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Kruka [31]
2 years ago
8

1. Block A with a mass of 1 kg and block B with a mass of 2 kg lie on a smooth table. You wish to apply a horizontal force to ea

ch ball to give them the same horizontal acceleration (and you may safely ignore friction). Which of the following best describes the relative magnitude of the forces you must apply?
a) The magnitude of the force on block A must be twice as much as the force on block B.
b) The magnitude of the force on block B must be twice as much as the force on block A.
c) Both forces must have the same magnitude.
d) More information is needed to determine the relative magnitudes of the forces.

2. Block A has a mass of 1 kg and block B has a mass of 2 kg. Which of following best describes the relative magnitudes of the gravitational force (due to the Earth) on each block?
a)The gravitational force on block A is twice as much as the gravitational force on block B.
b)The gravitational force on block B is twice as much as the gravitational force on block A.
c)Both gravitational forces have the same magnitude.
d)More information is needed to determine the relative magnitudes of the forces.

3. You hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other. You release them both from the same height above the ground at the same time. (Air resistance can safely be ignored as they fall.) Which of the following describes and explains the motion of the two objects after you release them?
a)Block A hits the ground before the block B. The same force of gravity acts on both objects, but because block A has a lower mass, it will have a larger acceleration.
b)Block B hits the ground before the block A. The larger force on block B causes it to have a greater acceleration than block A.
c)The two blocks hit the ground at the same time, because they both accelerate at the same rate while falling. The greater gravitational force on block B is compensated by its greater mass, giving it the same acceleration.
d)The two blocks hit the ground at the same time, because they both move at the same constant speed while falling. The forces on the blocks are irrelevant once they start falling.
Physics
1 answer:
goblinko [34]2 years ago
7 0

The gravitational force on block B is twice as much as the gravitational force on block A.

<h3>What is Force?</h3>

Force is the product of mass and acceleration. If the mass of A is 1Kg and the mass of B is 2Kg, In order to give the masses the same horizontal acceleration, the magnitude of the force on block B must be twice as much as the force on block A.

Given that the mass of A is 1Kg and the mass of B is 2Kg, the gravitational force on block B is twice as much as the gravitational force on block A.

If I hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other  and release both from the same height above the ground at the same time, the two blocks hit the ground at the same time, because they both accelerate at the same rate while falling. The greater gravitational force on block B is compensated by its greater mass, giving it the same acceleration.

Learn more about force:brainly.com/question/12528243

#SPJ1

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You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
Maxwell shoots a rubber band at his friend Jimmy. Which type of energy is converted into kinetic energy?
Leokris [45]

A stretched rubber band is storing <em>elastic potential energy. (A)</em>

3 0
3 years ago
A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500
mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

                                   Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

                                   Dm = Tm*Vm

                                   Dm = 20*60*3

                                   Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

                                   Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

                                   Dw = Tw*Vw

                                   Dw = 15*60*4

                                   Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

                                   s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

            dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

                                   2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

                                   s*ds/dt = (dm + dw)*(Vm + Vw)

                                   ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

                 ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

                 ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]  

                ds/dt = 6.98 ft/s

                 

           

7 0
3 years ago
A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of
nexus9112 [7]

Answer:

F_1=366.67Hz

Explanation:

From the question we are told that:

Frequency of 3rd harmonics F_3=1100

Frequency of 5th harmonics F_3=1833

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

 \lambda_3=\frac{4}{3}l

Therefore

 F_3=\frac{3v}{4l}

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

 \lambda_1=\frac{4}{1}l

Therefore

 F_1=\frac{v}{4l}

Generally the equation for the frequency of the first harmonic is mathematically given by

 F_1=\frac{F_3}{3}

 F_1=\frac{1100}{3}

 F_1=366.67Hz

7 0
3 years ago
What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude
Bas_tet [7]

Answer:

The fraction of kinetic energy to the total energy is \frac{K}{T}=\frac{8}{9}.

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)

The total energy is

T =0.5 mw^2A^2 ..... (2)

Divide (1) by (2)

\frac{K}{T}=\frac{8}{9}

5 0
3 years ago
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