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Nezavi [6.7K]
3 years ago
5

A skateboarder with a mass of 45 kilograms is riding on a skateboard with a mass of 2.5 kilograms. What should be the velocity o

f the skateboarder if she has to attain a momentum of 264 kilogram meters/second?. . Blaaah Physics.-.
Physics
2 answers:
DanielleElmas [232]3 years ago
6 0
Given momentum (P) and mass (m), the velocity of the object is dividing the momentum by the mass. The total mass of the skateboarder and the skateboard is 47.5 kg. Performing the operation,
                              v = P / m
                             v = (264 kg m/s  / 47.5 kg) = 5.56 m/s
Therefore, the velocity would be 5.56 m/s. 

klasskru [66]3 years ago
5 0
So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.

Since we know momentum and mass we can now solve.

Velocity = 264/(45+2.5)
              = 5.56 m/s
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The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

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Answer:

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