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faltersainse [42]
3 years ago
8

What is the force between two charged objects called ?

Physics
2 answers:
Ierofanga [76]3 years ago
7 0

Answer:

Coulomb force

Explanation:

The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force. The force is along the straight line joining the two charges.

Travka [436]3 years ago
4 0
Coulomb force I think
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In 1965, a group of students wore black arm bands to school in protest of American policies in Vietnam. Administrators banned th
Ede4ka [16]

Answer:

protected under students first amendment rights

Explanation:

did the studyisland :)

4 0
3 years ago
How much time will it take to perform 440 joule of work at a rare of 11 w?​
kifflom [539]

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

7 0
2 years ago
A person throws a ball from height of 6 feet with an initial vertical velocity of 48 feet per second. Use the vertical motion mo
True [87]

Explanation:

Recall the equation for time is distance divided by speed. Here you can use that to solve for "t".

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3 years ago
A water line with an internal radius of 5.29 x 10-3 m is connected to a shower head that has 15 holes. The speed of the water in
fiasKO [112]

Answer:

(a) 3.44 x 10^-3 m^3/s

(b) 8.4 m/s

Explanation:

area of water line, A = 5.29 x 10^-3 m

number of holes, N = 15

Speed of water in line, V = 0.651 m/s

(a) Volume flow rate is given by

V = area of water line x speed of water in water line

V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s

(b) area of one hole, a = 4.13 x 10^-4 m

Let v be the velocity of water in each hole

According to the equation of continuity

A x V = a x v

5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v

v = 8.4 m/s  

5 0
3 years ago
How do you do this problem?
kvasek [131]

Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

Δy = h

v = 0

a = -g

Find: v₀

v² = v₀² + 2aΔy

(0)² = v₀² + 2(-g)(h)

v₀ = √(2gh)

Now find the height reached if the projectile is launched at a 45° angle.

Given:

v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

v = 0

a = -g

Find: Δy

v² = v₀² + 2aΔy

(0)² = √(gh)² + 2(-g)Δy

2gΔy = gh

Δy = h/2

5 0
3 years ago
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