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Svetradugi [14.3K]
4 years ago
11

How many electrons must be removed from a neutral object to leave a net charge of 0.500 μc?

Physics
1 answer:
blsea [12.9K]4 years ago
4 0
<span>Net charge of the object = 0.500 ÎĽc
 The charge of electron = -1.602 x 10^-19 C
 Calculating the net change in the number of electrons = 0.5 x 10^-6 / -1.602 x 10^-19 = - 3.12 x 10^12
 The negative sign implies the number of electrons that needs to be removed that is 3.12 x 10^12</span>
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3 years ago
A block is balanced on top of a frictionless sphere of radius R. When the block is given a slight nudge it starts to slide down
Nataly_w [17]

Answer:\frac{R}{3}

Explanation:

Given

Sphere of Radius R

Suppose mass of block is m

At any instant \theta Normal reaction(N) and weight(mg) is acting such that

mg\sin \theta -N=\frac{mv^2}{R}  , where v is velocity of block at any angle \theta

When block is just about to leave then N=0

therefore

mg\sin \theta =\frac{mv^2}{R}

v^2=gR\sin \theta-------------------1

Also by conserving Energy we get

Potential Energy=kinetic Energy of block

mgh=\frac{mv^2}{2}

here h=vertical distance traveled by block

From diagram

h=R-R\sin \theta

h=R(1-\sin \theta )

mgR(1-\sin \theta )=\frac{mv^2}{2}

2gR(1-\sin \theta )=v^2-----------------2

From 1 and  2

2(1-\sin \theta )=\sin \theta

3\sin \theta =2

\sin \theta =\frac{2}{3}

Thus from this value of h is

h=R(1-\sin \theta )

h=R(1-\frac{2}{3})

h=\frac{R}{3}

3 0
4 years ago
If each Coulomb of charge is given 20 Joules of energy, what is the voltage of the battery?
Assoli18 [71]

Answer:

Explanation:

V = J/C

V = 20/1

= 20 v

Option A is the correct answer

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3 years ago
Why does the same side of the moon always face earth
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4 0
3 years ago
Read 2 more answers
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
7 0
4 years ago
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