light-year is the distance traveled by the light through a vacuum in one year. (a) Calculate the intensity of the emitted light at distance d (in nW/m2). nW/m2 (b) What is the power of the emitted light intercepted by the Earth (in kW)? (The radius of Earth is 6.37 x 10 m.) kw What If?
Answers
(a) Intensity of emitted light at distance d is approximately 1.95 nW/m². (b) Power of emitted light intercepted by Earth is approximately 3.67 kW.
How did you calculate the intensity of emitted light in part (a)?The intensity of emitted light at distance d was calculated using the inverse square law, which states that intensity is inversely proportional to the square of the distance. I used the equation I = P/(4πd²) to calculate the intensity, where P is the power output of the star and d is the distance to the star.
How did you calculate the power of emitted light intercepted by Earth in part (b)?To calculate the power of emitted light intercepted by Earth, I first calculated the surface area of a sphere with a radius equal to the distance between the Earth and the star. Then, I multiplied the intensity of the emitted light at that distance by the surface area of the sphere to get the total power of emitted light intercepted by Earth.
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If the universe were to suddenly begin shrinking rather than continue expanding, how would it affect the cosmic microwave background radiation?
A. It would decrease in temperature.
B. It would blue-shift.
C. It would red-shift.
D. It would increase in temperature.
Answers
C. It would red-shift. The cosmic microwave background radiation (CMB) is the leftover radiation from the Big Bang.
What is radiation?Radiation is the process by which energy is emitted from a source and travels through a medium or space. It is a form of energy that can be found in a variety of forms, such as light, heat, and sound. Radiation can also be used to refer to the release of particles or electromagnetic waves from an atomic nucleus during radioactive decay. These particles and waves can be harmful to humans and animals if they are too strong or long lasting. Radiation is also used in medical, industrial, and scientific applications to diagnose and treat diseases, sterilize products, and to provide energy for power plants.
It is composed of a low-energy form of light called microwaves. If the universe were to start shrinking, this microwaves would be stretched out (red-shifted) as the universe gets smaller. This red-shifting would cause the CMB to become less energetic, and therefore decrease in temperature.
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. If the same quantity of is supplied to P and Q and P has a temperature rise twice that of Q and has a mass that is half of the mass of Q. Find the ratio of the specific heat capacity of P to Q
Answers
The ratio of the specific heat capacity of P to Q is 1:2, or cP/cQ = 1/2.
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It is a physical property that helps characterize the thermal behavior of materials.
Let's use the formula for heat energy:
Q = mcΔT
where Q is the heat energy supplied, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
We can write two equations for P and Q using this formula:
For P: Q = (1/2)mP * cP * 2ΔT
For Q: Q = mQ * cQ * ΔT
We know that both P and Q receive the same amount of heat energy, so we can equate the two equations:
(1/2)mP * cP * 2ΔT = mQ * cQ * ΔT
Simplifying this expression, we get:
cP/cQ = mQ/(2mP)
We are given that P has half the mass of Q, so mP = (1/2)mQ. Substituting this into the expression above, we get:
cP/cQ = mQ/(2(1/2)mQ) = 1/2
Therefore, the ratio of the specific heat capacity of P to Q is 1:2, or cP/cQ = 1/2.
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One elevator arrangement includes the passenger car, a counterweight, and two large pulleys, as shown in Figure 11-50. Each pulley has a radius of 1.2 m and a moment of inertia of 380 kg•m². The top pulley is driven by a motor. The elevator car plus passengers has a mass of 3100 kg, and the counterweight has a mass of 2700 kg. If the motor is to accelerate the elevator car upward at 1.8 m/s², how much torque must it generate?
Answers
The torque generated by the motor, accelerating with 1.8 m / Of the elevator is, 18372 Nm.
What is torque?
Torque is the term for the force that rotates an object about any axis. It is equivalent to the product of force and perpendicular distance in mathematical terms.
Given: Radius of pulley (r) = 1.2 meters;
Mass of counterweight (W) = 2700 kg;
Mass of elevator plus passenger (M) = 3100 kg;
Moment of inertia of pulley (I) = 380 kg
To calculate torque:
Now if the rope tension is then
T₁ = 3100 × (9.8 + 1.8)
T₁ = 35960 N
Now Angular acceleration (α) = a / r
α = 1.8 / 1.2
α = 1.5 rad / sec
If T₂ is the counterweight tension then
T₂ = 2700 × (9.8 - 1.8) N
T₂ = 21600 N
If T is the total torque then
T = 2 × 380 × 1.5 + (35960 - 21600) ×1.2
T = 18372 Nm
Therefore, the total torque that is generated through the given measurement is, 18372 Nm.
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In which of the following scenarios is the left hemisphere of the brain primarily needed?
Answers
The left hemisphere of the brain is primarily needed in scenario, Solving a complex mathematical problem. Option a is correct.
The brain is divided into two hemispheres, left and right, and they are specialized for different cognitive functions. The left hemisphere of the brain is primarily responsible for language processing, logical reasoning, and analytical thinking. Solving a complex mathematical problem involves logical reasoning, analytical thinking, and the use of language, all of which are primarily controlled by the left hemisphere of the brain.
Mathematical problems often require precise calculations, sequencing of steps, and the use of symbols and formulas, all of which require a strong left-brain function. In contrast, appreciating a work of art, listening to music, and recognizing facial expressions are all more complex perceptual and emotional processes that involve the right hemisphere of the brain. Option a is correct.
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--The complete question is, In which of the following scenarios is the left hemisphere of the brain primarily needed?
a. Solving a complex mathematical problem.
b. Appreciating a work of art
c. Listening to music
d. Recognizing facial expressions.--
v=√gr tan 31.0 grados
Answers
The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination.
v = √rg tanθ
tanθ=v²/rg
The relation gives the angle of banking of the cyclist going round the curve. Here v is the speed of the cyclist, r is the radius of the curve, and g is the acceleration due to gravity.
The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination. The normal force acting on the car while travelling through such a curving road has a horizontal component. The centripetal force needed to prevent skidding is provided by this component.
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The Mako rollercoaster at Sea World has a starting drop of 61m from the ground. What is the velocity of the 500kg passenger cart if it passes over a second hump that is 20m off the ground?
Answers
mgh1 = (1/2)mv^2 + mgh2
v = sqrt(2gh1 - 2gh2)
Substituting the values we get:
v = sqrt(29.861 - 29.820) ≈ 29.3 m/s
Therefore, the velocity of the passenger cart at the top of the second hump is approximately 29.3 m/s.
In this problem you will go through a simplified version of Rutherford’s calculation of the size of the gold nucleus. Suppose a piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all -particles go through undeflected. The density of gold is 19,300 kg/m3. Determine the radius of the gold nucleus. Hint: first calculate the total number of gold atoms in the foil
Answers
A simplified version of Rutherford’s calculation of the size of the gold nucleus. A piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all particles go through undeflected. The density of gold is 19,300 kg/m3.
Rutherford's experiment involved firing alpha particles (helium nuclei) at a thin sheet of gold foil to study the structure of atoms. Based on the results of this experiment, Rutherford was able to deduce that atoms have a small, dense nucleus at their center.
In this problem, we will go through a simplified version of Rutherford's calculation of the size of the gold nucleus.
First, we need to calculate the total number of gold atoms in the foil. We know that the foil is 0.010 cm thick and has an area of 1 cm x 1 cm, so its volume is
V = thickness x area = 0.010 cm x (1 cm x 1 cm) = 0.010 [tex]cm^{3}[/tex]
The density of gold is 19,300 kg/[tex]m^{3}[/tex], which is equivalent to 19.3 g/[tex]cm^{3}[/tex]Therefore, the mass of the gold foil is
m = density x volume = 19.3 g/[tex]cm^{3}[/tex] x 0.010 [tex]cm^{3}[/tex] = 0.193 g.
The molar mass of gold is 197 g/mol, so the number of gold atoms in the foil is
N = (0.193 g) / (197 g/mol) x (6.022 x [tex]10^{23}[/tex] atoms/mol) = 1.86 x [tex]10^{21}[/tex] atoms
Next, we need to determine the fraction of alpha particles that are deflected by the gold nucleus. We are told that 99.93% of all alpha particles go through undeflected, which means that only 0.07% of the alpha particles are deflected. This is a very small fraction, which suggests that the size of the gold nucleus must be very small compared to the size of the atom.
Assuming that the alpha particles are deflected only by the gold nucleus and not by the electrons, we can use the principle of conservation of momentum to estimate the size of the gold nucleus. When an alpha particle approaches the gold nucleus, it experiences a repulsive electrostatic force that causes it to change direction. The magnitude of this force is given by Coulomb's law
F = k[tex]q_{1}[/tex][tex]q_{2[/tex] / [tex]r^{2}[/tex]
Where k is Coulomb's constant, [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the charges of the alpha particle and gold nucleus, respectively, and r is the distance between them. Since the alpha particle has a positive charge and the gold nucleus has a positive charge, the force is repulsive.
If we assume that the alpha particle is initially moving directly toward the center of the gold nucleus, then at the point of closest approach, the alpha particle will have a velocity v that is perpendicular to the direction from the alpha particle to the gold nucleus. At this point, the force on the alpha particle will be perpendicular to its velocity, which means that it will change only the direction of the alpha particle's velocity, not its magnitude.
Using conservation of momentum, we can relate the angle of deflection θ to the distance of closest approach r.
m[tex]v^{2}[/tex] / r = k[tex]q_{1}[/tex][tex]q_{2[/tex] / [tex]r^{2}[/tex]
Where m is the mass of the alpha particle. Solving for r, we get
r = k[tex]q_{1}[/tex][tex]q_{2[/tex] / m[tex]v^{2}[/tex]
To estimate the size of the gold nucleus, we assume that the alpha particles are deflected by a single, stationary gold nucleus at the center of the atom. In reality, the gold nucleus is not stationary, but this assumption gives us a rough estimate of its size.
Hence, the alpha particles are undeflected with a probability of 0.9993, we can assume that they do not interact with the gold nucleus and that their path is a straight line through the foil.
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Jeremy is developing an experiment and is concerned about the accuracy of his data.
Which step can he take to best ensure accuracy?
conduct more trials
compare his results to other scientists’ results
record only the data that matches the correct value
make exact measurements and follow the procedure exactly
Answers
Answer:
make exact measurements and follow the procedure exactly
Explanation:
accurate doesn't mean correct. It just means that the results from the experiment are the result of following the procedure
Use the heating curve (Figure 1) to answer the question.
Temperature (C)
200
150
100
50
0
-50
-100
-150
0 200 400 600 800 1000 1200
Energy (J)
What do the horizontal sections represent?
They represent where no energy was added.
They represent where phase changes occur.
MacBook Air
They represent where particles moved faster.
They represent where the temperature changed.
Answers
They represent where phase changes occur.
A current of 1.4A when flowing through a circuit for 15 minutes dissipates 200 KJ of energy calculate (a) the p.d (b) power dissipated and (c) the resistance of the circuit.
Answers
When a current of 1.4A when flowing through a circuit for 15 minutes dissipates 200 KJ of energy then he p.d is 158 Volt , Power dissipated and the resistance of the circuit is 222 W and 112Ω resp.
Electric circuit, a channel for carrying electric current. An electric circuit consists of a device that provides energy to the charged particles that make up current, such as a battery or a generator, as well as equipment that consume current, such as lights, electric motors, or computers, and the connecting wires or transmission lines. Ohm's law and Kirchhoff's rules are two fundamental laws that quantitatively define the behaviour of electric circuits.
Given,
Current I = 1.4 A
Energy Dissipated E = 200 kJ
time t = 15 minute = 900s
Power dissipated in the circuit,
P = E/t = 200000/900 = 222 W
Power is a voltage times current,
P =VI
222 = 1.4 × V
V = 158 Volt
according to ohms law
V = RI
R = V/I = 158/1.4 = 112Ω.
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25.
The sensory hair cells of the semicircular canals convert vibrations into electrical impulses.
O True
False
Answers
It is a true statement that the sensory hair cells of the semicircular canals convert vibrations into electrical impulses.
How does sensory hair cells of the semicircular canals convert vibrations?The sensory hair cells located in the semicircular canals of the inner ear are responsible for converting mechanical vibrations caused by head movements into electrical impulses that are then transmitted to the brain for processing. This process helps the body maintain balance and spatial orientation.
The semicircular canals are part of the vestibular system in the inner ear, which plays a crucial role in detecting changes in head position and movement. The sensory hair cells in the semicircular canals are embedded in a gelatinous structure called the cupula, which moves in response to the flow of endolymphatic fluid inside the canals.
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Please help with the physics problem(single-loop circuits), provide step-by-step solution
Answers
Resistance of the resistors, R₁ and R₂ are 80Ω and 200Ω respectively.
V(A) = 12 V
ΔV(B) = 2 V
ΔV(C) = 5 V
R₃ = 200Ω
Since, the total voltage-drop along the upper branch must be 12 V, according to loop rule, the voltage-drop across resistor 3 is 5.0V
So, current through the loop,
I = V(C)/R₃
I = 5/200
I = 25 x 10⁻³A
a) Therefore, the resistance,
R₁ = V(B)/I
R₁ = 2/25 x 10⁻³
R₁ = 80Ω
b) Resistor 2 has the same voltage-drop as resistor 3. So, its resistance, R₂ is 200Ω.
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A 2.99 x 10-6 C charge is moving
perpendicular (90°) to the Earth's
magnetic field (5.00 x 10-5 T). If the
force on it is 2.14 x 10-8 N, how fast is
it moving?
[?] m/s
Velocity (m/s)
Enter
Answers
The charged particle is moving at a speed of 1.424 x 10³ m/s.
What is magnetic field?In the vast expanse of space, a magnetic force can be detected within a region known as a magnetic field. This field is generated by either a magnet, an electric charge in motion, or an electric field undergoing change. To visually represent this force, it is depicted through directional field lines which illustrate the direction of the force on an imaginary magnetic pole situated at any point in space.
The intensity of this magnetic field is quantified by its measurement in units called Tesla (T).
Equation:The force on a charged particle moving perpendicular to a magnetic field is given by the equation:
F = Bqv
where F is the force on the charge, B is the magnetic field strength, q is the charge of the particle, and v is its velocity.
Rearranging this equation, we can solve for v:
v = F / (B*q)
Substituting the given values, we get:
v = (2.14 x 10⁻⁸ N) / (5.00 x 10⁻⁵T * 2.99 x 10⁻⁶ C)
v = 1.424 x 10³ m/s
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Answer:142
Explanation: on acellus, and maybe on what ever else you are on, but i do know it's right on acellus!! :)
32.
Passive sonar uses underwater microphones called hydrophones.
True
False
Answers
Answer:
True
Explanation:
"Passive sonar uses specialized transducers called hydrophones (or underwater microphones) to listen to sounds in the ocean. These hydrophones convert received sounds into electrical signals that are then sent to a computer for a sonar operator to look at, listen to, and analyze."
no use of bots . anything bot will be reported
Answers
Period, T [tex]\Longrightarrow T=\frac{60}{14} \Longrightarrow T \approx 4.286 \ s[/tex]
Using the equation, [tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex], find l. ("g" is the acceleration from gravity, 9.8 m/s^2)
[tex]T=2 \pi \sqrt{\frac{l}{g} } \Longrightarrow 4.286=2 \pi \sqrt{\frac{l}{9.8} } \Longrightarrow \frac{4.286}{2 \pi} =\sqrt{\frac{l}{9.8} }[/tex]
[tex]\Longrightarrow (\frac{4.286}{2 \pi})^2 =\frac{l}{9.8} \Longrightarrow l =9.8(\frac{4.286}{2 \pi})^2 \Longrightarrow l \approx 4.560 \ m[/tex]
Thus, the radio booth is 4.560 m from the field.
Period, T [tex]\Longrightarrow T=\frac{60}{14} \Longrightarrow T \approx 4.286 \ s[/tex]
Using the equation, [tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex], find l. ("g" is the acceleration from gravity, 9.8 m/s^2)
[tex]T=2 \pi \sqrt{\frac{l}{g} } \Longrightarrow 4.286=2 \pi \sqrt{\frac{l}{9.8} } \Longrightarrow \frac{4.286}{2 \pi} =\sqrt{\frac{l}{9.8} }[/tex]
[tex]\Longrightarrow (\frac{4.286}{2 \pi})^2 =\frac{l}{9.8} \Longrightarrow l =9.8(\frac{4.286}{2 \pi})^2 \Longrightarrow l \approx 4.560 \ m[/tex]
Thus, the radio booth is 4.560 m from the field.
A 6.47 mm high firefly sits on the axis of, and 12.3 cm in front of, the thin lens A, whose focal length is 5.77 cm. Behind lens A there is another thin lens, lens B, with a focal length of 20.9 cm. The two lenses share a common axis and are 58.1 cm apart.What is the height of this image? Express the answer as a positive number.
Answers
A 6.47 mm high firefly sits on the axis of, and 12.3 cm in front of, the thin lens A, whose focal length is 5.77 cm, the height of the image is 2.61 mm.
We can use the thin lens equation to find the image distance and then use the magnification equation to find the height of the image.
Let's call the distance between the firefly and lens A "d1", the distance between the lenses "d2", the image distance from lens B "d3", and the height of the firefly "h1".
Using the thin lens equation for lens A:
1/fA = 1/d1 + 1/d3
Since the firefly is very small, we can assume that the rays of light from it are parallel to the axis of the lenses, so d1 = 12.3 cm.
Solving for d3, we get:
1/d3 = 1/fA - 1/d1
1/d3 = 1/5.77 cm - 1/12.3 cm
d3 = -23.46 cm
The negative value for d3 indicates that the image is formed on the same side of lens B as the firefly, which means it is a virtual image.
Now we can use the magnification equation:
m = -d3/d2
where m is the magnification of the image. The negative sign indicates that the image is inverted.
Using the distance between the lenses, d2 = 58.1 cm, we get:
m = -(-23.46 cm) / 58.1 cm
m = 0.403
This tells us that the image is smaller than the firefly, and its height is:
h2 = m * h1
h2 = 0.403 * 6.47 mm
h2 = 2.61 mm
Therefore, the height of the image is 2.61 mm.
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Suppose studentsA,B,C, and D are running a 100meter race student A takes12.5 seconds ,student B takes 10 seconds , students C takes 14 seconds in student D takes 15 seconds . calculate their respective velocities
Answers
Answer:
Va=8m/s Vb =10m/s Vc=7.14m/s Vs=6.67
Explanation:
#student A runs 100m in 12.5 seconds so in average he run 8 m in every second
#student B runs 100m in 10 seconds those means I'm average the student run 10m in each second.
#student C runs 100m in 14seconds so in average he/she run approximately 7.14m in every second.
#student D run 100m in 15secends so in average the student run approximately 6.67m in each second.
11. A car, starting from rest, accelerates in a straight-line path at a constant rate of 2,5 m/s2 . How far will the car travel in 10 seconds?
(a) 180 m
(b) 30 m
(c) 125 m
(d) 4.8 m
Answers
Answer: 125 m
Explanation:
[tex]V_{i}[/tex] = 0
[tex]V_{f}[/tex] = no se conoce
a = 2,5 [tex]\frac{m}{s^{2} }[/tex]
t = 10 s.
d= x no los piden
Formula:
[tex]d=V_{i} .t + \frac{a.t^{2} }{2}[/tex]
Remplazamos:
[tex]d=0.(10) + \frac{2,5.10^{2} }{2}[/tex] = 0 + [tex]\frac{2,5.(100)}{2}[/tex] = 2,5.(50) = 125 m
f q1 has the same magnitude as before but is negative, in what region along the x-axis would it be possible for the net electric force on q3 to be zero? (a) x , 0 (b) 0 , x , 2 m (c) 2 m , x
Answers
(b) 0, x, 2m
Explanation:
Let's first draw a diagram to help visualize the situation. Let's say that q1 is located at the point x=2m and q3 is located at the origin (x=0).
q1 q3
|-------------------|
2 meters
Since the magnitude of q1 is the same as before, the magnitude of the electric force that it exerts on q3 is also the same as before.
However, since q1 is now negative, the direction of the electric force is also reversed.
To find the region along the x-axis where the net electric force on q3 is zero, we need to find the point where the electric force due to q1 is equal in magnitude but opposite in direction to the electric force due to q3.
This means that the electric field vectors due to q1 and q3 must be pointing in opposite directions at this point.
Since the electric field vectors due to q1 and q3 both point radially away from the point charges, the point where they point in opposite directions must lie somewhere along the line that passes through q1 and q3.
Using our diagram, we can see that the point where the electric field vectors due to q1 and q3 point in opposite directions must lie somewhere between x=0 and x=2m, since this is the region where the line passing through q1 and q3 intersects the x-axis.
12) The notion of task-appropriate processing implies that if you are preparing for a quiz you
should...
Answers
2. Make a study group
3. Practice remembering everything
Which of the following is not an application of ultrasonic waves?
acoustic amplification
medical imaging
O echolocation
nondestructive testing
Answers
Acoustic amplification is not an application of ultrasonic waves.
What are Ultrasonic waves used for?Ultrasonic waves are used in many different applications such as medical imaging, where they are used to create images of the internal structures of the human body.
Echolocation, which is used by animals such as bats and dolphins to navigate their environment, also relies on the use of ultrasonic waves. Additionally, ultrasonic waves are used in nondestructive testing to detect flaws or defects in materials without damaging them.
Acoustic amplification, on the other hand, involves the use of sound waves to amplify or enhance the sound produced by a musical instrument or a speaker. It does not involve the use of ultrasonic waves.
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A particle was moving in a straight line with a constant acceleration. If the particle
covered 17 m in the 2nd second and 46 m in the 9th and 10th seconds, calculate its
acceleration a and its initial velocity vo.
Answers
Answer:
Acceleration (a) = 3 m/s^2
Initial velocity (vo)= 7 m/s.
Explanation:
We can use the equations of motion to solve this problem. Let's start with the second equation of motion:
d = vot + (1/2)at^2
where d is the displacement, vo is the initial velocity, a is the acceleration, and t is the time.
Using this equation for the 2nd second, we have:
17 = vo(2) + (1/2)a(2^2)
17 = 2vo + 2a
Using the same equation for the 9th and 10th seconds, we have:
46 = vo(10) + (1/2)a(10^2) - vo(9) - (1/2)a(9^2)
46 = 10vo + 50a - 9vo - 40.5a
46 = vo + 9.5a
Now we have two equations with two unknowns (vo and a). We can solve for one variable in terms of the other and substitute into the other equation. For example, we can solve the first equation for vo:
2vo = 17 - 2a
vo = (17 - 2a)/2
Now we can substitute this expression for vo into the second equation:
46 = [(17 - 2a)/2] + 9.5a
Solving for a, we get:
a = 3 m/s^2
Now we can use the expression for vo to find its value:
vo = (17 - 2a)/2
vo = (17 - 2(3))/2
vo = 7 m/s
Therefore, the acceleration of the particle is 3 m/s^2 and its initial velocity is 7 m/s.
Answer:
Acceleration (a) = 3 m/s^2
Initial velocity (vo)= 7 m/s.
Explanation:
We can use the equations of motion to solve this problem. Let's start with the second equation of motion:
d = vot + (1/2)at^2
where d is the displacement, vo is the initial velocity, a is the acceleration, and t is the time.
Using this equation for the 2nd second, we have:
17 = vo(2) + (1/2)a(2^2)
17 = 2vo + 2a
Using the same equation for the 9th and 10th seconds, we have:
46 = vo(10) + (1/2)a(10^2) - vo(9) - (1/2)a(9^2)
46 = 10vo + 50a - 9vo - 40.5a
46 = vo + 9.5a
Now we have two equations with two unknowns (vo and a). We can solve for one variable in terms of the other and substitute into the other equation. For example, we can solve the first equation for vo:
2vo = 17 - 2a
vo = (17 - 2a)/2
Now we can substitute this expression for vo into the second equation:
46 = [(17 - 2a)/2] + 9.5a
Solving for a, we get:
a = 3 m/s^2
Now we can use the expression for vo to find its value:
vo = (17 - 2a)/2
vo = (17 - 2(3))/2
vo = 7 m/s
Therefore, the acceleration of the particle is 3 m/s^2 and its initial velocity is 7 m/s.
Explanation:
i need too get them right because me and my friend are challeging each other
Answers
2. Hooke saw tiny openings in cork and named them "cells", 3. Schleiden and Schwann discovered that all "organisms" are made of cells, 4. Schleiden and Schwann used microscopes to determine that the cell is the "basic unit" of life, 6. Water - do not mix with lipids, 7. Nucleic acids - contain instructions, 8. Proteins - some help break down nutrients, 9. Lipids - do not mix with water, and 10. Carbohydrates - sugar is one.
Water is a polar molecule, meaning it has a positive and negative end. Because of this polarity, water does not mix well with non-polar substances such as lipids. This property allows lipids to form membranes that can control the movement of substances in and out of cells.
Nucleic acids are large biomolecules that contain instructions for the development, function, and reproduction of all living organisms. The two main types of nucleic acids are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), which store and transmit genetic information.
Proteins are complex molecules made up of amino acids that perform a variety of functions in the body. Some proteins, called enzymes, help break down nutrients into smaller molecules that the body can use for energy and other processes.
Lipids are a group of biomolecules that are insoluble in water but are soluble in nonpolar solvents. They include fats, oils, and waxes, and are important for energy storage, insulation, and cell membrane structure.
Carbohydrates are biomolecules that contain sugars, such as glucose and fructose. They are an important source of energy for the body and are found in many foods such as fruits, vegetables, and grains.
Therefore, The correct answers are 2. Hooke saw tiny openings in cork and named them "cells", 3. Schleiden and Schwann discovered that all "organisms" are made of cells, 4. Schleiden and Schwann used microscopes to determine that the cell is the "basic unit" of life, 6. Water - do not mix with lipids, 7. Nucleic acids - contain instructions, 8. Proteins - some help break down nutrients, 9. Lipids - do not mix with water, and 10. Carbohydrates - sugar is one.
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Please help me in this exercise.
Answers
A. We can see here that the girl has kinetic energy with respect to the escalator.
B. The kinetic energy doesn't depend on the chosen reference.
What is kinetic energy?Kinetic energy is the energy that an object possesses due to its motion. Any object that is in motion has kinetic energy, which is determined by its mass and velocity. The formula for kinetic energy is KE=1/2mv^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity or speed of the object.
Kinetic energy is a scalar quantity, meaning it has only magnitude and no direction. The unit of kinetic energy is Joules (J) in the International System of Units (SI).
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A bag of sand weighing 5kg is suspended from the lower end of a rope. The bag is initially at rest. A 20.0 g bullet is fired at the bag with horizontal velocity of 650 ms¹, strikes the block, and exits with 100 ms. To what vertical height will the block be raised?
Answers
Let's define the positive direction as upward. Then, the initial momentum of the bullet is:
p_i = m_bullet * v_bullet = 0.02 kg * 650 m/s = 13 kg m/s
After the collision, the bullet and bag move together as a single system. Let's assume that the final velocity of the bullet-bag system is v_f, and the mass of the bullet and bag combined is:
m_total = m_bullet + m_bag = 0.02 kg + 5 kg = 5.02 kg
Using the conservation of momentum, we can equate the initial and final momenta:
p_i = p_f
m_bullet * v_bullet = m_total * v_f
Solving for v_f, we get:
v_f = (m_bullet * v_bullet) / m_total
v_f = (0.02 kg * 650 m/s) / 5.02 kg
v_f = 2.6 m/s
The final velocity of the bullet-bag system after the collision is 2.6 m/s. To find the height to which the bag is raised, we can use the principle of conservation of energy. The initial energy of the system is all gravitational potential energy stored in the bag, and the final energy of the system is the sum of the gravitational potential energy of the raised bag and the kinetic energy of the bullet-bag system:
m_bag * g * h = (1/2) * m_total * v_f^2
where g is the acceleration due to gravity (9.81 m/s^2), and h is the height the bag is raised.
Solving for h, we get:
h = [(1/2) * m_total * v_f^2] / (m_bag * g)
h = [(1/2) * 5.02 kg * (2.6 m/s)^2] / (5 kg * 9.81 m/s^2)
h = 0.337 m
Therefore, the bag is raised to a vertical height of about 0.337 meters when hit by the bullet.
2) The velocity of three particles of masses 20g, 30g and 50g are 2i, 10j and 10k respectively. The velocity of the centre of mass the three particle) is
Answers
v_COM = (m1v1 + m2v2 + m3v3) / (m1 + m2 + m3)
where m1, m2, and m3 are the masses of the particles, and v1, v2, and v3 are their velocities.
Substituting the given values, we get:
v_COM = (0.02 kg)(2i m/s) + (0.03 kg)(10j m/s) + (0.05 kg)(10k m/s) / (0.02 kg + 0.03 kg + 0.05 kg)
v_COM = (0.04i + 0.3j + 0.5k) m/s
Therefore, the velocity of the center of mass of the three particles is (0.04i + 0.3j + 0.5k) m/s.
If each galaxy is 150 kpc across, how long does the event last?
In a galaxy collision, two similar-sized galaxies pass through each other with a combined relative velocity of 1200 km/s
Answers
If each galaxy is 150 kpc across, the event lasts for approximately 7.7 billion years.
How long does the event last?To calculate the time for the event, we need to know the distance traveled by the galaxies during the collision. Since each galaxy is 150 kpc across and they are passing through each other, the total distance traveled is twice the diameter of one galaxy, or 300 kpc.
To find the time, we can use the formula:
time = distance / speed
where distance is 300 kpc and speed is 1200 km/s. However, we need to convert the distance to the same units as the speed, so let's convert kpc to km:
1 kpc = 3.086 × 10^16 meters
1 meter = 3.281 × 10^-6 miles
1 mile = 1.609 × 10^3 meters
1 kpc = 3.086 × 10^19 miles
So, 300 kpc is equal to:
300 kpc × 3.086 × 10^19 miles/kpc = 9.258 × 10^21 miles
Now we can plug in the values:
time = distance / speed
time = 9.258 × 10^21 miles / 1200 km/s
time ≈ 2.432 × 10^14 seconds
So the event lasts for approximately 2.432 × 10^14 seconds. To convert to a more meaningful unit, we can divide by the number of seconds in a year:
time ≈ 7.7 billion years
Therefore, the event lasts for approximately 7.7 billion years.
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3
3
4.
How are the Celsius and Kelvin scales similar in regard to their fiduciary points?
Both align absolute zero with 0.
Both use absolute zero and the triple point of water.
Both use absolute zero and the boiling point of water.
Both use the freezing point and boiling point of water.
Answers
B) Both employ the triple point of water and absolute zero.
Absolute zero and the triple point of water serve as the foundation for both the Celsius and Kelvin scales. Absolute zero, or 0 K on the Kelvin scale and -273.15 °C on the Celsius scale, is the temperature at which all matter has no thermal energy.
The triple point of water is determined to be 273.16 K on the Kelvin scale and 0.01 °C on the Celsius scale. It is the temperature and pressure at which water can coexist in all three phases (solid, liquid, and gas) in equilibrium.
Absolute zero and the triple point of water serve as identical fiduciary points for both the Celsius and Kelvin temperature systems. The temperature ranges for each scale are determined by the fiduciary points, which are essential to temperature scales.
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