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NeX [460]
3 years ago
9

Transfer of heat through conduction can occur in which of the following?

Physics
2 answers:
lianna [129]3 years ago
8 0

Answer:d I think

Explanation:

Sav [38]3 years ago
8 0

Answer:

<em><u>D:All of the </u></em><em><u>above</u></em>

Explanation:

<em><u>Solid </u></em><em><u>is </u></em><em><u>the </u></em><em><u>best </u></em><em><u>in </u></em><em><u>conduction</u></em><em><u> </u></em><em><u>whereas</u></em><em><u> </u></em><em><u>liquids</u></em><em><u> and</u></em><em><u> gases</u></em><em><u> are</u></em><em><u> </u></em><em><u>poor</u></em><em><u> </u></em><em><u>in </u></em><em><u>conduction</u></em>

<em><u>Hope </u></em><em><u>this </u></em><em><u>helps</u></em>

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You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through
Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0
3 years ago
Crews at the International Space Station are researching the effects of the weightlessness of space on ________.
andrezito [222]

A: Human Body

C is wrong because they don’t have the tools to test it on another planet

6 0
3 years ago
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A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a
klemol [59]

Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

7 0
3 years ago
Read 2 more answers
In subduction, _____.
user100 [1]

The answer is the less dense plate slides over the denser plate.

4 0
3 years ago
A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
Leya [2.2K]

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

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8 0
4 years ago
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