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Ivahew [28]
4 years ago
10

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop

. She drops a wooden block with a mass of 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. What’s the acceleration of the wooden block when it hits the sensor? Use F = ma.
A.
2.45 m/s2
B.
4.4 m/s2
C.
5.4 m/s2
D.
9.8 m/s2
Physics
1 answer:
Butoxors [25]4 years ago
4 0

Answer:

D!

Explanation:

Using the formula F = ma, you plug in 4.9 for F (force), and 0.5 for m (mass), then solve for a (acceleration).

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On Mars, the acceleration due to gravity is 3.77 m / s 2 . 3.77 m/s2. How far would a 15 g 15 g rock fall from rest in 2.5 s 2.5
Zinaida [17]

Answer:

s = 11.78 m

Explanation:

given,

acceleration due to gravity, g = 3.77 m/s²

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s = ut +\dfrac{1}{2}at^2

initial speed = 0 m/s

s = \dfrac{1}{2}at^2

s = \dfrac{1}{2}\times 3.77 \times 2.5^2

s = 11.78 m

distance traveled by the rock is equal to 11.78 m.

7 0
3 years ago
Read 2 more answers
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

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Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
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Answer:

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3 years ago
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