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3 years ago

What is the electric potential at a distance of 1.2 m from a 7.5 UC point charge?

Physics

2 answers:

grandymaker [24]3 years ago

7 0

Answer:

A. 5.6x10^4

Explanation:

Big Brain

ANTONII [103]3 years ago

7 0

Answer:

5.6x10^4

Explanation:

use the equation V=kq/d

k is the constant 8.99x10^9

V=(8.99x10^9)q/d

q is the charge, 7.5 micro coulombs, but to get coulombs, multiply by 10^-6

V=(8.99x10^9)(7.5x10^-6)/d

And from the problem, we know that the distance is 1.2 meters

V=(8.99x10^9)(7.5x10^-6)/1.2

This simplifies to 5.6x10^4

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V = at + V0

where v0 is the initial speed, a is the acceleration and t is the time.

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v = 5m/s^2*5s + 25m/s = 50 m/s

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A spaceship is traveling at 24,000 m/sec. At T=5 sec, the rocket trusts are turned on. At T=55 sec, the spaceship reaches a spee

slava [35]

Answer:

Explanation:

Acceleration is the change in velocity of a body with respect to time;

Acceleration = change in velocity/change in time

change in velocity = 29,500 - 24,000

change in velocity= 5,500

Change in time = 55 - 5

change in time = 50secs

Substitute into the formula;

spaceships acceleration = 24000/50

spaceships acceleration = 480 m/s²

<em>Hence the spaceships acceleration is 480m/s²</em>

8 0

3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$