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WINSTONCH [101]
3 years ago
11

What is the Ultraviolet Catastrophe?​

Physics
1 answer:
Sholpan [36]3 years ago
7 0

The ultraviolet catastrophe was the prediction of late 19th century/early 20th century classical physics that an ideal black body (also blackbody) at thermal equilibrium will emit radiation in all frequency ranges, emitting more energy as the frequency increases.

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A flask with a tap has a volume of 200cm3 when full of air, tye flask has a mass of 30.98g. the flask is connected to vacuum pum
fgiga [73]
1. 30.98g
2 D = M / V = 30.98/200 = 0.1549
7 0
2 years ago
Help plz anyone ?????
Alexus [3.1K]

Answer:

Most likely, it will be harder to get strong magnets to change phase because they have more density.

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3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How w
Ksivusya [100]

It will be approximately equal.

<h3>How will the final kinetic energy change?</h3>

We can infer that all of the energy in the electron is Potential energy (PE) because the energy provided by the photon is hardly enough to outweigh the work function.

It will gain kinetic energy (KE) as it advances in the direction of the anode because it is moving through an electric field. All of the PE will have been transformed to KE by the time it reaches the anode.

According to the question

K = hf - W

W = Work function

The energy of photons is comparable. After conversion, there was only a little amount of KE remaining.

Therefore, PE (W) essentially equals KE (K).

It will about be equal.

Learn more about work function here:

brainly.com/question/19595244

#SPJ4

3 0
2 years ago
10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m
Mariulka [41]

Work formula:

W = Fd\cos(\theta)

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

W = (50N)(1.0)(0) = 0

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

W = (50 N)(1.0m)(1)

W = 50

total work = 50 joules

6 0
2 years ago
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