Answer:
49.3 m/s
Explanation:
The momentum is defined as the product of the object velocity and its mass.
So the momentum of the truck is

For the car to have the same momentum, its speed must be

Newton's 2nd law of motion is: <em>Force</em> = (mass) x (acceleration) .
Force is the only way to change an object's velocity.
Answer:
%
Explanation:
We have given cooler temperature that is
= 6°C =273+6=279 K
And the warm temperature of water that is 
The maximum possible efficiency is given by 
So maximum efficiency
%
An ant can have more momentum than an elephant when the elephant is standing still.
Answer: A
Explanation
The momentum is the quantification of the movement done by an object.
It is found to be dependent on the mass of the object and the velocity with which it is moving.
In the present case, the ant has negligible mass compared to elephant so the momentum can be more for ant only when the velocity with which the elephant is moving tends to be zero.
As the velocity of elephant will be zero, the momentum of elephant will be zero so in this criteria, the moving ant will be having more momentum compared to elephant with zero velocity.
So an elephant with zero velocity means the elephant is standing still.
Thus, the condition in which the ant will be having more momentum compared to elephant is when the elephant stands still.
Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be
and force on +Q charge y axis due to -Q charge on x-axis be
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge =
units
= Coulomb constant


Net force on +Q charge on y-axis is:




![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)