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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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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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Answer:

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Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

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