Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Answer:
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We know the equation
weight = mass × gravity
To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.
So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)
Therefore,
100N = mass × 10
mass= 100N/10
mass= 10 kg
Now, all we need are the moon's gravitational field strength and to apply this to the equation
weight = 10kg × (gravity on moon)
