Answer:

Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

The friction forces are computed by


Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]

Simplifying

Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)

The Coulomb force is equal to the constant k times the product of charge one and charge two over radius.
F=k((q1q2)/r)
Answer:
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Explanation:
This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
1 sin θ₁ = 1.33 sin θ₂
θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Answer:
d ≈ 7,6 g/cm³
Explanation:
d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³
V = l³ = (1.74cm)³ ≈ 5,27 cm³
Answer:

Explanation:
<u>Free Fall Motion</u>
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:

Where 
Similarly, the distance y the object has traveled is calculated as follows:

If we know the height h from which the object was dropped, we can solve the above equation for t:

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

Therefore, it has traveled down a distance:

Thus, the height of the pen is:
