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
Answer:
d) v1 = v2 = v3
Explanation:
This can be answered using conservation of energy. We calculate the mechanical energy E=K+U (sum of kinetic and gravitational potential energies) at the original and final points, and impose they are equal.
At the original point we have, for the three balls:
At the final point we have, for the three balls:
Since we have 
, and 
is the same for all balls, then 
is the same for all balls, which means that 
, the final velocity, is the same for all balls.
Answer: Macroscoptic Output
Explanation:
Answers to the rest:
1. B) macroscopic outputs.
2.A) a microscopic change creating a macroscopic output
3.B) Because the energy levels of the electrons in different metals are usually not the same, different metals usually emit different colors of visible light.
4.A) Heat is applied to a solid, causing its molecules to move quickly.
5.A) strontium, sodium, copper, potassium
answer:
2.5 m/s²
Explanation:
120 km/h = 120 ÷ 3.6 = 100/3 ≈ 33 m/s
a = (v2 - v1)/∆t = (33m/s - 0)/ 13s = 33/13 m/s²≈ 2.5 m/s²
