Answer:
No, the sound produced by screeching brakes is not a useful form of energy, because it does not contribute to stopping the car.
The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
Answer:
Speed - scalar
Velocity - vector
Displacement - vector
Distance - scalar
Measurement - scalar
Measurement and direction - vector
60 m north - vector
100 m west - vector
200 m/s - scalar.
Answer:
<u>0.04 °C⁻¹</u>
Explanation:
First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.
<u />
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Finding Linear Expansivity :
⇒ α = Final length - Original length / (Original length × ΔT)
⇒ α = 9 - 4 / (4 × 70 - 20)
⇒ α = 5 / 5 × 50
⇒ α = <u>0.02</u>
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Finding Area Expansivity :
⇒ Area Expansivity = 2 × Linear Expansivity
⇒ β = 2 × α
⇒ β = 2 × 0.02
⇒ β = <u>0.04 °C⁻¹</u>