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3 years ago

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A flat coil having 40 turns, each one of cross-sectional area 12.0 cm^2, is oriented with its plane perpendicular to a uniform m

agnetic field. The field varies steadily from 0.00 T to 1.20 T in 20.0 ms. What emf is induced in the coil during this time?

1 answer:

iragen [17]3 years ago

4 0

what means emf ? what it means

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A 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same.

Pavlova-9 [17]

As the temperature of the lead and helium is the same. Thus the average kinetic energy is also the same for lead and helium.

Reason:

It is given that a 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same. It is required to compare the average kinetic energy of the lead atoms and helium atoms.

The average kinetic energy is calculated as, $K=\frac{3}{2} \frac{R}{N} T$ .

Here K is the average kinetic energy, R is the gas constant, N is the Avogadro's number, and T is the temperature.

As the temperature is the same for both lead and helium. As a result, the average kinetic energy is also the same for lead and helium.

Learn more about average kinetic energy here,

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2 years ago

21. If the Sun's rays were at 45° to a vertical pillar, how would

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Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

Explanation:

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3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

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Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

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