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pickupchik [31]

3 years ago

5

A wrestler weighs in for the first match on the moon. will the athleates wiegh more or less on the moon than he will on the eart

h?

1 answer:

Evgesh-ka [11]3 years ago

4 0

The athletes weight will be less on the moon than the Earth. His mass, though, will be the same on both because that is the amount of matter in an object. He will weigh less on the moon because there is less gravity, causing this force to be less.

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Answer each question yes or no. Assume the motions and currents mentioned are along the x axis and fields are in the y direction

Sergeeva-Olga [200]

(a) Does an electric field exert a force on a stationary charged object? YES ( F = Eq)

(b) Does a magnetic field do so?- NO ( F= qvB)

(c) Does an electric field exert a force on a moving charged object? YES

(d) Does a magnetic field do so? YES ( F = qvB)

(e) Does an electric field exert a force on a straight current-carrying wire? ( NO)

(f) Does a magnetic field do so? Yes

(g) Does an electric field exert a force on a beam of moving electrons? Yes

(h) Does a magnetic field do so? YeS

To know more about magnetic field visit : brainly.com/question/10353944

#SPJ4

6 0

2 years ago

A 905 kg test car travels around a 3.25 km circular track. if the magnitude of the force that maintains the car's circular motio

Verizon [17]

Link provided in other answer is a SCAM, don’t click on the link !!!!

8 0

3 years ago

Through which of the following would sound travel fastest?

mixas84 [53]

Answer: ??? D? dont quote me on it tho <3

Explanation:

4 0

2 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

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

or

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago