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gavmur [86]
3 years ago
7

A typical student listening attentively to a physics lecture has a heat output of 100 w . how much heat energy does a class of 1

70 physics students release into a lecture hall over the course of a 50 min lecture?
Physics
1 answer:
nexus9112 [7]3 years ago
7 0

Since each student emits 100 W, so 170 students will emit:

total heat = 100 W * 170 = 17,000 W

 

Convert minutes to seconds:

time = 50 min * (60 s / min) = 3000 s

 

The energy is therefore:

E = 17,000 W * 3000 s

<span>E = 51 x 10^6 J = 51 MJ</span>

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9.8m/s^2 down  (option C)

Explanation:

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The figure shows two springs (k1 = 10 N/m and k2 = 20 N/m ) attached to a block that can slide on a frictionless surface. In the
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3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

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