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:
35.7kJ
Explanation:
we can calculate the amount of heat energy required , using this formula
Q = mcθ
where.
Q = heat energy (Joules, J)
m = mass of a substance (kg)
c = specific heat capacity (units
)
θ = change in temperature (Celcius,C or Kelvin K)
Assume Specific heat capacity (c) of water =
mass =0.1 kg

We have that K.E and work done is
From the Question we are told that
mass= 80.0 kg
Velocity= 7.50 m/s.
vertical distance of 5.20 m
Slowed to 3.00 m/s.
Generally the equation for Kinetic energy is mathematically given as


B)

Therefore


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Answer:
2.1 × 10⁻⁵ T
Explanation:
Given:
Inner radius, r = 4 mm = 0.004 m
Outer radius, R = 25 mm = 0.025 m
Current, I = 4 A
Distance of the point from the center, a = 17 mm = 0.017 m
μ₀ = 4π × 10⁻⁷ T·m/A
Now,
For the hollow cylinder magnetic field (B) is given as:
on substituting the respective values, we get
or
B = 2.1 × 10⁻⁵ T