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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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A.) Carbon

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4 0
3 years ago
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The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g
Bumek [7]

Answer:

For outer points of shell

E = \frac{k_eQ}{r^2}

Now for inner point of shell

E = 0

Explanation:

As we know that out side the shell electric potential is given as

V = \frac{K_e Q}{r}

inside the shell the electric potential is given as

V = \frac{K_e Q}{R}

now we know the relation between electric potential and electric field as

E = - \frac{dV}{dr}

so we can say for outer points of the shell

E = -\frac{dV}{dr}

E = - \frac{d}{dr}(\frac{K_eQ}{r})

E = \frac{k_eQ}{r^2}

Now for inner point again we can use the same

E = - \frac{dV}{dr}

E = - \frac{d}{dr}(\frac{K_eQ}{R})

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6 0
3 years ago
A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia
Bingel [31]

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

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v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

5 0
3 years ago
An artillery shell is launched on a flat, horizontal field at an angle of α = 31.7° with respect to the horizontal and with an i
kompoz [17]

Answer:193.90 m/s

Explanation:

Given

launch angle \theta =31.7^{\circ}

launch velocity v_0=202 m/s

Horizontal velocity of the shell after t=18.96 s

time of flight of Projectile T=\frac{2u\sin \theta }{g}

T={\frac2\times 202\sin 31.7}{9.8}

T=21.66 s

i.e. projectile is declining as t>\frac{T}{2}

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Horizontal component of velocity is

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8 0
3 years ago
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If a rock on a cliff starts with 3,000 J of GPE and 0 J of KE and ends with 0 J of GPE and 5,000 J of KE, what law of physics wo
alexgriva [62]

The law of physics that this scenario would violate is the law of conservation of energy.

<h3>What is the law of conservation of energy?</h3>

The law of conservation of energy states that energy possessed by an object or physical body can neither be created nor destroyed but can only be transformed from one form to another.

This ultimately implies that, the energy possessed by the rock at the beginning must be equal to the energy possessed by the rock at the end in accordance with the law of conservation of energy.

Read more on energy here: brainly.com/question/1242059

6 0
2 years ago
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