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asambeis [7]
3 years ago
8

How long does it take for a rotating object to speed up from 15.0 to 33.3 rad/s?

Physics
1 answer:
Kitty [74]3 years ago
8 0
(18.3) / (its angular acceleration). seconds.
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What would happen to moon if gravity no longer affected it?
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Answer: The moon would continue to move on its orbital path.

Explanation:

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"It is impossible to devise a process which may convert heat, extracted from a single
Vladimir [108]

Answer:

According to the second law of thermodynamics, we are unable to use the heat of the ocean and the atmosphere because we do not have a reservoir that has a temperature lower than the ocean or the atmosphere.

Explanation:

As you already know, the ocean and atmosphere have a lot of thermal energy, however, we are unable to convert this energy into mechanical energy that would be useful for our activities. This can be explained by the second law of thermodynamics, since it states that the presence of two bodies with different temperatures is necessary for it to be possible to transform heat into work.

In this case, to transform the thermal energy of the ocean and the atmosphere into mechanical energy we would need the existence of a thermal motor, which is only possible to be established when there is a body with high thermal energy and a sink, a reservoir, with low thermal energy, which will be the place where the heat will be expelled, to be converted into work. We do not have a reservoir with less thermal energy than the ocean and the atmosphere, so we cannot use their energy.

7 0
3 years ago
Compare and contrast inference and critical thinking
fgiga [73]
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5 0
3 years ago
Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression
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Answer:

F_{xtotal} = F_{x} - \mu N

Explanation:

Let the total force be F

The force that produces motion in the x direction is given as, F_{x}

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F_{fric} = \mu N

Thus, the net force is given as:

F_{xnet}  = F_{x} - \mu N

7 0
3 years ago
The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge
Eva8 [605]

Answer:

a) v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}

We can write the mass of a proton in MeV/c².

m_{p}=938.28 MeV/c^{2}

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.04

v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) Linac (400 MeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.71

v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) Booster (8 GeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.994

v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) Main ring or injector (150 Gev)

\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.999

v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) Tevatron (1 TeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.9999

v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Have a nice day!

4 0
4 years ago
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