Answer:
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Explanation:
I have done this before and got it correct
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;

I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Answer:
(a) 328 Nm
(b) 79.35 Nm
Explanation:
N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T
A = side^2 = 0.175^2 = 0.030625 m^2
(a) Torque = N x i x A x B x Sinθ
For maximum torque, θ = 90 degree
Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90
Torque = 328 Nm
(b) θ = 14 degree
Torque = 150 x 42 x 0.030625 x 1.7 x Sin 14
Torque = 79.35 Nm
Answer:

Explanation:
Given



Required
Determine the impulse
The impulse is calculated as follows:

Substitute values for Force and Time


<em>Hence, the impulse experienced is 8.0Ns</em>
Answer:
the impulse experienced by this object is 4 Ns
Explanation:
Given;
mass of the object, m = 3 kg
velocity of the object, v = 4 m/s
resistive force, F = 20 N
duration of impact, t = 0.2 s
The impulse experienced by this object is calculated as follows;
J = F x t
J = 20 x 0.2
J = 4 Ns
Therefore, the impulse experienced by this object is 4 Ns