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:
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
so:
where 
is the moment of inertia of the merry-go-round, 
is the initial angular velocity of the merry-go-round, 
is the moment of inertia of the merry-go-round and the child together and 
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where 
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
Finally we replace all the data:
Solving for :
Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7
<span>force applied causes movement of an object in the same direction as the applied force.</span>
