Physics

Question 1

A man walks 30 m east and 40 m North. What is the difference between his traveled distance and his displacement?

Answer 1

Using the Pythagorean theorem, calculate the magnitude of the mans displacement. Ends up being 50 m. His total distance traveled is equal to 30+40=70m therefore the difference between these two is 20m

Question 2

A 1000kg rocket ship, traveling at 100 m/s is acted upon by an average force of 20kN applied in the direction of its motion for 8s. What is the change in velocity of the rocket?

Answer 2

160 m/s

The average force on the rocket equals its mass times the average acceleration; the average acceleration equals the change in velocity divided by the time over which the change occurs. So the change in velocity equals the average force tone the time divided by the mass:
F=ma——> a=F/m
a= delta v/delta t

DeltaV =a•Delta t=Fdeltat/m

Question 3

A car is traveling at 40 km/hr and the driver puts on the brakes, bringing the car to rest in a time of 6s. What is the magnitude of the average acceleration of the car?

Answer 3

24,000 km/hr^2

The magnitude of the average acceleration is the change in velocity divided by the time. The velocity changes by -40km/hr because the car comes to rest.

The time in hours is 6s x [1hr/3600s]=1/600hr
The average acceleration is then a=deltaV/delta t = (-40 km/hr)/(1/600hr)=-24,000 km/hr^2
This question asks for magnitude of this acceleration which is just 24,000

Question 4

What is the relationship between the maximum tension in the elevator cable and the maximum weight of the elevator while the elevator is moving upward?

Answer 4

The forces in the elevator are the tension upward and the weight downward, so the net force on the elevator is the difference between the two. For the elevator to accelerate upwards the tension in the cable will have to be greater than the maximum weight so that there is a new force directed upwards.

Question 5

A firefighter jumps horizontally from a burning building with an initial speed of 1.5 m/s. At what time is the angle between his velocity and acceleration vectors the greatest?

Answer 5

The instant he jumps.
The mans acceleration is always directed downward, whereas his velocity starts out horizontally and gradually rotated downward as his downward velocity increases. Therefore as time progresses, the angle between his velocity and acceleration decreases, which means that the maximum angle occurs at the instant he jumps

Question 6

A 10kg wagon rests on an inclined plane. The plane makes an angle of 30 degrees with the horizontal. Approximately how large is the force required to keep the wagon from sliding down the plane?

Answer 6

The static force of friction acts parallel to the plane and is in the opposite direction from the parallel component of gravity in this setup. Because the wagon is in equilibrium these two forces are equal in magnitude. The parallel component of gravity is given by mgsinø.
So both the parallel component of gravity and static force of friction must be equal to (10kg)(9.8m/s^2)(sin30)=49 N

Question 7

Sin and cos of 0 degrees

Answer 7

(1,0)
(Cos,sin)
(X,y)

Question 8

Sin and cos of 30 degrees

Answer 8

(Square root of 3/2, 1/2)

Cos, sin

Question 9

Sin and cos of 60 degrees

Answer 9

(1/2, square root of 3/2)

Cos, sin

Question 10

Cos and sin of 90

Answer 10

(0,1)

Cos, sin

Question 11

Cos and sin of 120

Answer 11

(-1/2, square root of 3/2)

Cos, sin

Question 12

Cos and sin of 180

Answer 12

(-1,0)

Cos,sin

Question 13

What are the correct SI units?

Answer 13

Mass is kg
Velocity is m/s
Time is a
Newton is a derived unit and is not considered a base unit of SI 
A Newton is equal to kg•m/s^2

Question 14

A 20kg wagon is released from rest from the top of a 15 m long lane which is angled at 30 degrees with the horizontal. Assuming that there is friction between the ramp and the wagon, how is this frictional force affected if the angle of the incline is increased?

Answer 14

The frictional force decreases.

The force of friction on an object sliding down an incline equals the coefficient of friction time the normal force. The normal force is equal in magnitude to the perpendicular component of gravity which is given by mgcosø. As ø increases, Cosø decreases. Therefore the normal force and frictional force decrease as the angle of the incline increases

Question 15

Equation for torque

Answer 15

In order for a seesaw to be balance the torque on the right must exactly match the torque on the left.

r1F1Sinø1=r2F2Sinø2

Question 16

A base jumper runs off a cliff with a speed of 3 m/s. Which of the following is closest to his speed after 0.5 seconds?

Answer 16

6 m/s

This is a projectile motion question. The horizontal component of the jumpers velocity will remain 3m/s throughout the jump. The vertical component of his velocity starts at 0m/s. After 0.5 seconds it will be
V=Vo+at

V=(-9.8 m/s^2)(0.5s)=-4.9 m/s

To get the overall velocity, consider both the vertical and horizontal components using vector analysis. Squareroot of [(3^2)+(4.9^2)]≈ squareroot of 34

Which is about 6

Question 17

Centrifugal force always

Answer 17

Acts anti parallel to the centripetal force vector. They must have equal magnitudes and opposite directions.