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Sidana [21]
3 years ago
6

A 1.4-µC point charge is placed between the plates of a parallel plate capacitor. The charge experiences a force of 0.59 N. What

is the magnitude σ of the charge density on either plate of the capacitor?
Physics
1 answer:
solmaris [256]3 years ago
4 0

Answer:

\sigma=3.72\times 10^{-6}\ C/m^2

Explanation:

Given that,

Charge, q=1.4\ \mu C=1.4\times 10^{-6}\ C

Force experienced by the charge, F = 0.59 N

We need to find the magnitude of charge density on either plate of the capacitor. Electric field on parallel plate capacitor is given by :

E=\dfrac{\sigma}{\epsilon_o}

E=\dfrac{F}{q}

\dfrac{F}{q}=\dfrac{\sigma}{\epsilon_o}

\sigma=\dfrac{F.\epsilon_o}{q}

\sigma=\dfrac{0.59\times 8.85\times 10^{-12}}{1.4\times 10^{-6}}

\sigma=3.72\times 10^{-6}\ C/m^2

So, the charge density on either plate of the capacitor is 3.72\times 10^{-6}\ C/m^2. Hence, this the required solution.

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A device known as Atwood's machine consists of two masses hanging from the ends of a vertical rope that passes over a pulley. As
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Answer:

a= \frac{(m_{1} -m_{2} )*g}{m_{1} +m_{2} }

T= \frac{m_{2}g(1+m_{1} -m_{2} ) }{m_{1}+m_{2}  }

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

M1 free body diagram : Look at the attached graphic

∑F = m₁*a

W₁ -T= m₁*a

W₁ - m₁*a = T  Equation 1

M2 free body diagram :Look at the attached graphic

∑F = m₂*a

T-W₂= m₂*a

W₂ + m₂*a = T  Equation 2

Equation 1  = Equation 2

W₁ - m₁*a = W₂ + m₂*a

W₁ - W₂ =  m₁*a + m₂*a

m₁*g -m₂*g = a* (m₁ + m₂)

a = (m₁*g -m₂*g) / (m₁ + m₂)

a= \frac{(m_{1} -m_{2} )*g}{m_{1} +m_{2} }

Calculation of the tension in the rope (T)

We replace a in the equation 2

W₂ + m₂*a = T

W₂ + m₂*g*(m₁ -m₂) /  (m₁ + m₂) = T

m₂*g + m₂*g*(m₁ -m₂) /  (m₁ + m₂) = T

m₂*g( (1+ (m₁ -m₂)) /  (m₁ + m₂) = T

m₂*g (1+m₁ -m₂) /  (m₁ + m₂) = T

T= \frac{m_{2}g(1+m_{1} -m_{2} ) }{m_{1}+m_{2}  }

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3 years ago
Felipe drives his car at a velocity of 28 m/s. He applies the brake, which slows the vehicle down at a rate of 6.4 m/s2 and caus
stiv31 [10]

Answer:

Explanation:

Use the one-dimensional equation

v_f=v_0+at where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time it takes to reach that final velocity. We are solving for t. Filling in the other values:

0 = 28 + (-6.4)t and

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t = 4.4 seconds

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3 years ago
MA seconds class lever always more than 1​
Hoochie [10]
It is highly helpful to know that the mechanical advantage (M.A.) of Class two levers is usually greater than one. It is because the overall length of the Effort Arm is higher than the overall length of Load Arm. It is easily known by MA is effort arm/load arm.

!! ⬇️

The mechanical advantage of a lever of the second order is always greater than one because its effort arm is always longer than the load arm i.e. Effort arm > Load arm.

Second class lever has mechanical advantage always more than one as load is in between fulcrum and effort making the effort arm longer than the load arm.

First Class Lever -- the effort and the load on either side of the fulcrum. Some examples would be a crowbar or a seesaw. The effort is only less than the load if the load is closer to the fulcrum. The lever then acts as a force magnifier and the mechanical advantage is greater than one.
8 0
3 years ago
An object with a mass M = 250 g is on a plane inclined at 30º above the horizontal and is attached by a string to a mass m = 150
malfutka [58]

Answer:

0.495 m/s

Explanation:

T = tension force in the string connecting the two objects

M = Mass of the object on inclined plane = 250 g = 0.250 kg

m = Mass of the hanging object = 150 g = 0.150 kg

a = acceleration of each object

From the force diagram, force equation for the motion of the object on the inclined plane is given as

T - Mg Sin30 = Ma\\T = Mg Sin30 + Ma

From the force diagram, force equation for the motion of the hanging object on the inclined plane is given as

mg - T = ma\\T = mg - ma

Using the above two equations

Mg Sin30 + Ma = mg - ma

(0.250)(9.8) Sin30 + (0.250) a = (0.150) (9.8) - (0.150)a

a = 0.6125 ms^{-2}

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v = Speed gained by the object

Speed gained by the object can be given as

v = sqrt(2ah)\\v = sqrt(2(0.6125)(0.20))\\v = 0.495 ms^{-1}

5 0
3 years ago
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