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Ilya [14]
3 years ago
14

A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

of the spacecraft is 250 m/s and the asteroid is pushed back –25 m/s, what is the mass of the asteroid? Assume there is no net force on the system.
A) 100 kg
B) 6,250 kg
C) 10,000 kg
D) 6,250,000 kg
Physics
2 answers:
Jet001 [13]3 years ago
3 0

Before the launch, the momentum of the (spacecraft + asteroid) was zero.  So after the launch, the momentum of the (spacecraft + asteroid) has to be zero.

Momentum = (mass) x (velocity)

Momentum after the launch:

Spacecraft:  (1,000 kg) x (250 m/s) = 250,000 kg-m/s

Asteroid: (mass) x (-25 m/s)

Their sum:  250,000 - 25(mass) .

Their sum must be zero, so  250,000 kg-m/s = (25 m/s) x (mass)

Divide each side by  25 :  10,000 kg-m/s = (1 m/s) x (mass)

Divide each side by (1 m/s) :  10,000 kg = mass


Deffense [45]3 years ago
3 0

Answer:

IT IS C EDGEINUTY

Explanation:

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Nonamiya [84]

Answer:

a_x=6\ \text{m/s}^2 and a_y=0\ \text{m/s}^2

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Explanation:

t_1=2\ \text{s}

v_x=1\ \text{m/s}

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t_2=2.5\ \text{s}

v_x=4\ \text{m/s}

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Average acceleration in the different axes

a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2

a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2

The components of the acceleration is a_x=6\ \text{m/s}^2 and a_y=0\ \text{m/s}^2

The magnitude of acceleration

a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2

Direction

\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}

The magnitude of accleration is 6\ \text{m/s}^2 and the direction is 0^{\circ}.

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2 years ago
3. If the net work done on an object is negative, then the object's kinetic energy
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Answer:

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Korolek [52]

Answer:

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

Explanation:

The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:

\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f} (1)

Where:

M - Mass of the merry-go-round, in kilograms.

m_{1}, m_{2}, m_{3} - Masses of the three children, in kilograms.

R - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.

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If we know that M = 182\,kg, m_{1} = 17.4\,kg, m_{2} = 28.5\,kg, m_{3} = 32.8\,kg, R = 1.60\,m and \dot n_{o} = 15.3\,\frac{rev}{min}, then the final angular speed of the system is:

\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)

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\dot n_{f} = 18.388\,\frac{rev}{min}

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

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