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ASHA 777 [7]
3 years ago
9

1. A student demonstrates electromagnetic induction using a

Physics
1 answer:
Katen [24]3 years ago
4 0

answer is :D it would be a great answer

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Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
With a force of 5 newtons, Amanda pushes the stack of books to the right. At the same time, Jeremy, her little brother, pushes t
Vaselesa [24]
Its letter C. 5N to the left. Since Jeremy's force in Newtons are higher than Amanda's (in newtons), and since Jeremy's force directs to the left, then the direction of the force will be to the LEFT. Then subtract the higher one to the lower one so that would be: 10N-5N=5N. So it is C. 5N to the left.
5 0
3 years ago
Read 2 more answers
A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.
Ugo [173]

Answer:

(A) 10052.2 m/s²

(B)  0.00678 seconds

Explanation:

From the question,

(A) Applying

V² = U²+2as..................... Equation 1

Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.

make a the subject of the  equation

a = (V²-U²)/2s........................ Equation 2

Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

Substitute these values into equation 2

a = (68²-0²)/(2×0.230)

a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

make t the subject of the equation

t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

t = (68-0)/10052.2

t = 0.00678 seconds

5 0
3 years ago
What is the wavelength of a sound wave<br> with a speed of 331 m/s and a frequency<br> of 500 Hz?
ludmilkaskok [199]

Answer:

0.777m

Explanation:

The sound wave has a wavelength of 0.773m.

Explanation:

To solve this problem we have to use the wave equation that is given below:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for  

λ :

λ = v f

Let's plug in our given values and see what we get!

λ = 340 m s

440 s − 1

λ = 0.773 m

Hope this helps, Mark as brainliest if u want

4 0
2 years ago
Applying the speed of Sound
solong [7]
Did you find the speed of sound from how far away the storm is
7 0
3 years ago
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