Answer:
The work done is equal to zero = 0
Explanation:
Let us remember that the definition of physical work is given by the product of force by the displacement of the body.
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Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
The blue car is moving with a velocity of -14 m/s using the red car as a reference frame.
Answer:
1. 0 J
2. 7500 J
3. 7500 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Final velocity (v₂) of car = 5 m/s
Original kinetic energy (KE₁) =?
Final kinetic energy (KE₂) =?
Work used =?
1. Determination of the original kinetic energy.
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Original kinetic energy (KE₁) =?
KE₁ = ½mv₁²
KE₁ = ½ × 600 × 0²
KE₁ = 0 J
Thus, the original kinetic energy of the car is 0 J.
2. Determination of the final kinetic energy.
Mass (m) of car = 600 Kg
Final velocity (v₂) of car = 5 m/s
Final kinetic energy (KE₂) =?
KE₂ = ½mv₂²
KE₂ = ½ × 600 × 5²
KE₂ = 300 × 25
KE₂ = 7500 J
Thus, the final kinetic energy of the car is 7500 J
3. Determination of the work used.
Original kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 7500 J
Work used =?
Work used = KE₂ – KE₁
Work used = 7500 – 0
Work used = 7500 J
