A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonconducting 2.50 cm wire. A uniform electric field of magnitude 1.85×108N/C is directed parallel to the wire, as shown in the figure. (Figure 1)
Part A
Find the tension in the wire.
Express your answer with the appropriate units.
Part B
What would the tension be if both charges were negative?
Express your answer with the appropriate units.

A +8.75 C Point Charge Is Glued Down On A Horizontal Frictionless Table. It Is Tied To A -6.50 C Point

Answers

Answer 1

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

first charge, q₁ = 8.75 μC second charge, q₂ = -6.5 μC  electric field, E = 1.85 x 10⁸ N/Cdistance between the two charges, r = 2.5 cm

(a)

The attractive force between the charges is calculated as follows;

[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N[/tex]

The force on the negative charge due to the electric field is calculated as follows;

[tex]F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N[/tex]

The tension on the wire is the resultant of the two forces and it is calculated as follows;

[tex]T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N[/tex]

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N[/tex]

The force on the first negative charge due to the electric field is calculated as follows;

[tex]F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N[/tex]

The force on the second negative charge due to the electric field is calculated as follows;

[tex]F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N[/tex]

The tension on the wire is the resultant of the three forces and it is calculated as follows;

[tex]T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N[/tex]

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Answer:

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The static electricity light our homes. This statement is false.

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The charge gets accumulated on them, due to which spark occurs.

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What is the magnitude of the electric field at a point 0.0075 m from a 0.0035
C charge?
Use E-
kq
and k - 9.00 x 10° N.m²/C2.
A. 1.9 x 1010 N
B. 4.2 x 10' N
O c. 5.6 x 1011 N
D. 5.5 x 1012 N

Answers

Explanation:

E = 9×10^9 × 0.0035 /(0.0075)²

= 9×10^9 × 3.5×10`³/(7.5×10`³)²

= 31.5×10^6 /(56.25 ×10^-6)

= 0.56 × 10^12

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help pls will give brainiest :)

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use a flight sim

You throw a rock up into the air as hard as you can. It stays in the air a total of 6.0 s. What
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Explanation:

Two students were pushing a heavy sofa. One student pushed with a force of 200.0 N to the right, while the other pushed with a force of 150.0 N to the right. The floor exerted a frictional force of 100.0 N. If the sofa's mass is 91.0 kg, what is its acceleration? Round your answer to three significant figures.

Answers

Answer:

Explanation:

F = ma

a = F/m

a = (200.0 + 150.0 - 100.0) / 91.0

a = 250.0/91.0

a = 2.7472527...

a = 2.75 m/s²

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Answers

The velocity of the object is 4.1 m/s. Recall that the centripetal acceleration is the acceleration of a body moving in a circular path.

Given that the centripetal acceleration is obtained using the formula;

ac = v^2/r

ac= centripetal acceleration

v = velocity

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Substituting values;

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v = 4.1 m/s

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How can people reconcile technology with environmental preservation?​

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1. Developing renewable energy technology

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2. Saving endangered wildlife

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3. Adopting a smarter lifestyle

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If you traveled 50m/s for 60 seconds, how far did you travel? Remember speed=distance/time
Question options:


300 m/s



500 m/s



3,000 m/s



300 km/h

for some reason my question got removed-_-

Answers

the answer is 3,000 m/s

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1. During an experiment in which you are investigating the acceleration changes due to force changes, what value must stay constant during these trials?

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Answers

Answer:

1. A. Force

2.

3. B. or C.

4. 1,000 N as well

Explanation:

1. Force is mass multiplied by acceleration. if force and mass stay constant, then according to that equation, the acceleration stays constant as well.

I couldn't figure out what number 2 is though. Sorry.

The value which must stay constant during these trial is:

A. Force

The best description of the motion of the animal is:

B. The animal will move to the right due to an unbalanced force on the right  

The thing which could be true about the difference between the rockets is:

C. One rocket engine produces more force than the other

The amount of force which the wall hits the sledgehammer is:

C. 1000 N

Force is simply defined as the mass of a body which is multiplied by the acceleration and according to the laws of physics, if there is a constant in the mass and force, then the acceleration would also remain constant.

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Answers

Explanation:

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Answer:

Explanation:yy

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Explanation

(m) is measured in kilograms (kg)

(F) is measured in newtons (N)

acceleration (a) is measured in metres per second squared (m/s²)

Q1) ~mass (m) is measured in kilograms (kg) 

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Hope this helped you- have a good day bro cya)

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Answers

Answer:

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Answer:

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Lotion - Semisolid

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Please mark as brianliest answer.

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Answers

The velocity of the package  after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package  after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

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To calculate the velocity of the package  after it has fallen for 3.0 s

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v = 29.4 m/s

Hence, the velocity of the package  after it has fallen for 3.0 s is 29.4 m/s

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A 10 kg box initially at a speed of 10 m/s accelerates uniformly to a speed of 20 m/s in 2
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Answer:

What Um Sorry Where's The Answer?....

When adding vectors, what cannot change?
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Answers

Answer:

A

Explanation:

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number A is right I think

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