Answer:
t = 96.1 nm
Explanation:
For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength
now we know that the path difference of two reflected light from thin liquid layer is given as
here we know that
t = thickness of layer
N = 0 (for minimum thickness of layer)
now we have
1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .
1 N = 1 kg-m/s² .
It's a force equal to roughly 3.6 ounces.
Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision
Answer:
maximum possible velocity = 
Explanation:
centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.
centripetal acceleration 
where v is the velocity of car and r is the radius of circle
maximum friction = umg
where u is the coefficient of static friction.
therefore
therefore maximum possible velocity =
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
