We want to study the impact of a sledgehammer and a wall.
Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.
When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.
This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.
Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.
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brainly.com/question/13952508
Answer:

Explanation:
y = Distance from the center point
d = Separation between slits = 0.3 mm
D = Distance between slit and screen = 1.5 m
= Wavelength = 700 nm
m = Order = 1
We have the relation

The distance from the screen at which the first bright fringe beyond the center fringe appear is
.
Answer:
Change in position of an object A vector quantity with unit of distance.
Explanation:
Where is the object going? Final - initial
Answer:0.061
Explanation:
Given

Temperature of soup 
heat capacity of soup 
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by



integrating From
to 


![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by

Fraction of the total heat that is lost by the soup can be turned is given by

![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)




Answer:
6495.19 Joule
Explanation:
F = Weight of the crate = 250 N
d = Distance the cart is pushed = 30 m
θ = Angle of inclination = 60°
The weight of the crate will be resloved into two components
Fdsinθ and Fdcosθ
Work done by the force of gravity is
W = Fdsinθ
⇒W = 250×30×sin60
⇒W = 6495.19 Joule
∴ The work done by the force of gravity is 6495.19 Joule