Answer:
150 Nm
Explanation:
Unit conversion:
5 cm = 0.05 m
8 cm = 0.08 m
The torque generated by such force would be the product of the force by the muscles and the moment arm of that force.
T = FR
Where T is the torque (Nm), F = 3000 N is the force and R = 0.05 m is the moment arm
T = 3000*0.05 = 150 Nm
Answer:
v ≈ 4.47
Explanation:
The Formula needed = <u>KE = </u>
<u> m v²</u>
<u></u>
Substitute with numbers known:
2000J = × 200kg × v²
Simplify:
÷100 ÷100 (Divide by 100 on both sides)
2000J = 100 × v²
= v²
20 = v²
√ √ (Square root on both sides)
√20 = √v²
4.472135955 = v (Round to whatever the question asks)
v ≈ 4.47 (I rounded to 2 decimal places or 3 significant figures, as that is what it usually is)
Answer:
v = 14 m/s
= 31.3 mph
The answer would be the same if the mass of the car were 2000 kg
Explanation:
Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.
V^2 = v^2 + 2×a×Δx
Now V, the final velocity is zero as the car stops
0 = v^2 + 2×a×Δx
v^2 = -2×a×Δx
v =√-2×a×Δx .....*
Now applying Newton's Second Law
Fnet = m×a
-Fk = m×a
-μ×N = m×a
-μ×m×g = m×a (The mass cancels out)
a = -μ×g
Substituting the value of a back to equation *
v = √-2×(-μ×g)×Δx
v = √-2×(-0.5×9.8)×20
v = 14 m/s
Therefore the speed the car was travelling with v = 14 m/s
which is equal to 31.3 mph
Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.
The force between two charges objects is related to separation.
