I am going to need a picture for this question
Answer:
horizontal component of normal force is equal to the centripetal force on the car
Explanation:
As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle
This force is due to friction force when car is moving in circle with uniform speed
Now it is given that car is moving on the ice surface such that the friction force is zero now
so here we can say that centripetal force is due to component of the normal force which is due to banked road
Now we have


so we have

so this is horizontal component of normal force is equal to the centripetal force on the car
Current = charge per second
2 Coulombs per second = 2 Amperes
Potential difference = (current)x(resistance) in volts.
That's (2 Amperes) x (2 ohms).
That's how to do it.
I think you can find the answer now.
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s