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:
distance = 33.124 meters
Explanation:
To solve this question, we will use one of the equations of motion which is:
s = ut + 0.5a * t^2
where:
s is the distance that we want to get
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/sec^2
t is the time = 2.6 sec
Substitute with the givens in the equation to get the distance as follows:
s = ut + 0.5a * t^2
s = (0)(2.6) + 0.5(9.8)(2.6)^2
s = 33.124 meters
Hope this helps :)
Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion.
https://www.khanacademy.org › stat...
Static and kinetic friction example (video) | Khan Academy
Answer a would be static friction
Answer b is fluid friction
(Air resistance is fluid friction. Fluid friction is the friction experienced by objects which are moving in a fluid and the air is a fluid.)
Answer c is static friction
ANSWER D IS KINETIC FRICTION
Hope this helps :D
Answer:
Final volume, V2 = 3 Litres
Explanation:
Given the following data;
Initial volume, V1 = 1 litre
Initial temperature, T1 = 20°C
Final temperature, T2 = 60°C
To find the final volume, we would use Charles' law;
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
V1/T1 = V2/T2
Making V2 as the subject formula, we have;
V1T2 = V2T1
V2 = (V1T2)/T1
Substituting into the formula, we have;
V2 = (1 * 60)/20
V2 = 60/20
Final volume, V2 = 3 Litres
Do you remember this formula for the distance traveled while accelerated ?
<u>Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)</u>
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice <em>D)</em> is the closest one.
