If an astronaut throws an object in space, the object’s speed will _____

1 answer:

7 0

The object's speed will not change.

In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
$\sum F=ma$
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.

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Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0

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Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

$m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}$

j-direction:

$m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}$

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

$\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J$

Answer: The system lost 500J worth of kinetic energy in the collision

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Bedrock that is closer to oceanic ridges is younger in age than bedrock that is farther away. What can best be concluded from th

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The shape of Earth's lithosphere is continually changing. Bedrock that is closer to oceanic ridges is younger in age than bedrock that is farther away. What can best be concluded from this information? It can cause pollution of water in lakes and rivers

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2. A liquid that is twice as dense as water is used in a barometer. With this barometer, atmospheric pressure would push the liq

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Answer:

The reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

Explanation:

We know that pressure, P = ρgh where ρ = density of material, g = 9.8 m/s and h = height.

Since density of liquid = 2 × density of water = 2 × 1000 kg/m = 2000 kg/m³

Since atmospheric pressure = 101500 N/m²,

The height of liquid for this pressure is h = P/ρg = 101500/(2000 × 9.8) = 5.18 m

For the same pressure, the height for water is h = P/ρg = 101500/(1000 × 9.8) = 10.36 m