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steposvetlana [31]
3 years ago
11

Pls help I will give brainliest

Physics
2 answers:
Firlakuza [10]3 years ago
8 0

Answer: Pretty sure the answer is B but this kind of looks like a test question.

Explanation:

Over [174]3 years ago
6 0

Answer: hi

Explanation: cool is me

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heat

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Calculate: a student lifts an apple to a height of 1 m. The apple weighs 1 N. How much work does the student do on the apple? Ty
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1J

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Two asteroids collide and stick together. The first asteroid has mass of 1.50 × 104 kg andis initially moving at 0.77 × 103 m/s.
KengaRu [80]

Answer:

Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.

Explanation:

This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction  (in the plane of the two velocity vectors) as y-axis. So, we have that:

p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}

And, since p=mv, we get:

m_1v_{1o}+m_2v_{2o}\cos\theta=(m_1+m_2)v_{fx}\\\\m_2v_{2o}\sin\theta=(m_1+m_2)v_{fy}

Solving for v_fx and v_fy, and calculating their values, we get:

v_{fx}=\frac{m_1v_{1o}+m_2v_{2o}\cos\theta}{m_1+m_2}\\\\\implies v_{fx}=\frac{(1.50*10^{4}kg)(0.77*10^{3}m/s)+(2.00*10^{4}kg)(1.02*10^{3}m/s)\cos20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=878m/s\\\\\\v_{fy}=\frac{m_2v_{2o}\sin\theta}{m_1+m_2}\\\\\implies v_{fy}=\frac{(2.00*10^{4}kg)(1.02*10^{3}m/s)\sin20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=199m/s

Now, the final speed can be calculated using the Pythagorean Theorem:

v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s

And the direction \beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°can be obtained using trigonometry:

\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°

That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.

7 0
3 years ago
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