Answer:
If the angle of incidence is bigger than the critical angle, the refracted ray will not emerge from the medium, but will be reflected back into the medium. This is called total internal reflection. The critical angle occurs when the angle of incidence where the angle of refraction
Work is = to force*displacement which give W=F*d so if 0 force is put on an object the W= 0*d = 0 work, so a is out. for b if force is put on the object but it does not move then W=F*0 (no displacement)= 0 so if there is no movement then no work is done so b is out. for D since W=F*d also says no force is put on the object so W=0*d which would equal 0 work so D is also out. but for c W=F*d and if we assume that F is not = to 0 and d is not equal to 0 then the equation W=F*d would give a number other than 0 so only c will result in a nonzero value for work. in all other cases no work is done. As a side note if a force is applied to an object and the object goes around the world and ends up and the same place that it started then since Work is equal to force * displacement there will still end up with no work being done since W=F*0=0 work (since the object ended up in the same place it started even though it moved the displacement is still 0, so work done will still be 0.
Answer:
For each 1.5 cm the input piston moves, then output piston will move 0.5 cm
Explanation:
Let cross-sectional area of the input piston = A
Let cross-sectional area of the output piston = 3A
pressure (P) = Force (F) / Area (A)
F = PA
For a constant gravitational force on the inlet and outlet piston;
P₁A₁ = P₂A₂
But pressure = ρgh
where;
ρ is density of water
g is acceleration due to gravity
h is the distance or height moved by the piston
(ρg)h₁A₁ = (ρg)h₂A₂
h₁A₁ = h₂A₂
Area of output piston = 3 times area of input piston
h₁A₁ = h₂(3A₁)
For each 1.5 cm the input piston moves, then output piston will move;
1.5A₁ = h₂(3A₁)
1.5 = 3h₂
h₂ = 1.5 / 3
h₂ = 0.5 cm
Thus, for each 1.5 cm the input piston moves, then output piston will move 0.5 cm
Answer:
x = 16 [m]
Explanation:
This problem can be solved using the following equation of kinematics.
![v_{f}=v_{o}+a*t](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At)
where:
Vf = final velocity [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration = 3 [m/s²]
t = time = 2 [s]
![v_{f}=5+3*(2)\\v_{f}=11[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D5%2B3%2A%282%29%5C%5Cv_%7Bf%7D%3D11%5Bm%2Fs%5D)
Now we can find the displacement using the following equation of kinematics.
![v_{f}^{2} =v_{o}^{2} +2*a*x\\11^{2}=5^{2} +2*3*x\\96=6*x\\x=16[m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3Dv_%7Bo%7D%5E%7B2%7D%20%2B2%2Aa%2Ax%5C%5C11%5E%7B2%7D%3D5%5E%7B2%7D%20%2B2%2A3%2Ax%5C%5C96%3D6%2Ax%5C%5Cx%3D16%5Bm%5D)
Answer:
D. 100 J
Explanation:
Given
Mass (m) = 2 kg
Height (h) = 5 m
Acceleration due to gravity (g) = 10 m/s²
Now
Work done(W)
= m * g * h
= 2 * 10 * 5
= 100 Joule
Hope it will help :)❤