Answer:
Observed time, t = 5.58 s
Explanation:
Given that,
Speed of light in a vacuum has the hypothetical value of, c = 18 m/s
Speed of car, v = 14 m/s along a straight road.
A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.
We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :
t is observed time.
So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.
4/3 pi (radius)^3
If the radius is in cm the volume will be in mL
Answer:
Top speed is 31.68 m/s
Explanation:
Speed of a body is the distance traveled by the body in unit time.
Speed is generally expressed in meter per second or kilometer per hour.
Here, the cheetah travels a distance of 274 m in 8.65 s.
Therefore, speed of cheetah is the distance traveled by it in one second.
For 8.65 s, distance traveled is 274 m
For 1 s, distance traveled will be 
meters per second. This value is called speed.
So, speed is given as the ratio of the distance traveled to that of the time taken.
∴ Top speed of cheetah is given as:
Here, 
is the top speed of cheetah.
Answer:
The free-body diagram of the cannonball is found in the attachment below
<em>Note The question is incomplete. The complete question is as follows:</em>
<em>A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.</em>
<em>Draw the free-body diagram of the cannonball.</em>
Explanation:
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.
In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:
F = force exerted by the cannon acting in the direction of angle of projection
Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon
Fw = weight of the cannonball acting in a downward direction
The free body diagram is as shown in the attachment below.
