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umka21 [38]
3 years ago
9

NEED HELP NOW DUE TODAY Which of these statements is most likely correct about Newton's law on gravity? (2 points)

Physics
1 answer:
yKpoI14uk [10]3 years ago
4 0
_______________________a
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A soccer ball collides with another soccer ball at rest. the total momentum of the balls
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I am pretty sure the answer is C.

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If a sound wave traveled from a solid to a liquid its speed would:.
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A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the
levacccp [35]

Tension in the rope due to applied force will be given as

F = 142 N

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

W = F d cos\theta

now we can plug in all values above

W = 142 * 6.1 * cos37

W = 691.8 J

So here work done to pull is given by 691.8 J


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2 years ago
A heater has a resistance of 20 ohms. If the heater operates on 120-V, what is the power created by the heater?
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6 0
2 years ago
bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc
USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

m_A = mass of car A;

v_{A1} = the initial velocity of car A;

v_{A2} = the final velocity of car A;

and

m_B = mass of car B;

v_{B1} = the initial velocity of car B;

v_{B2} = the final velocity of car B.

Then, the law of conservation of momentum demands that

m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}

And the conservation of kinetic energy says that

\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2

These two equations are solved for final velocities  v_{A2} and v_{B2} to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

\boxed{v_{A2} = -1.05m/s}

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

\boxed{v_{B2} = 3.49m/s}

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

4 0
3 years ago
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