The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.
If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero.
Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.
The 53N of friction is (the weight) x (the coefficient of kinetic friction).
53N = (490N) x (coefficient).
Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082 .
Rounded to the nearest hundredth, that's <em>0.11 </em>. (choice 'd')
The kinetic energy of the electron is
where
is the mass of the electron and v its speed. Since we know the value of the kinetic energy,
, we can find the value of the speed v:
Answer:
D
Explanation:
19.6 newtons
A 2.00-kilogram object weighs 19.6 newtons on Earth.
Answer:
3secs
Explanation:
Given the following parameters
height H= 81.3m
Velocity v = 12.4m/s
Required
Time it take to reach the ground
Using the equation of motion
H = ut+1/2gt²
81.3 = 12.4t + 1/2(9.8)t²
81.3 = 12.4t + 4.9t²
4.9t² + 12.4t - 81.3 = 0
Using the general formula to find t
t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)
t = -12.4±√153.76+1593.48/2(4.9)
t = -12.4±√1747.24/9.8
t = -12.4+41.8/9.8
t = 29.4/9.8
t = 3secs
Hence it took 3secs to reach the ground
The given question is incomplete. The complete question is as follows.
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 
, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.
Explanation:
We will calculate the work done as follows.
W = 
= 
= ![[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}](https://tex.z-dn.net/?f=%5B14000x%20%2B%205000x%5E%7B2%7D%20-%208666.7x%5E%7B3%7D%5D%5E%7B0.54%7D_%7B0%7D)
= 7560 + 1458 - 1364.69
= 7653.31 J
or, = 7.65 kJ (as 1 kJ = 1000 J)
Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.
