Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed 
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:
The heat liberated by the aluminium container would be equal to heat absorbed by the water.
According to the first law of thermodynamics for a closed system the total energy of the system is always conserved. Therefore the energy liberated by the aluminium container will completely absorbed by the water untill they both come to thermal equilibrium.
Answer:
Explanation:
Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .
So the given process of isothermal compression must have been done at the temperature of 300K , keeping the temperature constant .
Work done on gas at isothermal compression is equal to heat transfer .
work done on gas = 80 x 10³ J
work done on gas = n RT ln v₁ / v₂
n is number of moles v₁ and v₂ are initial and final volume
molecular weight of gas = 28.97 g
1.5 kg = 1500 / 28.97 moles
= 51.77 moles
work done on gas = n RT ln v₁ / v₂
Putting the values in the equation above
80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2
ln v₁ / .2 = .62
v₁ / .2 = 1.8589
v₁ = 0.37 m³
Answer:
Explanation:
Let 'F₁' and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.
Now, the horizontal and vertical components of these forces are:
As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,
Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.
At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.
Therefore, the net torque by the forces 
will be zero. This gives,
But, 
Therefore,
We know,
∴
The answer is the diameter of the cam shaft.
This is used to compute for the area of the circle so you can multiply it with the rocker ratio to get maximum amount of weight the valve can lift.
