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VMariaS [17]
2 years ago
11

How much heat must be removed

Physics
2 answers:
Cerrena [4.2K]2 years ago
5 0

Answer:95400

Explanation:

Liula [17]2 years ago
3 0

Answer:

95,400

Explanation:

3 sig figs

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A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t
emmainna [20.7K]
Efficiency =  Work Output / Work Input

92%  =  Work Output / 100

0.92 =   Work Output / 100

Work Output = 0.92 * 100

Work Output  = 92 joules.
8 0
3 years ago
Eat the majority of your calories in the evening to fuel your sleep hou<br> True<br> False
Nataliya [291]

Answer: false

Explanation:

because

7 0
3 years ago
Read 2 more answers
What's the airplane velocity when it flies 100 miles in 20 seconds<br><br>​
Allushta [10]

Answer : 5m/s

Explanation:the formular for velocity is distance /time or you can say displacement /time. Then it would then be

100/20 =5m/s

3 0
3 years ago
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A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the
exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

(√(0.320/K))² = (0.38197)²

0.320/K = 0.1459

K = 2.19 N/m

4 0
3 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
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