From the question, The kinetic energy of the fired arrow is equal to the work done by the bale of hale in stopping the arrow.
We make use of the following formula
mv²/2 = F'd................... Equation 1
Where
- m = mass of the arrow
- v = velocity of the arrow
- F' = average stopping force acting on the arrow
- d = distance of penetration
Make F' the subject of the equation
F' = mv²/2d.................. Equation 2
From the question,
Given:
- m = 20 g = 0.02 kg
- v = 60 m/s
- d = 40 cm = 0.4 m
Substitute these values into equation 2
Hence, The average stopping force acting on the arrow is 90 N
Learn more about average stooping force here: brainly.com/question/13370981
Answer:
0
Explanation:
Weight = mass x gravity
if gravity = 0 then Weight =0 as well
Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge

where
Distance between the two charges

negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be

Now it will add with force due to 1 charge
Thus net force will be
![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
here we can say that there is no external force on fisherman and dock
so here we will use momentum conservation theory
As per momentum conservation
initial momentum of fisherman + boat = final momentum of fisherman + boat

now we will have



so the speed of boat and fisherman will be 1.16 m/s
An example of a high specific heat is water’s specific heat, which requires 4.184 joules of heat to increase the temperature of 1 gram of water 1 degree Celsius. Scientifically, water’s specific heat is written as: 1 calorie/gm °C = 4.186 J/gm °C.