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guapka [62]
3 years ago
14

The mass of the Earth is 5.98 × 1024 kg . A 11 kg bowling ball initially at rest is dropped from a height of 2.63 m. The acceler

ation of gravity is 9.8 m/s 2 . What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Answer in units of m/s.
Physics
1 answer:
jekas [21]3 years ago
3 0

Answer:

v = 7.18_m/s

Explanation:

Velocity of the earth towards the ball is = velocity of the ball moving towards earth

For object in free fall, we have

Where

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

S = height of ball above ground

v^2 = u^2 - 2×g×(-S)

= 0 + 2×9.8×2.63 = 51.55_m^2/s^2

Velocity of the ball just before it hits the ground is

v = 7.18_m/s

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A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), Fig. 5–39.
Trava [24]

Answer:

1742.24106 revolutions per day

Explanation:

v = Velocity

d = Diameter = 1.1 km

r = Radius = \dfrac{d}{2}=\dfrac{1.1}{2}=0.55\ km

g = Acceleration due to gravity = 9.81 m/s²

g = 0.9 g

The centrifugal force will balance the gravitational force

F_c=mg\\\Rightarrow \dfrac{mv^2}{r}=m0.9g\\\Rightarrow v=\sqrt{\dfrac{0.9gmr}{m}}\\\Rightarrow v=\sqrt{0.9gr}\\\Rightarrow v=\sqrt{0.9\times 9.81\times 0.55\times 10^3}\\\Rightarrow v=69.68464\ m/s

\dfrac{1}{T}=\dfrac{v}{2\pi r}\\\Rightarrow \dfrac{1}{T}=\dfrac{69.68464}{2\pi 0.55\times 10^3}\times 24\times 60\times 60\\\Rightarrow \dfrac{1}{T}=1742.24106\ rev/day

The rotation speed is 1742.24106 revolutions per day

6 0
3 years ago
Read 2 more answers
Which describes an image that a concave mirror can make? Which describes an image that a concave mirror can make?
Shkiper50 [21]

Answer:

the image can be rather real or virtual

3 0
2 years ago
If momentum is conserved, what happened to velocity when mass changes?
kifflom [539]

Explanation:

The momentum of an object is given in terms of its mass and its velocity. The law of conservation of momentum says that the net momentum of a system remains conserved for an isolated system.

Momentum is given by :

p = m × v

If p is conserved, velocity is inversely proportional to the mass of an object. When mass changes, velocity also changes but inversely.

5 0
3 years ago
How many revolutions per minute would a 20 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the
timurjin [86]
The general equation for the forces acting on the passengers at the topmost point of the ferris wheel is
mg - R = m \omega^2 r
where
mg is the weight of the passengers
R is the normal reaction of the cabin
m \omega^2 r is the centripetal force

In order to feel weightless, the normal reaction felt by the passengers should be zero. Therefore, the equation becomes:
mg=m \omega^2 r
or
g=\omega^2 r
where \omega is the angular frequency of the wheel and r is its radius. Since we know its radius, 
r= \frac{20 m}{2}=10 m
we can calculate the angular frequency:
\omega= \sqrt{ \frac{g}{r} } = \sqrt{ \frac{9.81 m/s^2}{10 m} } =0.99 rad/s
From which we find the frequency at which the ferris wheel should rotate:
f= \frac{\omega}{2 \pi}= \frac{0.99 rad/s}{2 \pi}=0.158 s^{-1}
This is the number of revolutions per second, so the number of revolutions per minute will be
f=0.158 s^{-1} \cdot 60 = 9.48 min^{-1}
6 0
3 years ago
An object with mass 3M is launched straight up. When it reaches its maximum height, a small explosion breaks the object into two
german

Answer:

x_{L} = 106\,m

Explanation:

Each piece is analyzed by using the Principle of Momentum Conservation and the Impulse Theorem:

Heavier object:

(2\cdot M)\cdot (0) + F\cdot \Delta t = (2\cdot M)\cdot v_{H}

Lighter object:

M\cdot (0) + F\cdot \Delta t = M\cdot v_{L}

After the some algebraic handling, the following relationship is found:

M\cdot v_{L} = 2\cdot M\cdot v_{H}

v_{L} = 2\cdot v_{H}

Given that both pieces have horizontal velocities only and both are modelled as projectiles, the horizontal component of velocity remains constant and directly proportional to travelled distance. Then:

\frac{v_{L}}{v_{H}} = \frac{x_{L}}{53\,m}

2 = \frac{x_{L}}{53\,m}

x_{L} = 106\,m

8 0
3 years ago
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