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PolarNik [594]
3 years ago
15

The acceleration due to gravity on the Moon is gM. Suppose an astronaut on the Moon drops an object from a height of H. The time

it would take the object to reach the Moon’s surface would be TM. The same object is dropped from the same height on Earth, where the acceleration due to gravity is gE. The time it takes the object to reach the Earth’s surface is TE. Which of the following is a correct mathematical relationship for the two times?
TE=gEgM−−−√TM

TE=gEgMTM
A

TE=gMgE−−−√TM

TE=gMgETM
B

TE=gEgMTM

TE=gEgMTM
C

TE=gMgETM

TE=gMgETM
D

TE=TM
Physics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

TE = sqrt(GM/GE)TM

Explanation:

To solve for this problem, you have to use the second kinematic equation and set the height equal to each other.  Because the heights are equal, 1/2GETE^2 = 1/2GMTM^2.  Rearrange the equation and you'll get the answer

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A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,
MakcuM [25]

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

T - mg cos\theta = \frac{mv^2}{R}

T = mg cos\theta + \frac{mv^2}{R}

now by energy conservation we have

\frac{1}{2}mv^2 = mg(R cos\theta)

mv^2 = 2mgR cos\theta

T = mg cos\theta + 2mgcos\theta

T = 3mg cos\theta

For string break down we have

T = 2mg = 3mgcos\theta

cos\theta = \frac{2}{3}

Now height from the ground is given as

h = R - Rcos\theta

h = 24 - 24(\frac{2}{3})

h = 8 cm

7 0
3 years ago
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
The work-energy theorem states that the work done on an object is equal to a change in which quantity?
Fynjy0 [20]

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

8 0
2 years ago
A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
3 years ago
A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor i
svet-max [94.6K]

Answer:

0.261\ \text{m}

Explanation:

\Delta f = Change in frequency = 2.1 Hz

f = Frequency of source of sound = 440 Hz

v_m= Maximum of the microphone

v = Speed of sound = 343 m/s

T = Time period = 2 s

We have the relation

\Delta f=2f\dfrac{v_m}{v}\\\Rightarrow v_m=\dfrac{\Delta fv}{2f}\\\Rightarrow v_m=\dfrac{2.1\times 343}{2\times 440}\\\Rightarrow v_m=0.8185\ \text{m/s}

Amplitude is given by

A=\dfrac{v_m T}{2\pi}\\\Rightarrow A=\dfrac{0.8185\times 2}{2\pi}\\\Rightarrow A=0.261\ \text{m}

The amplitude of the simple harmonic motion is 0.261\ \text{m}.

4 0
2 years ago
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