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bulgar [2K]
3 years ago
13

A charge Q is uniformly distributed along the x axis from x = a to x = b. If Q = 45

Physics
1 answer:
luda_lava [24]3 years ago
3 0

Answer:

The potential at the point 8 meters is approximately 48.98  Joules/C

Explanation:

Notice that we need to find the potential at a point aligned to the rod's axis (see attached figure), that is located 6 meters from one end of the rod. Notice as well, that the length (L) of the rod is 5 meters.

Since we have a uniformly distributed charge, the charge density per unit of length is defined as:

\lambda=\frac{Q}{L} \\\lambda=\frac{45}{5}\,\frac{10^{-9}\,C}{m} \\\lambda=9\,\frac{10^{-9}\,C}{m}

In order to find the contribution of each little segment dx of charge

dq=\lambda\,dx

to the potential at the requested point, we need to perform an integral:

V=\int\limits^{11}_{6} {} \, dV \\V=\int\limits^{11}_{6} {} \, k\,\frac{dq}{x}  \\V=\lambda \,*k\,\int\limits^{11}_{6} {} \, \frac{dx}{x}  \\V=8.98*9  \, \,ln(\frac{11}{6}) \,\frac{J}{C} \\V=80.82\,* \,0.606 \,\frac{J}{C}\\V=48.98 \frac{J}{C}

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Answer:

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7 0
3 years ago
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What type of organization is used in a paragraph that lists similarities between two objects?
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<span>A. Comparison 

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3 0
3 years ago
An integrated circuit is a ___.
EastWind [94]
HI buddy!

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5 0
3 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel
BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

t=\frac{6.22}{2}=3.11 s

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

Solving the  formula, we find

s=\frac{1}{2}(9.8)(3.11)^2=47.4 m

7 0
3 years ago
A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st
wel

Answer:

the tension T2 when the rock is completely immersed is T2 =  29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where  ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V=  T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced  

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V  =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 =  29.05 N

3 0
3 years ago
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