Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play
Answer:

Explanation:
Those kind of problems of electric physics is about capacitors, so the normal questions are:
What are the charge on each electrode?
Solve this you can get other information required in the problem or write down the other questions you need


Capacitance





The charge is find by the equation



Answer:
D) No, since kinetic energy is not conserved.
Explanation:
Since momentum is always conserved in all collision
so in Y direction we can say


Now similarly in X direction we will have


now final kinetic energy of both puck after collision is given as


initial kinetic energy of both pucks is given as


since KE is decreased here so it must be inelastic collision
D) No, since kinetic energy is not conserved.
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Okay, here is my stab at this, I hope it helps!
You know the bullet's initial velocity, V₀ = 450 m/s
You know the final velocity, V = 220 m/s
You also know how long the bullet accelerates (actually decelerates), 14cm, or .14 m
With this information, you learn that you need this equation.
V² = V₀² + 2a (x - x₀), because we have all the information except a, which is the acceleration. So putting it into the equation, it looks like this.
(220m/s)² = (450m/s)² +2a(.14m - 0m)
I'll let you solve the rest, but here are some hints. Your answer will be really big because the bullet slows down really quickly in a really small distance, and you answer will be negative, because this acceleration is causing the bullet to go slower, which is also called deceleration. Hope that helps!