The work done by Joe is 0 J.
<u>Explanation</u>:
When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.
The energy is transferred from one state to another and the stored energy is equal to the work done.
W = F . D
where F represents the force in newton,
D represents the distance or displacement of an object.
Force = 0 N, D = 20 cm = 0.20 m
W = 0
0.20 = 0 J.
Hence the work done by Joe is 0 J.
The vertical component of the initial velocity is ![v_0_y = \frac{y}{t} + \frac{1}{2} gt](https://tex.z-dn.net/?f=v_0_y%20%3D%20%5Cfrac%7By%7D%7Bt%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt)
The horizontal component of the initial velocity is ![v_0_x = \frac{x}{t}](https://tex.z-dn.net/?f=v_0_x%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
The horizontal displacement when the object reaches maximum height is ![X = \frac{xy}{gt^2} + \frac{x}{2}](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7Bxy%7D%7Bgt%5E2%7D%20%2B%20%5Cfrac%7Bx%7D%7B2%7D)
The given parameters;
the horizontal displacement of the object, = x
the vertical displacement of the object, = y
acceleration due to gravity, = g
time of motion, = t
The vertical component of the initial velocity is given as;
![y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt](https://tex.z-dn.net/?f=y%20%3D%20v_0_yt%20-%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Cv_0_yt%20%3D%20y%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Cv_0_y%20%3D%20%5Cfrac%7By%7D%7Bt%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt)
The horizontal component of the initial velocity is calculated as;
![x = v_0_xt\\\\v_0_x = \frac{x}{t}](https://tex.z-dn.net/?f=x%20%3D%20v_0_xt%5C%5C%5C%5Cv_0_x%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
The time to reach to the maximum height is calculated as;
![T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bv_f_y%20-v_0_y%7D%7B-g%7D%20%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B-v_0_y%7D%7B-g%7D%20%5C%5C%5C%5CT%20%3D%20%5Cfrac%7Bv_0_y%7D%7Bg%7D%20%5C%5C%5C%5CT%20%3D%20%20%5Cfrac%7B1%7D%7Bg%7D%20%20%28v_0_y%29%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B1%7D%7Bg%7D%20%28%5Cfrac%7By%7D%7Bt%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%29%5C%5C%5C%5CT%20%3D%20%5Cfrac%7By%7D%7Bgt%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20t)
The horizontal displacement when the object reaches maximum height is calculated as;
![X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}](https://tex.z-dn.net/?f=X%3D%20v_0_x%20%5Ctimes%20T%5C%5C%5C%5CX%3D%20%5Cfrac%7Bx%7D%7Bt%7D%20%5Ctimes%20%28%5Cfrac%7By%7D%7Bgt%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20t%29%5C%5C%5C%5CX%20%3D%20%5Cfrac%7Bxy%7D%7Bgt%5E2%7D%20%2B%20%5Cfrac%7Bx%7D%7B2%7D)
Learn more here: brainly.com/question/20689870
If each side is 0.1 feet extra,
The volume will be 5.1*2.1*1.1= about 11.781.
Perhaps this helps.
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, ![V_{Hall} = 20\ mV = 0.02\ V](https://tex.z-dn.net/?f=V_%7BHall%7D%20%3D%2020%5C%20mV%20%3D%200.02%5C%20V)
Magnetic Field, B = 0.10 T
Hall emf, ![V'_{Hall} = 69\ mV = 0.069\ V](https://tex.z-dn.net/?f=V%27_%7BHall%7D%20%3D%2069%5C%20mV%20%3D%200.069%5C%20V)
Now,
Drift velocity, ![v_{d} = \frac{V_{Hall}}{B}](https://tex.z-dn.net/?f=v_%7Bd%7D%20%3D%20%5Cfrac%7BV_%7BHall%7D%7D%7BB%7D)
![v_{d} = \frac{0.02}{0.10} = 0.2\ m/s](https://tex.z-dn.net/?f=v_%7Bd%7D%20%3D%20%5Cfrac%7B0.02%7D%7B0.10%7D%20%3D%200.2%5C%20m%2Fs)
Now, the expression for the electric field is given by:
(1)
And
![E_{Hall} = V_{Hall}d](https://tex.z-dn.net/?f=E_%7BHall%7D%20%3D%20V_%7BHall%7Dd)
Thus eqn (1) becomes
where
d = distance
(2)
(a) When ![\theta = 90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2090%5E%7B%5Ccirc%7D)
![B = \frac{0.069}{0.2\times sin90} = 0.345\ T](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B0.069%7D%7B0.2%5Ctimes%20sin90%7D%20%3D%200.345%5C%20T)
(b) When ![\theta = 60^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2060%5E%7B%5Ccirc%7D)
![B = \frac{0.069}{0.2\times sin60} = 0.398\ T](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B0.069%7D%7B0.2%5Ctimes%20sin60%7D%20%3D%200.398%5C%20T)
T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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