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Harlamova29_29 [7]
3 years ago
9

A 3 kg stone is dropped from a height of 4 m what is its momentum as it strikes the ground

Physics
1 answer:
Vitek1552 [10]3 years ago
7 0

Okay first you have to recognize that the maximum Gravitational potential energy will equal the maximum kinetic energy. The maximum GPE will equal when the stone is at its highest point and the max KE will be right before the stone hits the ground. So

GPE max = KE max

mgh = 1/2mv^2

Mass cancels as it is on both sides

gh = 1/2v^2

Multiply by 2

2gh = v^2

Square root

v = √2gh

Now plug in

v = √2(9.8)(4)

v = 8.85 m/s

Now use the mass to calculate the momentum just before it hits the ground as this is the speed right before it hits the ground

p = mv

p = (3)(8.85)

p = 26.55 kgm/s

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5km

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The Earth's biosphere is consists of
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After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume t
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Read 2 more answers
Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
nikklg [1K]

Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2} (14)

a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

4 0
3 years ago
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