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Umnica [9.8K]
2 years ago
9

Concept Simulation 2.3 offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to

determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 17.4 m/s and measures a time of 12.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Physics
1 answer:
Alexandra [31]2 years ago
7 0

Answer:

1.40 m/s^2

Explanation:

Given data

Velocity= 17.4 m/s

time= 12.4 seconds

We want to find the acceleration of the rock

We know that

acceleration = velocity/time

Substitute

acceleration= 17.4/12.4

acceleration=1.40 m/s^2

Hence the acceleration is 1.40 m/s^2

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Physics is confusing :(
Ket [755]

Answer:

30's-40's

Explanation:

4 0
3 years ago
A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of
Vikentia [17]

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)

mv^2 = mv_t^2 + 2mgR(1 + cos \theta)

we know that,

N - mgcos \theta = \dfrac{mv^2}{R}

N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}

N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)

N  = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)

N  = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)

N = 26.59 N

3 0
3 years ago
Photoelectrons with a maximum speed of 7.00 · 105 m/s are ejected from a surface in the presence of light with a frequency of 8.
lord [1]
This is just testing your ability to recall that kinetic energy is given by: 

<span>k.e. = ½mv² </span>

<span>where m is the mass and v is the velocity of the particle. </span>

<span>The frequency of the light is redundant information. </span>

<span>Here, you are given m = 9.1 * 10^-31 kg and v = 7.00 * 10^5 m/s. </span>
<span>Just plug in the values: </span>

<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
<span>k.e. = 2.23 * 10^-19 J 
so it will be d:2.2*10^-19 J</span>
4 0
3 years ago
A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler
natka813 [3]

Answer:

9.8 m/s^2, downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity (downward)

According to Newton's second law,

F=ma

where F is the net force on the object and a is its acceleration. Rearranging for a,

a=\frac{F}{m}

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

a=\frac{mg}{m}=g

Therefore, the ball has a constant acceleration of 9.8 m/s^2 downward for the entire motion.

5 0
3 years ago
During an experiment, Ellie records a measurement of 0.0034 m. How would
Goshia [24]

Answer:

(A)   She needs to move the decimal point by 3 places

8 0
3 years ago
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