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Effectus [21]
3 years ago
5

A positive charge of 0.00047 C is 15 m from a negative charge of 0.00089 C. What is the force of one of the charges due to the o

ther charge in units of Newtons?
Physics
1 answer:
Lady_Fox [76]3 years ago
8 0

Answer:

16.732 N

Explanation:

Given:

q1 = 0.00047 C = 4.7 x 10^-4 C

q2 = 0.00089 C = 8.9 x 10^-4 C

d = 15 m

k = 9 x 10^9 N m^2 / C^2

To Find:

F = ?

Solution:

F = k x q1 x q2/d^2

F = 9 x 10^9 x 4.7 x 10^-4 x 8.9 x 10^-4 / 15 x 15

F = 9 x 4.7 x 8.9 x 10^9 x 10^-4 x 10^-4 / 225

F = 9 x 4.7 x 8.9 x 10^9 x 10^-8 / 225

F = 9 x 4.7 x 8.9 x 10 / 225

F = 418.3/25

F = 1673.2/100

Therefore, F = 16.732 N

PLZ MARK ME AS BRAINLIEST!!!

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What property do protons can electrons have that neutrons do not ?
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Protons and electrons are charged particles. Neutrons have no charge.

4 0
3 years ago
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Andru [333]

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

3 0
3 years ago
The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the
Liono4ka [1.6K]

Answer: 4.17m

Explanation:

The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.

If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.

So

d2 - d1 = (n *lamba)/ 2

Where n=1,3,5

lamda=v/f =349/62.8

lamda=5.56m

d2= d1 + nlamda/2

d2= 1 + 5.56/2

d2= 3.78m

X'= 1 cos 60= 0.5m

Y= 1 sin60= 0.866m

X"^2 + Y^2 =d2^2

X" =√(y^2 - d2^2)

X"=√(3.78^2 - 0.886^2)

X"= 3.67m

So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67

4.17m

3 0
3 years ago
Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi
valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

x=vt=(15.1m/s)(13.49s)=203.69m   (1)

Next you calculate the distance that teacher has traveled for 13.6s:

x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m  (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

x_{min}=692.88m-203.69m=489.19m

5 0
3 years ago
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