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zavuch27 [327]
2 years ago
10

What happens if : . The test charge is not tiny.

Physics
1 answer:
docker41 [41]2 years ago
3 0

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

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since centripetal acceleration is always towards the center of the circle

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3 years ago
A photon has 3.4 × 10–18 joules of energy. Planck’s constant is 6.63 × 10–34 J•s.
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3 years ago
A refrigerator having a power rating of 350W operates for 12 hours a day. calculate the cost of electrical energy to operate it
rodikova [14]

Answer:

See the answers below.

Explanation:

The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.

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The fuse can be calculated by knowing the amperage.

P=V*I

where:

P = power = 350 [W]

V = voltage = 240 [V]

I = amperage [amp]

Now clearing I from the equation above:

I=P/V\\I=350/240\\I=1.458[amp]

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3 0
3 years ago
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

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4 0
3 years ago
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one
alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

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If a wire has greater resistance then,the wire will be greater in length.

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Total resistance of copper  wire=Sum of resistance of two piece of wires

r+7r=13

8r=13

r=\frac{13}{8}ohm

Resistance of long piece of wire=7\times\frac{13}{8}=\frac{91}{8}\Omega

Resistance of short piece of wire =\frac{13}{8}\Omega

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

R=\frac{\rho L}{A}=\frac{\rho}{A}L=KL

\frac{R}{L}=K

Where K=\frac{\rho}{A}=Constant

Using the formula

\frac{13}{L}=\frac{\frac{13}{8}}{L'}

\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}

L'=\frac{L}{8}

Length of short piece of wire=L'=\frac{L}{8}

Length of long piece of  wire=L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L

% of length of short piece of   wire=\frac{\frac{L}{8}}{L}\times 100=12.5%%

The resistance of the short piece=\frac{13}{8}\Omega

The resistance of the long piece=\frac{91}{8}\Omega

8 0
3 years ago
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