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drek231 [11]
3 years ago
8

As you take your leisurely tour through the solar system, you come across a 1.45 kg rock that was ejected from a collision of as

teroids. Your spacecraft's momentum detector shows a value of 1.93 x 108 kg.m/s for this rock. What is the rock's speed?
Physics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

1.3310\times 10^8 m/s is the rock's speed.

Explanation:

Momentum is defined as motion possessed by the moving body's mass. Mathematically it a product of mas and velocity of the body.

Momentum(P)=Mass(m)\times velocity(v)

Given : Mass of rock = m = 1.45 kg

Velocity of the rock = v

Momentum of the rock = P =1.93\times 10^8 kg m/s

1.93\times 10^8 kg m/s=1.45 kg\times v

v=\frac{1.93\times 10^8 kg m/s}{1.45 kg}=1.33\times 10^8 m/s

1.3310\times 10^8 m/s is the rock's speed.

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Read 2 more answers
Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
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Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

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