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aev [14]
2 years ago
11

A test charge gains 10 joules of potential energy as it moves through an electric field. It starts its movement at point 1 and e

nds its movement at point 2. The potential difference between the two points is 4 volts. How many coulombs of charge are carried by the test charge
Physics
1 answer:
Nataliya [291]2 years ago
8 0

2.5 C coulombs of charge are carried by the test charge.

<h3>What is test charge?</h3>

A test charge is a charge of such a modest size that it has little impact on the field surrounding the site where it is placed. To determine the strength and direction of an electric field, a positive test charge is a unit positive charge.

U = qV

where,

U = 10 j

V = 4V

then, q = U/V

q = 10/4

q = 2.5 C

to learn more about test charge go to - brainly.com/question/21830343

#SPJ4

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What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
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Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing a=76m^2

We know that density of air d=1.29kg/m^3

Speed at top surface v_2=290m/sec and speed at bottom surface v_1=150m/sec

According to Bernoulli's principle force is given by

F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N

4 0
3 years ago
Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1). Graph △XYZ and its image after a rotation of 180° about (2, –3).
Alla [95]

The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).

If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

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Ndividuals living in highly populated areas are more inclined to _______.
V125BC [204]

Answer:

B. Social Interaction.

Explanation:

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3 years ago
One hundred turns of insulated copper wire are wrapped into a circular coil of crosssectional area 1.20⇥103 m2. The two ends of
arsen [322]

Answer:

236.3  x 10^-^3 C

Explanation:

Given:

B(0)=1.60T and B(t)=-1.60T

No. of turns 'N' =100

cross-sectional area 'A'= 1.2 x 10^-^3m²

Resistance 'R'= 1.3Ω

According to Faraday's law, the induced emf is given by,

ℰ=-NdΦ/dt

The current given by resistance and induced emf as

I = ℰ/R

I= -NdΦ/dtR

By converting the current to differential form(the time derivative of charge), we get

\frac{dq}{dt}=  -NdΦ/dtR

dq= -N dΦ/R

The change in the flux dФ =Ф(t)-Ф(0)

therefore, dq = \frac{N}{R} (Ф(0)-Ф(t))

Also, flux is equal to the magnetic field multiplied with the area of the coil

dq = NA(B(0)-B(t))/R

dq= (100)(1.2 x 10^-^3)(1.6+1.6)/1.3

dq= 236.3  x 10^-^3 C

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vova2212 [387]
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2 years ago
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