We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:
Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.
(1/2)*m*v²=m*g*h, masses cancel out and we get:
(1/2)*v²=g*h, and we multiply by 2 both sides of the equation
v²=2*g*h, and we take the square root to get v:
v=√(2*g*h)
v=99.04 m/s
So the package is moving with the speed of v= 99.04 m/s when it hits the ground.
Answer:
They will move the fridge if they all push in the same direction, but it will not move with constant velocity
Explanation:
The maximum static friction force is
(negative sign since its direction is opposite to the push applied by the people)
Sam can apply a force of 130 N, while Amir and Andre can apply a push of 65 N each, so the total force that they can apply, if they push in the same direction, will be:

This force is larger than the frictional force, so the fridge will start moving.
However, the net force on the fridge will be:

And according to Newton's second law,

where m is the mass of the fridge and a its acceleration, since the net force is not zero, then the fridge will have a non-zero acceleration, so it will not move with constant velocity.
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
Velocity(v) = frequency(f) × wavelength
f = 0.3165
Wavelength = 2×length(L)
L = 157cm
Convert the length in centimetres to metre = 1.57m
v = 2×1.57 × 0.3165
v = 0.99m/s
Approx. 1m/s
Explanation:
The velocity of a wave is the product of its frequency and it's wavelength. The frequency is already known. The wavelength is the distance between two successive wave crests which is formed by sloshing water back and forth in the bath tub. Sloshing water to one end of the tub will produce a wave crest first at that end then the other completing a cycle. The wavelength will be twice the length of the bath tub as it is the distance that both crests are formed.
Wave crest is the highest point of a wave, and in this case is where the water rises to a high point in the bath tub
Answer:
Time take to fill the standing wave to the entire length of the string is 1.3 sec.
Explanation:
Given :
The length of the one end
, frequency of the wave
= 2.3 Hz, wavelength of the wave λ = 1 m.
Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.
We know,
∴
λ
Where
speed of the standing wave.
also, ∴ 
where
time take to fill entire length of the string.
Compare above both equation,
⇒
sec

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.