Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
B :)
Explanation:
:) JUST TRUST ME I GOT IT CORRECT
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.
Answer:
bounce up and down
Explanation:
Buoys are used for two main reasons, one is to let the people on land know of a big incoming wave, while the second reason is to generate electricity. When a big wave is approaching the buoy starts to bounce up and down with the strength of the smalled previous waves and then bounce very strongly up as the bigger wave passes by. This movement is combined with pistons within the buoy in order to conduct electricity.
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:

Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,






The hall voltage is given by

Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,

