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balu736 [363]
3 years ago
6

given a 60uC point charge located at the origin, find the total electric flux passing through: a) that portion of the sphere r=2

6 cm bounded by 0
Physics
1 answer:
Ann [662]3 years ago
5 0
<span>my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2

</span><span>NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc</span>
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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
4 years ago
A force vector has a magnitude of 599 newtons and points at an angle of 40.8° below the positive x axis. What is the x scalar co
ASHA 777 [7]

Answer:

F_{x}=453.44N

Explanation:

Given data

Force F=599 N

Angle α=40.8°

To find

x scalar component

Solution

The Scalar x component can be found by

F_{x}=FCos\alpha  \\F_{x}=599Cos(40.8)\\F_{x}=453.44N

The Scalar y component can be found by

F_{y}=-FSin\alpha  \\F_{y}=-599Sin(40.8)\\F_{y}=-391.4N

3 0
3 years ago
A +26.3 uC charge qy is repelled by a force
Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0
3 years ago
What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2
kkurt [141]

Answer:

4.3\cdot 10^{-12} J

Explanation:

The fusion reaction in this problem is

4^1_1H \rightarrow ^4_2He +2e^+

The total energy released in the fusion reaction is given by

\Delta E = c^2 \Delta m

where

c=3.0\cdot 10^8 m/s is the speed of light

\Delta m is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

m(^1_1H)=1.007825u is the mass of one nucleus of hydrogen

m(^4_2 He)=4.002603u is the mass of one nucleus of helium

So the mass defect is:

\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u

The conversion factor between atomic mass units and kilograms is

1u=1.66054\cdot 10^{-27}kg

So the mass defect is

\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg

And so, the energy released is:

\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J

6 0
3 years ago
When you drop a strong magnet through the center of a copper pipe, what happens to the copper piper and the magnet?
Sedbober [7]
Copper is; unlike iron and steel; not ferromagnetic, but diamagnetic. This means that induced magnetic fields in copper will counter the applied force.

When you drop a strong magnet through a copper pipe, the moving magnetic field will induce currents (Lenz’ Law). These currents will now induce their own magnetic field. This magnetic field counters the falling magnetic.

Result: the magnet will fall way slower than if it was falling through a plastic pipe.
8 0
4 years ago
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