Answer:
the nucleus is the size of an apple, approximately 5 cm of radius e, the atom has a radius of R = 5 cm 104 = 50000 cm = 50 km
Explanation:
In the Rutherford experiments it was proved that the atomic nucleus has the volume 10-4 the volume of the atom.
If we make a scale design in which the nucleus is the size of an apple, approximately 5 cm of radius e, the atom has a radius of R = 5 cm 104 = 50000 cm = 50 km
This shows that almost the entire volume of the atom is empty.
Based on the law of conservation of energy, which statement is correct?
A.
Energy is always being added to all parts of the Universe.
B.
Energy is often destroyed in some parts of the Universe.
C.
Energy in a closed system cannot change forms.
D.
Energy in an isolated system remains constan
Answer:
D
Explanation:
Nothing can enter or leave so it remains constant
Give reason:
a) In 'coin on card' experiment a smooth card is used.
Answer:
Please mark as brainliest!!
Explanation:
In coin card experiment smooth card is used so that the card can slide easily from glass.
Answer:
In coin card experiment smooth card is used so that the card can slide easily from glass.
A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satellite's orbital speed. 9.82278e7 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (b) Find the period of its revolution. h (c) Find the gravitational force acting on it.
Answer:
a) v = 5.59x10³ m/s
b) T = 4 h
c) F = 1.92x10³ N
Explanation:
a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:
[tex] F_{c} = F_{G} [/tex]
[tex]\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}[/tex]
[tex] v = \sqrt{\frac{gr^{2}}{r+h} [/tex]
Where:
g is the gravity = 9.81 m/s²
r: is the Earth's radius = 6371 km
h: is the satellite's height = r = 6371 km
[tex]v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s[/tex]
b) The period of its revolution is:
[tex] T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h [/tex]
c) The gravitational force acting on it is given by:
[tex] F = \frac{GMm}{(r + h)^{2}} [/tex]
Where:
M is the Earth's mass = 5.97x10²⁴ kg
m is the satellite's mass = 782 kg
G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²
[tex] F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N [/tex]
I hope it helps you!
A stunt person jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following best describes why no injury occurs?
A. The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.
B. The bag increases the amount of time the force acts on the person and reduces the change in momentum.
C. The bag reduces the impulse to the person.
D. The bag provides the necessary force to stop the person.
E. The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person.
Answer:
A stunt person jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following best describes why no injury occurs?
The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.
ANSWER: A
Answer:
A. The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.
During stunts, people usually put an-air filled bag on the ground for safe landing and to prevent injuries.
Momentum is calculated by m* v where m is for mass and v is for velocity.
Force is directly proportional to momentum
F∝M
The bag in this scenario helps to increases the amount of time the force acts on the person. This implication means that there is a reduction in the change of momentum. Since Force and momentum have a directly proportional relationship then the Force taken to hit the floor is greatly reduced.
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Look at this model of an atom. Where are the protons located and how many are there?
Answer:
protons are in the nucleus .
Explanation:
there are 6 protons
A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and
the coefficient of static friction between tires and dry road, Ms = 0.5. Calculate the maximum
speed the car can have and still make the turn successfully
The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have
• net vertical force:
∑ F = N - W = 0
• net horizontal force:
∑ F = Fs = m a
where
N = magnitude of normal force
W = car's weight
Fs = mag. of static friction
m = car's mass
a = v ²/R = mag. of the centripetal acceleration
v = car's speed
R = radius of curve
Now,
• compute the car's weight:
W = m g = (1500 kg) (9.8 m/s²) = 14,700 N
• solve for the mag. of the normal force:
N = 14,700 N
• solve for the mag. of the friction force, using the given friction coefficient:
Fs = 0.5 N = 7350 N
• solve for the (maximum) acceleration:
7350 N = (1500 kg) a → a = 4.9 m/s²
• solve for the (maximum) speed:
4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s
A rectangular block measures 4.1cm by 2.8cm by 2.1cm. calculate its volume given you answer to an appropriate number of significant figures?
Answer:
Volume = 24 cm^3
Explanation:
We recall that the volume of the box is calculated via the formula:
Volume = length * height * width
and that in a product, the number of significant figures for the result should coincide with the number of significant figures of the factor that has the least of them.
In this case, all measures have the same number of significant figures: two. so we calculate the product, and then limit the answer value to exactly two significant figures:
Volume = 4.1 cm * 2.8 cm * 2.1 cm = 24.108 cm^3, which must be rounded to two significant figures as: 24 cm^3
Answer:
Volume = 24.108cm³
Explanation:
Volume = length × breadth × height
Volume = 4.1 × 2.8 × 2.1
Volume = 24.108
Using the law of conservation of energy, what is the kinetic energy at e?
Answer:
Send the pic so I can see
In the Sun, fusion reactions create helium nuclei. To form each helium
nucleus, four hydrogen nuclei fuse. The four hydrogen nuclei have a greater
total mass than the newly formed helium nucleus. Which statement explains
this difference in mass?
O A. Some of the mass burned and was transformed into gases.
O B. Mass was destroyed and disappeared.
O c. Some of the mass of the four hydrogen nuclei was converted into
energy
D. Some of the mass was transformed into protons.
Answer:
C. Some of the mass of the four hydrogen nuclei was converted into
energy
Explanation:
a p e x , just took the quiz
Some of the mass of the four hydrogen nuclei was converted into energy.
When four hydrogen nuclei fuse to form one helium nuclei, mass defect, some mass is converted into energy, that is the reason of energy of the sun.
What is fusion?"When two nuclei form a big nuclei, the phenomenon is known as fusion."
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Calculate the potential energy of a coffee cup that is resting on a 24.5 meter ledge, and
weighs 5.4 Newtons
Answer:
E_{pot} = 132.3 [J]
Explanation:
In order to solve this problem, we must use the definition of the potential energy which can be calculated by means of the following formula.
[tex]E_{pot}=W*h[/tex]
where:
Epot = potential energy [J]
W = weight = 5.4 [N]
h = elevation = 24.5 [m]
[tex]E_{pot}=5.4*24.5\\E_{pot}=132.3[J][/tex]
2. A car accelerates at a rate of 1.4 m/s. Find the mass of the car if a 2250 N net force is
required to produce this acceleration.
Answer:
1607.14 kgExplanation:
The mass of the car can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{2250}{1.4} \\ = 1607.14285...[/tex]
We have the final answer as
1607.14 kgHope this helps you
Evidence supporting the theory of continental drift includes:
a.
b.
c.
d .
Answer:
The apparent fit of the eastern coastline of South America and western coastline of Africa
Similarities of plants and animal fossils in South America and some parts of African continent which were separated by a vast ocean
Similarities in the sequence of rock layers of opposite sides of the Atlantic Ocean
What is the magnitude of the Box's Acceleration?
The Box's Acceleration : g sin θ
Further explanationNewton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object
∑F = m. a
F = force, N
m = mass = kg
a = acceleration due to gravity, m / s²
We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.
Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then
[tex]\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta[/tex]
A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleration α, the torque on the child is τ. The child moves to a position half way between the center and edge of the merry-go-round, and the angular acceleration increases to 2α. The torque on the child is now
Answer:
The torque on the child is now the same, τ.
Explanation:
It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:[tex]I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)[/tex]
So, τ = 3/2*m*r²*α (2)When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:[tex]I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)[/tex]
Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
Prob. 3: Manifestation of quantum phenomena (total 25 points) (a)-(e) 5 points each.
Please provide some experimental demonstration or explanation of natural observations or characteristics of application which confirm(s) / display(s) the following notion of quantum physics. In either experiments or observation, please attach a brief explanation and justification to your statement.
3-(A) Vibration of molecules can be treated by quantum mechanics using a parabolic potential. In classical cases, we can use mass(es) attached to a spring. In quantum mechanics, energy levels are quantized, unlike classical oscillators. How can you demonstrate that?
3-(B) The ground state energy of classical oscillator is zero, while that of quantum oscillators has finite (zero-point) energy. How can you demonstrate that (not by calculations but by some experiments or observations)?
3-(C) Electrons can have both orbital and spin angular momentum, and associated magnetic moment. There are two spin quantum numbers (+1/2 or -1/2) for an electron. Describe some quantum phenomena or observation which demonstrate(s) that an electron has a spin 1/2.
3-(D) Laser light is a coherent light. Which property of laser depends on this feature? What application is related to the coherence?
3-(E) Quantum particle(s) can tunnel through a potential barrier.
Answer:
3A. This phenomenon can be seen in the discrete emission of the molecules.
3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.
3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams
3D. This is due to the stimulated emission
3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)
Explanation:
This problem asks for some experimental explanations of various quantum phenomena.
3A. This phenomenon can be seen in the discrete emission of the molecules.
In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed
In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms
3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.
The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen
3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.
3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .
This is widely used for holographic and interference work
3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.
Answer:
Explanation:
Please provide some experimental demonstration or explanation of natural observations or characteristics of application which confirm(s) / display(s) the. Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature. Quantum mechanics arose gradually from theories to explain observations. By M Arndt · 2009 · Cited by 239 — Wave-particle duality of light: A paradigmatic example for this fact is the dual nature of light that manifests itself in Young's famous double-slit diffraction.
What is meant by child rights?
Ultimate frisbee relies upon good sportsmanship since there are no referees and players must self-officiate the game. What is this known as?
Group of answer choices
Spirit of the Team
Spirit of the Frisbee
Spirit of Sportsmanship
Spirit of the Game
Answer:
spirit of the team / game
A 1.13 kg skateboard is coasting along the pavement at a speed of 4.28 m/s when a 0.93
kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the
skateboard-cat combination?
By conservation of momentum,
Pinitial = Pfinal
m1v1 + m2v2 = (m1 + m2)*vf
m1 = mass of skateboard = 1.13 kg
m2 = mass of cat = 0.93 kg
v1 = initial velocity of skateboard = 4.28 m/s
v2 = initial velocity of cat = 0 m/s
vf = final velocity of skateboard-cat combo
So plug in the values and solve for vf,
1.13(4.28) + 0.93(0) = (1.13 + 0.93)vf
vf = 2.35 m/s
A 0.60-kg mass at the end of a spring vibrates 3.0 times per second with and amplitude of 0.13m. Determine
(a) The velocity when it passes the equilibrium point,
(b) The velocity when it is 0.10 m from equilibrium
(c) The total energy of the system, and
(d) The equation describing the motion of the mass, assuming the x was a maximum at t = 0.
Answer:
a) The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.
b) The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.
c) The total energy of the system is 1.802 joules.
d) The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].
Explanation:
a) If all non-conservative forces can be neglected and spring has no mass, then the mass-spring system exhibits a simple harmonic motion (SHM). The kinematic formula for the position of the system ([tex]x(t)[/tex]), measured in meters, is:
[tex]x(t) = A\cdot \sin(\omega \cdot t +\phi)[/tex] (1)
Where:
[tex]A[/tex] - Amplitude, measured in meters.
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
The kinematic equation for the velocity formula of the system ([tex]v(t)[/tex]), measured in meters per second, is derived from (1) by deriving it in time:
[tex]v(t) = \omega\cdot A\cdot \cos (\omega\cdot t+\phi)[/tex] (2)
The velocity when it passes the equilibrium point occurs when the cosine function is equal to 1 or -1. Then, that velocity is determined by following formula:
[tex]v = \pm \omega\cdot A[/tex] (3)
The angular frequency is calculated by this expression:
[tex]\omega = 2\pi\cdot f[/tex] (4)
Where [tex]f[/tex] is the frequency, measured in hertz.
If we know that [tex]f = 3\,hz[/tex] and [tex]A = 0.13\,m[/tex], then the velocity when it passes the equilibrium point, which is the maximum and minimum velocities of the mass:
[tex]\omega = 2\pi\cdot (3\,hz)[/tex]
[tex]\omega \approx 18.850\,\frac{rad}{s}[/tex]
[tex]v = \pm \left(18.850\,\frac{rad}{s} \right)\cdot (0.13\,m)[/tex]
[tex]v = \pm 2.451\,\frac{m}{s}[/tex]
The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.
b) First, we need to determine the spring constant of the system ([tex]k[/tex]), measured in newtons per meter, in terms of the angular frequency ([tex]\omega[/tex]), measured in radians per second, and mass ([tex]m[/tex]), measured in kilograms. That is:
[tex]k = \omega^{2}\cdot m[/tex] (5)
If we know that [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]m = 0.60\,kg[/tex], then the spring constant is:
[tex]k = \left(18.850\,\frac{rad}{s} \right)^{2}\cdot (0.60\,kg)[/tex]
[tex]k = 213.194\,\frac{N}{m}[/tex]
Lastly, we determine the velocity when the mass is 0.10 meters from equilibrium by the Principle of Energy Conservation:
[tex]U_{k} + K = K_{max}[/tex] (6)
[tex]\frac{1}{2}\cdot k\cdot x^{2} + \frac{1}{2}\cdot m\cdot v^{2} = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (7)
Where:
[tex]U_{k}[/tex] - Current elastic potential energy, measured in joules.
[tex]K[/tex] - Current translational kinetic energy, measured in joules.
[tex]K_{max}[/tex] - Maximum translational kinetic energy, measured in joules.
[tex]v[/tex] - Current velocity of the system, measured in meters per second.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{max}[/tex] - Maximum velocity of the system, measured in meters per second.
If we know that [tex]k = 213.194\,\frac{N}{m}[/tex], [tex]x = 0.10\,m[/tex], [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the velocity of the mass-spring system is:
[tex]\frac{k}{m} \cdot x^{2} + v^{2} = v_{max}^{2}[/tex]
[tex]v^{2} = v_{max}^{2}-\frac{k}{m}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{max}^{2}-\frac{k\cdot x^{2}}{m} }[/tex] (8)
[tex]v = \sqrt{\left(\pm 2.451\,\frac{m}{s} \right)^{2}-\frac{\left(213.194\,\frac{N}{m} \right)\cdot (0.10\,m)^{2}}{0.60\,kg} }[/tex]
[tex]v \approx \pm 1.567\,\frac{m}{s}[/tex]
The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.
c) The total energy of the system ([tex]E[/tex]), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:
[tex]E = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (9)
If we know that [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the total energy of the system is:
[tex]E = \frac{1}{2}\cdot (0.60\,kg)\cdot \left(\pm 2.451\,\frac{m}{s}\right)^{2}[/tex]
[tex]E = 1.802\,J[/tex]
The total energy of the system is 1.802 joules.
d) Given that initial position of the mass-spring system is a maximum, then we conclude that the equation of motion has the following parameters: ([tex]A = 0.13\,m[/tex], [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]\phi = 0.5\pi\,rad[/tex])
From (1) we obtain the resulting formula:
[tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex] (10)
The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].
Objects in free fall are weightless.
Please select the best answer from the choices provided
T
F
Answer:
[tex]\boxed{True}[/tex]
Explanation:
Objects are weightless in free fall because their is no other force pushing them rather than the gravitational force.
Hope it helps!<3
Answer:
It's ture hope this answer help have a great day .!Explanation:
Katniss everdeen applies 20 n of force back on her bow what happens to the arrow when she lets go?
ASAP DUE TO DAY PLS HELP MEEEEEE
1. Kinetic energy is the energy of:
A. potential
B. objects at rest
C. objects in motion
D. accelerating objects
--------------------------------------------------------------------------------------------
A mass m is located at the origin; a second mass m is at x = d. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass?
Answer:
√3 * Gm²/d²
Explanation:
m1 = m, m2= m, distance = d. hence:
F = Gm²/d²
Let the origin be A, the point x = d be B and the point above the first two is C.
The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]
Let j represent the vertical component and i the horizontal component. Hence:
[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]
Answer:
The net gravitational force on the third mass = [tex]3^{0.5} * \frac{Gm^2}{d^2}[/tex]Explanation:
For equilateral triangle,
[tex]\theta = 30^o[/tex]
Force between masses,
[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]
Therefore,
[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]
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a 2 kg block is attached to a horizontal ideal spring with a spring constant 200N/m. when the
Which lists three organic biological molecules?
O carbohydrates, salts, metals
O salts, proteins, minerals,
O proteins, lipids, carbohydrates
O lipids, metals, minerals
Answer:
B
Explanation:
I'm learning it in science.
Answer:
its not b i just took the test and b was wrong
Explanation:
In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m and 0.6 m, respectively. The wavelength is 630 nm. Suppose that the radius of the aperture can be increased from 0.5 mm, determine: (a) The first two radii when the center intensity at the reception screen is maximum. (b) The first two radii when the center intensity is minimum.
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the value of the resistance. Resonance occurs when omega squared space equals 1 divided by space L C At resonance, the power is maximum and the impedance is minimum. At resonance, R squared space equals space (X subscript L minus X subscript c )squared The impedance Z is always larger than the resistance R.
Answer:
True b and c
Explanation:
In an RLC circuit the impedance is
[tex]Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ] }[/tex]
examine the different phrases..
a) False. The maximum impedance is the value of the resistance
b) True. Resonance occurs when
(wL)² + (1 / wC)² = 0
w² = 1 / LC
c) True. In resonance the impedance is the resistive part and the power is maximum
d) False. In resonance the inductive and capacitive part cancel each other out
e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal
Many curves have banked turns, which allow the cars to travel faster around the curves than if the road were flat. Acutally, cars could also make turns on these banked curves if there were no friction at all. Explain this statement using the free-body diagram shown in the figure.
Answer:
the normal force that is perpendicular to the surface has a component towards the center of the curve.
Explanation:
When curves with superelevation or inclination, the normal force that is perpendicular to the surface has a component towards the center of the curve. This component is what maintains the vehicle and gives it centripetal acceleration.
Consequently, the vehicle does not need the friction force since it does not have a tendency to slide and this is zero.
Can I get help on this question please I don’t understand
Answer:
A
Explanation:
Weight = 55 * 9.8 = 539 N
Scale = 686 N
Since the scale reading is larger than the lady’s weight, the elevator must be accelerating as it moves upward.
Net upward force = 686 – 539 = 147 N
147 = 55 * a
a = 147 ÷ 55
The acceleration is approximately 2.67 m/s^2.
You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.
Answer:
Explanation:
When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed
At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.
F = ma
Acceleration is the change in the velocity of an object per change in time of motion.
At constant velocity, the acceleration of an object is zero.When acceleration of an object is zero, the force on the object is zero.A constant speed (magnitude only) and change in the direction of the object, implies a change in velocity of the object.at changing velocity, the acceleration on an object is positive, and hence net force acts on the object.Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
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