Well, if the bag is moving across the floor at a constant velocity. Then I would assume that the Fnet force is equal to 0, and thus the value of acceleration would also be equal to 0.
Answer:
B
Explanation:
Speed is the magnitude of the velocity vector, so it can never be negative.
Answer: = 5.75 × 10 -6
Explanation:
= 5.75 × 10-6
(scientific notation)
= 5.75e-6
(scientific e notation)
= 5.75 × 10-6
(engineering notation)
(millionth; prefix micro- (u))
= 0.00000575
(real number)
Work = Force * distance
Force = 70 N
Work = 3500 J
3500 = 70d
d = 3500/70 = 50 m
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
