4 years ago

Which of the following is an example of an object that could have a net force greater than zero acting on it?

Physics

2 answers:

mario62 [17]4 years ago

5 0

Net force is basically the force an object has when changing direction, so the answer would be D.

il63 [147K]4 years ago

3 0

An object could have a net force greater than zero acting on it is a toy car moving east at a constant velocity because of the direction it moves.

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You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with

Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$

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3 years ago

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iVinArrow [24]

Answer:

something

Explanation:

Something

8 0

3 years ago

Please help me!! I already did A,B and C but I don’t know D but if you want to help me with all then it would be great so I can

Inessa [10]

Answer:

350cm^3 =350milliters

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3 years ago

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3 years ago

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An object with a mass of 2 kg is moving with a velocity of 15 m/s.

7nadin3 [17]

KE= 1/2MV^2 - equation
KE= 1/2 (2 kg)(15 m/s)^2 - plug it to the equation
KE= (1 kg)(225 m/s) - multiply
KE= 225 J - Answer (letter D)
Hope this helps

8 0

4 years ago