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)

Answer:
<u>According </u><u>to </u><u>second </u><u>law </u><u>of </u><u>motion</u><u>,</u><u>t</u><u>he acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</u>
<em>So </em><em>simply</em><em>,</em><em> </em><em>it </em><em>can </em><em>be </em><em>affected </em><em>due </em><em>to </em><em>increasing </em><em>force </em><em>as </em><em>there </em><em>is </em><em>close </em><em>relationship </em><em>between </em><em>momentum.</em>
Explanation:
<em>The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.</em>
<em>I </em><em>hope </em><em>it </em><em>was </em><em>helpful </em><em>for </em><em>you </em><em>:</em><em>)</em>
Answer:

ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:

Where:


Now:

Where
is the mass of the car,
is the initial velocity of the car,
is the mass of train,
is the final velocity of the car and
is the final velocity of the train.
Replacing data:

Solving for
:

Changed to cm/s, we get:

b) The kinetic energy K is calculated as:
K = 
where M is the mass and V is the velocity.
So, the initial K is:



And the final K is:




Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As

For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Answer:

Explanation:
Average velocity of oxygen molecule at given temperature is

now we have
M = 32 g/mol = 0.032 kg/mol
T = 27 degree C = 300 K
now we have


now for de Broglie wavelength we know that


