There are no correct choices on the list you provided. Any unit of acceleration is (a unit of length) divided by (a unit of time, squared).
The part of the cerebrum that controls Voluntary movements is located in the Frontal lobe.
Answer:
215955.06 m/s^2
Explanation:
length of barrel, s = 0.89 m
initial velocity of the bullet, u = 0 m/s
Final velocity of the bullet, v = 620 m/s
Let a be the acceleration of the bullet in the barrel
Use third equation of motion, we get


a = 215955.06 m/s^2
Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.
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:
it is 2.2 m
Explanation:
because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2