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Anna35 [415]
3 years ago
12

A 50 kg boy runs and jumps with a forward velocity of 1.5 m/s into a 125 kg stationary boat.

Physics
2 answers:
Dafna11 [192]3 years ago
8 0

Answer:

0.43 m/s, forward

Explanation:

i just took the test

Vesnalui [34]3 years ago
4 0

Answer:

A. 0.43 m/s, forward

Explanation:

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The Liquid property
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consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
andriy [413]

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Moment of inertia of the solid sphere:

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Friction force equation
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It’s FF= μ•Fn
Ff stands for friction force
The weird symbol is your coefficient of friction which has no units
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3 years ago
In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied
fiasKO [112]

B. cart B

Explanation:

The acceleration of each cart is given by Newton's second law:

F=ma

a=\frac{F}{m}

where F is the force applied, a is the acceleration and m is the cart's mass.

The force F applied is the same for the two carts, however the mass of cart A (mA) is twice than the mass of cart B (mB), so we can rewrite the two accelerations:

a_A = \frac{F}{m_A}=\frac{F}{2 m_B}

a_B = \frac{F}{m_B}

we see that the acceleration of cart B is twice the acceleration of cart A, therefore cart B will move faster and will win the race.


7 0
3 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
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