A 240 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg tr
1 answer:
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Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = ( - v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
Answer:
1.45544 J
Explanation:
See attachment
Answer:
F = 69.5 [N]
Explanation:
We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.
where:
N = normal force [N]
miu = friction coefficient
f = friction force = 22 [N]
Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
where:
F = force exerted [N]
f = friction force [N]
m = mass = 95 [kg]
a = acceleration = 0.5 [m/s²]
Now replacing: