Answer:
A box sits stationary on a ramp
Explanation:
Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.
Static force of friction is calculated as follows:
F= μη
F is static force of friction.
μ is the coefficient of static friction.
η is the normal force.
Answer:
The distance between the two slits is 40.11 μm.
Explanation:
Given that,
Frequency 
Distance of the screen l = 88.0 cm
Position of the third order y =3.10 cm
We need to calculate the wavelength
Using formula of wavelength

where, c = speed of light
f = frequency
Put the value into the formula


We need to calculate the distance between the two slits


Where, m = number of fringe
d = distance between the two slits
Here, 
Put the value into the formula



Hence, The distance between the two slits is 40.11 μm.
The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

where <em>x</em> is measured in m.
The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

Solve for <em>x</em> :

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
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