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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?

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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A 45kg skater and a 60kg skater are standing still, holding hands on frictionless ice. They push away from each other in opposit

frutty [35]

Answer:

The 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.

Explanation:

The velocity of the 60 kg skater can be found by conservation of linear momentum:

$P_{i} = P_{f}$

$m_{a}v_{i_{a}} + m_{b}v_{i_{b}} = m_{a}v_{f_{a}} + m_{b}v_{f_{b}}$ (1)

Where:

$m_{a}$ : is the mass of the first skater = 45 kg

$m_{b}$ : is the mass of the second skater = 60 kg

$v_{i_{a}}$ : is the initial speed of the first skater = 0 (he is standing still)

$v_{i_{b}}$ : is the initial speed of the second skater = 0 (he is standing still)

$v_{f_{a}}$ : is the final speed of the first skater = 3.5 m/s

$v_{f_{b}}$ : is the final speed of the second skater =?

By replacing the above values into equation (1) and solving for $v_{f_{b}}$ we have:

$0 = m_{a}v_{f_{a}} + m_{b}v_{f_{b}}$

$v_{f_{b}} = \frac{-m_{a}v_{f_{a}}}{m_{b}} = \frac{-45 kg*3.5 m/s}{60 kg} = -2.63 m/s$

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I hope it helps you!

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3 years ago

What creates a bubble of hot ionized gas?

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3 years ago

Ablok slides on ahorizonted sur fors which has been lubricated with heavy oil such that the blok soffers a viscous resistance th

svetoff [14.1K]

Answer:

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$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

Explanation:

Given

$f(v) =- cv^\frac{1}{2}$

To start with, we begin with

$F = ma$

Substitute the expression for F

$-cv^\frac{1}{2} = ma$

$-ma = cv^\frac{1}{2}$

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$a = \frac{dv}{dt}$

So, the expression becomes:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

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Velocity (v) is:

$v = \frac{dx}{dt}$ --- distance/time

$v * \frac{dv}{dx}= \frac{dx}{dt}* \frac{dv}{dx}$

$v * \frac{dv}{dx}= \frac{dv}{dt}$

$\frac{dv}{dt} = v * \frac{dv}{dx}$

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So, we have:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

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Divide both sides by $v^\frac{1}{2}$

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$mv^{\frac{1}{2}} * \frac{dv}{dx} = -c$

Divide both sides by m

$v^{\frac{1}{2}} * \frac{dv}{dx} = -\frac{c}{m}$

$v^{\frac{1}{2}} * dv = -\frac{c}{m} * dx$

Integrate:

$\int\limits^v_{v_0} {v^{\frac{1}{2}}} \, dv = -\frac{c}{m}\int\limits^x_0 {}} \, dx$

$\frac{2}{3}v^{\frac{3}{2}}|\limits^v_{v_0} = -\frac{c}{m}x|\limits^x_0$

$\frac{2}{3}(v^{\frac{3}{2}} - v_0^{\frac{3}{2}} ) = -\frac{cx}{m}$

$v^{\frac{3}{2}} - v_0^{\frac{3}{2}} = -\frac{3cx}{2m}$

$v^{\frac{3}{2}} = v_0^{\frac{3}{2}}-\frac{3cx}{2m}$

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

Next, is to get the maximum velocity by distance x.

To do this, we find the derivation by x

$\frac{dv}{dx} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2}{3} - 1} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2-3}{3}} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{3c}{2m} = 0$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{c}{m} = 0$

Divide both sides by $-\frac{c}{m}$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} = 0$

Take cube roots of both sides

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{-1} = 0$

$v_0^{\frac{3}{2}}-\frac{3cx}{2m} = 0$

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7 0

3 years ago

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +15.2 m/s in a time of 22.3 s.

scZoUnD [109]

Answer:

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3 years ago