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belka [17]
2 years ago
9

* A 5 kg ball is dropped from a height of 20 m. How much kinetic energy will it have 10 m above the ground? a. 490 J b. 392 J C.

480 J d. 100 J​
Physics
1 answer:
Oksi-84 [34.3K]2 years ago
6 0

Answer:

A. 490

Explanation:

soln

mass = m = 5kg

Height = h = 10m

Acceleration due to gravity = g = 9.8ms²

K.E = 1/2 × mass × (velocity)²

Recall from equations of motion

v² = u² + 2gh

Therefore,

K.E = 1/2 × mass × ( u² + 2gh)

K.E = 1/2 × 5 × ( 0² + 2×10×9.8)

K.E = 1/2 × 5 × 196

K.E = 1/2 × 980

K.E = 490 Joules

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A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule
saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

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Since, initial momentum is zero. So,

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8 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall buildin
OlgaM077 [116]

Solution :

The motion in the y direction.

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7 0
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