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VikaD [51]
3 years ago
7

We define the specific heat of a material as the energy that must be transferred to 1.0 kg of that material in order to cause it

to warm 1.0∘C. Part A What happens to the specific heat if we transfer twice that much energy?
Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

No change in the specific heat.

Explanation:

Specific heat is an intrinsic property that it has not depend upon on the amount of substance or energy added, it depends upon the material.

So, when twice the amount of energy is transferred, specific heat of the material does not change rather the energy that is twice in the amount to 1 kg of that material cause it to warm 2.0° C.

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If a neutral material like fur is rubbed on another neutral object like a rubber ball what is the resulting charges of the objec
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B is the right answer I hope I helped
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Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, stick together and move off after
Natali [406]

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3 years ago
Read 2 more answers
A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has
Pani-rosa [81]

Answer:

Part a)

EMF = 0.38 V

Part b)

\frac{dA}{dt} = 0.43 m^2/s

Explanation:

Part a)

Initial value of magnetic flux is given as

\phi_1 = BAcos\theta

\phi_1 = (2.1)(0.35 \times 0.55) cos65

so we have

\phi_1 = 0.17 Wb

Final flux through the loop is given as

\phi_2 = 0

now EMF is given as

EMF = \frac{\phi_1 - \phi_2}{\Delta t}

EMF = \frac{0.17 - 0}{0.45}

EMF = 0.38 V

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

E = Bcos65 \times \frac{dA}{dt}

0.38 = 2.1 cos65(\frac{dA}{dt})

\frac{dA}{dt} = 0.43 m^2/s

5 0
3 years ago
Part A By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on e
aalyn [17]

Answer:

Part A: <em>4.74×10⁴ m</em>

P<em>art </em>B<em>: </em>0.000211 C

Explanation:

Part A:

From coulombs law,

F  = kq²/r²........................ Equation 1

Note: The two object are carrying equal charges.

Where F = Electric force, q = charge, r = distance of separation, k = proportionality constant

<em>Given: F = 4.0 N, q = 1.0 C, </em>

<em>Constant: k = 1/4πe₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1,</em>

<em>4 = 9×10⁹(1²)/r²</em>

<em>r² = (9×10⁹)/4</em>

<em>r² = 2.25×10⁹</em>

<em>r = √(2.25×10⁹)</em>

<em>r = 4.74×10⁴ m.</em>

<em>Part 2:</em>

<em>F = kq²/r²</em>

<em>Making q the subject of the equation</em>

<em>q = √(Fr²/k)........................... Equation 2</em>

Where: F = 4.0 N, r = 1.0 m, k = 9×10⁹ Nm²/c²

Substituting these values into equation 2

q = √(4×1²/(9×10⁹ )

q = 2.11/10⁴

q = 0.000211 C.

6 0
3 years ago
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