Answer:
Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.
Explanation:
what are the scales of measurement of temperature?
Answer:
There are three temperature scales in use today, Fahrenheit, Celsius and Kelvin. Fahrenheit temperature scale is a scale based on 32 for the freezing point of water and 212 for the boiling point of water, the interval between the two being divided into 180 parts
Answer: I have don't understand this question
Explanation:
If the speed of an object in uniform circular motion is constant and the radial distance is doubled, by what factor does the magnitude of the radial acceleration decrease?
Answer: Half
Explanation:
Given
The object is in uniform circular motion with constant speed.
Radial acceleration of the object is given by
[tex]\Rightarrow a_r=\dfrac{v^2}{r}[/tex]
Where, [tex]a_r=\text{Radial acceleration}[/tex]
[tex]v=\text{speed}\\r=\text{distance from the axis of rotation}[/tex]
If the radial distance [tex]r[/tex] is doubled, i.e. [tex]2r[/tex]
Radial acceleration is
[tex]\Rightarrow a'_r=\dfrac{v^2}{2r}\\\\\Rightarrow a'_r=\dfrac{a_r}{2}[/tex]
Radial acceleration becomes half of the initial value.
Learn more: https://brainly.in/question/21217248
A doctor examines a mole with a 15.0 cm focal length magnifying glass held 12.4 cm from the mole.
a. What is its magnification?
b. Where is the image?
c. How big is the image of a 5.00 mm diameter mole?
Answer:
a. Magnification = 6.1
b. The image formed is virtual, and on the same side of the lens as the object.
c. Image size = 119.8 squared millimetres
Explanation:
Magnification = [tex]\frac{Image distance}{Object distance}[/tex]
But, focal length, f = 15.0 cm, and object distance, u = 12.4 cm. Let the image distance be represented by v.
a. Applying the lens formula, we have;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{15}[/tex] = [tex]\frac{1}{12.4}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{15}[/tex] - [tex]\frac{1}{12.4}[/tex]
= -[tex]\frac{13}{930}[/tex]
v = -75.1538
The image distance, v = -75.2 cm
Magnification = [tex]\frac{75.2}{12.4}[/tex]
= 6.0645
Magnification = 6.1
b. The image formed is virtual, and on the same side of the lens as the object.
c. Given that diameter of mole = 5.00 mm.
Its radius = [tex]\frac{diameter}{2}[/tex] = [tex]\frac{5.0}{2}[/tex]
= 2.5 mm
Thus, the area of the mole would be;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](2.5)^{2}[/tex]
= 19.643
A = 19.64 square millimetres.
Thus, the size of the image can be determined by;
Magnification = [tex]\frac{Image size}{Object size}[/tex]
Image size = Magnification x object size
= 6.1 x 19.64
= 119.804
The size of the image is 119.8 squared millimetres.
The earliest mineral observed to showmagnetic properties is called
A leadstone
Blodestone
Cloadstone
Dnone of the above
E all of the above
Answer:
B: lodestone
Explanation:
Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.
Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.
An example of a short-term goal is
a.) becoming fluent in Japanese
b.) finishing the annual report in two weeks
c.) earning a college degree
d.) advancing to vice-president of the company
Answer:
a
Explanation:
all the other ones take years and hard work but learning japanese is pretty easy and you dont need to work to hard
Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➡ 150hrs.
[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]
➡ Time = distance × speed
➡ Time = 15*10
➡ Time = 150hrs ans.
If the knee has coordinates of (1.2,2.7) and the ankle has coordinates of (1.5,2.1), with the shank weighing 3.2kg, what is the moment of inertia of the shank about the proximal end point? (%k=0.528)
Answer:
The value is [tex]I = 0.48 \ kg m^2[/tex]
Explanation:
From the question we are told that
The coordinate of the knee is [tex]knee = (1.2,2.7)[/tex]
The coordinate of the ankle is [tex]ankle = (1.5,2.1)[/tex]
The mass of the shank is [tex]m = 3.2 \ kg[/tex]
Generally the length between the knee and the ankle is mathematically represented as
[tex]D = \sqrt{( 1.5 - 1.2)^2+ [ 2.1- 2.7]^2 }[/tex]
=> [tex]D =0.67 \ m[/tex]
Generally the moment of inertia of the shank about the proximal end point is mathematically represented as
[tex]I= m * \frac{D^2}{3}[/tex]
=> [tex]I = 3.2 * \frac{0.67^2}{3}[/tex]
=> [tex]I = 0.48 \ kg m^2[/tex]
A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3. B) 0.00337 m3. C) 0.00457 m3. D) 0.00766 m3 E) 0.00980 m3
Answer: B) 0.00337 m3.
Explanation:
Given data:
Mass of the ball = 10kg
Weight of the ball in air = 98N
Weight of the ball in water = 65N
Solution:
To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.
= 98 - 65 / 9.8
= 33 / 9.8
= 3.37kg
The volume of the ball is 3.37kg
The density of water is 1kg per Liter.
So 3.37 kg of water would have a volume of 3.37 Liters.
Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).
You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over. How can you achieve this?
Determine if the weight of the football is needed for you to prevent the football from rolling over. The coefficient of static friction of the plane is 0.067.
Answer:
You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.
Explanation:
spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.
Let any rolling ball of mass (m ) is traveling with velocity v ,
common effect on ball (N) = mg
because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.
hence,
fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)
Answer:
80kg = 133 Newtons I'm pretty sure this is right.
The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Weight of the object is force applied due to gravity, So force = mass.gravity .
on the moon gravity is 1/6 of that on the earth.
so weight or force = 80*(9.8/6)
= 130 Newtons
The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.
To learn more about force refer to the link:
brainly.com/question/13191643
#SPJ6
SI unit differ from one country to another . true or false
Answer:
false ..........false
Answer:
FALSE
Explanation:
You accidentally drop a book out of the window of a tall building. Assuming
no air resistance, how fast will the book be moving after 2.8 seconds?
Answer:
depends on how talll the building is but lets say its 100 ft tall 12MPH
Explanation:
Answer:
around 9.81m/s i think
Explanation:
2 Magnetism comes from the word
A Magnesia
B Magnentia
C Magnesium
D Magenta
E None of the above
Answer:
A Magnesia
Explanation:
The word magnetism comes from the word Magnesia, which is the name for a region in Asia Minor, where fragments of Fe3O4 ore (magnetite) were found, which attracts other metal objects.
Magnetism is a physical phenomenon by which we describe the attractive or repulsive force between materials.
This phenomenon has been known for thousands of years.
hi! hina can you help me with those 2 questions
which three statements about gravity in the formation of the solar system are true
The answer is A, B, D.
Answer:
The answer is A, B, D.
Explanation:
ap3x verified
Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the crate if the mass of the crate is 20 kg?
5.5 m/s2
20 m/s2
17.5 m/s2
72 m/s2
Answer:
a = 5.5 [m/s²]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.
ΣF =m*a
where:
F = Force [N] (units of Newtons)
m = mass = 20 [kg]
a = acceleration [m/s²]
Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.
[tex]230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ][/tex]
What is the energy contained in a 0.950 m3 volume near the Earth's surface due to radiant energy from the Sun?
PRACTICE
z
1.
What happens to the force of gravity between two objects if the distance between them is doubled?
Answer:
If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
Explanation:
Two speeding lead bullets, one of mass 15.0 g moving to the right at 295 m/s and one of mass 7.75 g moving to the left at 375 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (Lead has a specific heat of 128 J/(kg K), a melting point of 327.3°C, and a latent heat of fusion of 2.45 104 J/kg.)
Answer:
The final temperature of the bullets is 327.3 ºC.
Explanation:
Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:
[tex]K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0[/tex] (1)
Where:
[tex]K_{A,o}[/tex], [tex]K_{A}[/tex] - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.
[tex]K_{B,o}[/tex], [tex]K_{B}[/tex] - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.
[tex]U_{A,o}[/tex], [tex]U_{A}[/tex] - Initial and final internal energies of the 15-g bullet, measured in joules.
[tex]U_{B,o}[/tex], [tex]U_{B}[/tex] - Initial and final internal energies of the 7.75-g bullet, measured in joules.
By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:
[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0[/tex] (2)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the 15-g and 7.75-g bullets, measured in kilograms.
[tex]v_{A,o}[/tex], [tex]v_{A}[/tex] - Initial and final speeds of the 15-g bullet, measured in meters per second.
[tex]v_{B,o}[/tex], [tex]v_{B}[/tex] - Initial and final speeds of the 7.75-g bullet, measured in meters per second.
[tex]c[/tex] - Specific heat of lead, measured in joules per kilogram-Celsius degree.
[tex]T_{o}[/tex], [tex]T[/tex] - Initial and final temperatures of the bullets, measured in Celsius degree.
Now we clear the final temperature of the bullets:
[tex](m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})][/tex]
[tex]T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex]
[tex]T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex] (3)
If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the final temperature of the collided bullets is:
[tex]T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]
[tex]T = 852.534\,^{\circ}C[/tex]
Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:
[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0[/tex] (4)
[tex]U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)[/tex]
If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]T = 327.3\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then latent heat received by the bullets during impact is:
[tex]U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)[/tex][tex]U = 331.872\,J[/tex]
The maximum possible latent heat ([tex]U_{max}[/tex]), measured in joules, that both bullets can receive during collision is:
[tex]U_{max} = (m_{A}+m_{B})\cdot L_{f}[/tex] (5)
Where [tex]L_{f}[/tex] is the latent heat of fusion of lead, measured in joules per kilogram.
If we know that [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex] and [tex]L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}[/tex], then the maximum possible latent heat is:
[tex]U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)[/tex]
[tex]U_{max} = 557.375\,J[/tex]
Given that [tex]U < U_{max}[/tex], the final temperature of the bullets is 327.3 ºC.
A 235-kg merry-go-round at the Great Escape in Lake George is in the shape of a uniform, solid, horizontal disk of radius 1.50 m. It is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s?
Answer:
The constant force to be exerted on the rope is 221.55 N
Explanation:
Given;
mass of the merry, m = 235 kg
radius, r = 1.5 m
number of revolution per second, = 0.4 rev/s
time of motion, t= 2.00 s
The angular acceleration is given by;
[tex]\alpha = \frac{0.4 \ rev}{s} *\frac{2\pi \ rad}{rev} *\frac{1}{2.0\ s} = 1.257 \ rad/s^2[/tex]
Torque is given by;
τ = F.r
Also torque in uniform solid disk is given by;
τ = ¹/₂mr²α
Thus, equating the two;
F.r = ¹/₂mr²α
F = ¹/₂mrα
F = ¹/₂(235)(1.5)(1.257)
F = 221.55 N
Therefore, the constant force to be exerted on the rope is 221.55 N
a 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?
Answer:
5,000
Explanation:
Vf = Vi + a * t
If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.)
-30° C
40° C
-20° C
30° C
Answer:
-20°C
Explanation:
The specific heat capacity of ice using the cgs system is 0.5cal/g°C
The enthalpy change is calculated as follows
ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.
10cal = 2g×0.5cal/g°C×∅
∅=10cal/(2g×0.5cal/g°C)
∅=10°C
Final temperature= -30°C+ 10°C= -20°C
Which current is produced in homes
Answer:
answer is C on edge 2021
Explanation:
If I travel 300 m east, then 400 m west, what is my distance &
displacement?
Answer:100m west
Explanation:
(iii) Calculate the distance travelled by the car in part Q.
Use the equation
distance travelled = average speed x time
(2)
distance travelled = ....... m
Answer:
distance travelled = 3000m
Explanation:
distance travelled = average speed x time
=30m/s*100s
=3000m
Which of the following is the main idea of Thomas Paine's "Common Sense"?
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?
Answer:half-life (T1/2) of this isotope =
Explanation:
The number of nuclei of any radioactive substance at a given time is expressed by
Nt = N0e⁻kt
Nt=decay of material at a time t, =3110 decays per minute
N=decays at t=0, 8255 decays per minute
k=constant
Nt=N0e−kt
3110= 8255 e⁻k(4.50)
3110/ 8255=e−k(4.50)
0.3767 = e−k(4.50)
In 0.3767 = -k (4.50)
0.976=-4.5k
k=0.976/4.5
=0.2159
Also we know that t 1/2= time that it takes half the original material to decay.it is related to the rate constant by
T₁/₂=ln 2 / k
Therefore half-life (T1/2) of this isotope
T₁/₂=ln 2/0.2159
T₁/₂=3.12 days
A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days
Answer:
= 720000 [k]
Explanation:
The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.
[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]
= 720000 [k]
A gene carries the blank for the trait