Answer: A
Explanation: isotopes of the same thing element have the same number of protons in the nucleus but differ in the number of neutrons.
The correct answer to the question is : 9375 N.
CALCULATION:
As per the question, the mass of the car m = 1500 Kg.
The diametre of the circular track D = 200 m.
Hence, the radius of the circular path R = 
= 
= 100 m.
The velocity of the truck v = 25 m/s.
When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.
Hence, the force acting on the car is centripetal force.
The magnitude of the centripetal force is calculated as -
Force F = 
= 
= 9375 N. [ANS}
The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.
Answer:
A 75.1 N and a direction of 152° to the vertical.
B 85.0 N at 0° to the vertical.
Explanation:
A) The interaction partner of this normal force has what magnitude and direction?
The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>
B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?
Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.
Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.
<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>