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olasank [31]
4 years ago
15

What’s the answer ???

Physics
1 answer:
dybincka [34]4 years ago
3 0

Answer:

i cant even see it

Explanation:

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A machine applies a force of 100 newtons over a distance of 10 meters to raise a 500 newton
kiruha [24]

Explanation:

work input is the work done by the machine in lifting the drum = Force × distance

100N × 10m

= 1000 J

5 0
3 years ago
In Houston, Earth's B→ field has a magnitude of 5.2 × 10−5 Tand points in a direction 57∘ below a horizontal line pointing north
Anastasy [175]

Answer:

F = 0.0034 N

Explanation:

Given:

B = 5.2*10^(-5) T\\Q = 57 degrees\\I_{wire} = 12 A\\L_{wire} = 10 m

The angle between B and wire = 90 - 57 = 33 degrees

Using formula:

F = B*I*L*sin (90-Q)\\F = (5.2*10^(-5)*(12)*(10)*sin (33)\\F = 0.0034 N

6 0
3 years ago
Please help with these i dont know how to do them
Svetradugi [14.3K]

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3 0
4 years ago
Explain why liquids flow but solids do not.
goldfiish [28.3K]
Because they have no dignified shape they take on the shape of whate ever container they are in and when you pour in out it can't really take any shape except tho lay flat on the table. and solids do not flow because they have a dignified shape. hope it helps!
7 0
3 years ago
Read 2 more answers
Find the de Broglie wavelength of the following. (a) a 4-MeV proton 14.3 Correct: Your answer is correct. fm (b) a 40-GeV electr
kherson [118]

a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

a)

The de Broglie wavelength of an object is given by

\lambda=\frac{h}{p} (1)

where

h is the Planck constant

p is the momentum of the particle

Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

m_0 = 938 MeV

And since 4MeV, this means that the proton is non-relativistic. So its kinetic energy is related to its momentum by

E=\frac{p^2}{2m}

which means

p=\sqrt{2Em}

where

E=4 MeV \cdot 10^6 eV/MeV \cdot 1.6\cdot 10^{-19] J/eV=6.4\cdot 10^{-13} J is the kinetic energy

m=1.67\cdot 10^{-27} kg is the proton mass

Substituting, we find

\lambda=\frac{h}{\sqrt{2Em}}=\frac{6.63\cdot 10^{-34}}{\sqrt{2(6.4\cdot 10^{-13})(1.67\cdot 10^{-27})}}=14.3\cdot 10^{-15} m = 14.3 fm

b)

In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

m_0 = 0.511 MeV

Since 40 GeV >> 0.511 MeV, the electron is ultra-relativistic: so we can rewrite its energy as

E = pc

The equation (1) can also be rewritten as

\lambda = \frac{hc}{pc}

where c is the speed of light. The quantity at the denominator is the energy, so

\lambda=\frac{hc}{E}

where:

E=40 GeV = 40\cdot 10^9 eV \cdot (1.6\cdot 10^{-19})=6.4\cdot 10^{-9} J is the energy of the electron

And substituting, we find:

\lambda=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{6.4\cdot 10^{-9}}=3.1 \cdot 10^{-17} m = 0.031 fm

Learn more about de Broglie wavelength:

brainly.com/question/7047430

#LearnwithBrainly

4 0
3 years ago
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