Answer:
0.3257 seconds
39.67 m/s
Explanation:
Speed = 263 km/h
Converting to m/s


Distance to travel = 23.8 m
Time = Distance / Speed

The time taken to go from one end of the court to the other is 0.3257 seconds
Time = 0.6 s
Speed = Distance / Time

The speed of the tennis ball is 39.67 m/s
The formula for the torque is
<span>τf = p F
where
</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon
If there are given values, substitute in the equation and solve for the torque.</span>
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25

= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer: 6.36
Explanation:
Given
Radius of grindstone, r = 4 m
Initial angular speed of grindstone, w(i) = 8 rad/s
Final angular speed of the grindstone, w(f) = 12 rad/s
Time used, t = 4 s
Angular acceleration of the grinder,
α = Δw / t
α = w(f) - w(i) / t
α = (12 - 8) / 4
α = 4/4 = 1 rad/s²
Number of complete revolution in 4s =
Δθ = w(i).t + 1/2.α.t²
Δθ = 8 * 4 + 1/2 * 1 * 4²
Δθ = 32 + 1/2 * 16
Δθ = 32 + 8
Δθ = 40 rad/s
40 rad/s = 40/2π rpm = 6.36 rpm
Therefore, the grindstone does 6.36 revolutions during the 4 s interval