A 1500-kg car drives at 30 m/s around a flat circular track 300 m in diameter. What are the magnitude and direction of the net f
1 answer:
Answer:
F=9000N
Direction is towards the center
Explanation:
Car is moving in a circular path so the force acting on the car is the centrifugal force which is always towards the center. In order to find that force we will use the following formula:
Where:
F is the centrifugal force
m is the mass of object
v is the velocity of object
r is the radius of circular path
Since diameter is given then r=diameter/2
r=300/2
r=150m
F=9000N
Direction is towards the center
So the magnitude of the force is 9000N and the direction is towards the center of circular track as car is moving in circular path.
You might be interested in
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
Answer:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
= 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!