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pochemuha
4 years ago
5

An astronaut on a distant planet wants to determine its acceleration due to gravity. the astronaut throws a rock straight up wit

h a velocity of +16 m/s and measures a time of 21.2 s before the rock returns to his hand. what is the acceleration (magnitude and direction) due to gravity on this planet? (indicate direction by the sign of the acceleration.)
Physics
1 answer:
Tamiku [17]4 years ago
7 0
Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t  = 21.2 s, total time of travel.

The vertical motion obeys the equation
s = ut - (1/2)gt²

When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²

Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.


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Read 2 more answers
A baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs
miskamm [114]

Answer:

a. v = \frac{mu_sm_2g\Delta t}{m_1}

b. 21.64 m/s

Explanation:

Let g = 9.81m/s2

a. The weight of the block is product of its mass and gravitational acceleration

W = m_2g = 7.25*9.81 = 71.1225N

which is also the normal force acting on the block from the floor so it stays balanced.

N = 71.225N

The static friction of the block is product of its normal force from the floor and the friction coefficient

F_s = \mu_sN = \mu_sW = mu_sm_2g

For the block to move, the force generated by the impact must be at least equal to the static friction.

F = F_s = mu_sm_2g

The impulse is product of this force and time duration of impact.

I = F\Delta t = mu_sm_2g\Delta t

As impulse is generated by change in momentum of the ball, which is product of its mass and velocity v

I = \Delta p = m_1\Delta v

mu_sm_2g\Delta t = m_1 v

v = \frac{mu_sm_2g\Delta t}{m_1}

b. v = \frac{mu_sm_2g\Delta t}{m_1} = \frac{0.74*7.25*9.81*0.185}{0.45} = 21.64 m/s

8 0
3 years ago
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