Answer:
0.699 L of the fluid will overflow
Explanation:
We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C
and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C
Substituting these values into the equation, we have
ΔV = V₀β(T₂ - T₁)
= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)
= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L
= 0.698832 L
≅ 0.699 L = 0.7 L to the nearest tenth litre
So, 0.699 L of the fluid will overflow
Answer:
The answer is A sorry if i'm wrong
Explanation:
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
Answer:C
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