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777dan777 [17]
3 years ago
12

A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

he x-axis at −8 cm . −10−8 −6 −4 −2 2 4 6 8 10 8 µ C 3 µ C − 3 µ C x → (cm) What is the magnitude of the total electrostatic force on a third particle with charge −3 µC placed on the x-axis at 2 cm
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

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Why won’t anyone help me please anybody help me I really need help .
irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

8 0
2 years ago
A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o
Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0
3 years ago
A force of 200 N stretches a spring 30 cm. What is the spring constant of the spring? How far would this spring stretch with a f
bija089 [108]

Hooke's Law

F = k. Δx

Δx = 30 cm = 0.3 m

200 = k . 0.3

\tt k =\dfrac{200}{0.3}= 666.6

the spring stretch for 100 N:

\tt \Delta x=\dfrac{100}{666.6}=0.15=15\:cm

4 0
2 years ago
How do determine which is Y2, Y1, X2, and X1 on a graph. <br><br> And how do find rise over run.
fredd [130]
One point will be X1,Y1 and the other will be X2,Y2. It does not matter which is which except that X1 and Y1 have to be the same point and X2 and Y2 have to be the same point. For example, let's say you were given (2,3) and (6,8). No matter which point is X1,Y1 and the other is X2,Y2, the slope will still be 5/4. 

The rise is the change in y from one point to the other. The run would be the change in x from one point to the other.
7 0
3 years ago
The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif
Neko [114]

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

7 0
3 years ago
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