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HACTEHA [7]
1 year ago
7

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is

the sphere’s potential energy?
Physics
1 answer:
vodka [1.7K]1 year ago
8 0

The sphere’s Electric potential energy is 1.6*10^{-6}J

Given,

q=6. 5 µc, V=240 v,

We know that sphere’s Electric potential energy(E) = qV=6.5*10^{-6} *240=1.6*10^{-6}J

<h3>Electric potential energy</h3>

The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.

In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?

Learn more about Electric potential energy here:

brainly.com/question/24284560

#SPJ4

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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo
makkiz [27]

Answer:

(a) The speed of the target proton after the collision is:V_{2f} =433(m/s), and (b) the speed of the projectile proton after the collision is: v_{1f}=250(m/s).

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}, and y axle:0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}. Now replacing the value given as: v_{1i}=500(m/s), \beta_{1}=+60^{o} for the projectile proton and according to the problem \beta_{1}and\beta_{2} are perpendicular so \beta_{2}=-30^{o}, and assuming that m_{1}=m_{2}, we get for x axle:500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2} and y axle: 0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}, then solving for v_{2f}, we get:v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f} and replacing at the first equation we get:500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}, now solving for v_{1f}, we can find the speed of the projectile proton after the collision as:v_{1f}=250(m/s) and v_{2f}=\sqrt{3}*v_{1f}=433(m/s), that is the speed of the target proton after the collision.

5 0
2 years ago
A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera
yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

\Delta V_{max} = 240 V

frequency, f = 50Hz

I_{max} = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, \omega =  2\pi f = 2\pi \times50 = 100\pi

a). Inductive reactance,  X_{L} is given by:

\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega

X_{L} = 47.12\Omega 

b). The capacitive reactance,  X_{C} is given by:

\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega

X_{C} = 0.636\Omega

c). Impedance, Z = \frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega

Z = 2400\Omega

d). Resistance, R is given by:

Z = \sqrt {R^{2} + (X_{L} - X_{C})}

2400^{2} = R^{2} + (47.12 - 0.636)^{2}

R = \sqrt {5757839.238}

R = 2399.5\Omega

e). Phase angle between current and the generator voltage is given by:

tan\phi = \frac{X_{L} - X_{C}}{R}

\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})

\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}

\phi = 1.11^{\circ}

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Flauer [41]

Answer:

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Explanation:

Given:

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Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

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F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

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The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

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