Im afraid i cant help u if we can’t see the image u r working with. could u provide an image of the question?
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:
do explain what u need help with?
Answer:
Zero
Explanation:
As we know that the force and the motion direction should always be perpendicular to each other due to which the work is done by static friction be zero
Therefore
F.dcos(theta) = F.d cos(90) = 0
Hence, the work done by static friction is zero
Therefore the same is to be considered
First one is leaves second one is dams
