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GarryVolchara [31]
3 years ago
6

A good way to measure your power is __________. A. to walk on a balance beam B. to jog a mile to loosen up C. to stretch joints

and muscles regularly D. to run 10 flights of stairs in several minutes
Physics
2 answers:
Kisachek [45]3 years ago
8 0

Answer:

C. to stretch joints and muscles regularly

Explanation:

solmaris [256]3 years ago
8 0

Answer:

d run 10 flights of stairs in several minutes.

Explanation:

im right :)

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
If Neptune’s mass were reduced, what could be done to maintain the same force of gravitational attraction between Neptune and th
Alenkinab [10]
Make Neptune closer to the sun because then it would have a stronger gravitation pull. Because the closer the objects are, they will have a stronger gravitaional force and when the object has more mass, the gravity is also stronger. So, if the mass is reduced, the gravity force would be reduced, but if you bring neptune closer, the gravity force would increase
5 0
2 years ago
A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
Which best describes a chemical reaction? the rearrangement of atoms by breaking and reforming chemical bonds a reaction that ab
Ede4ka [16]

Answer:

A

Explanation:

5 0
3 years ago
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At which of the following air temperatures will the
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The answer is: B. -487 ..............

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