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Leni [432]
3 years ago
12

PLEASE ANSWER IN COMPLETE SENTENCE!!!

Physics
2 answers:
Ostrovityanka [42]3 years ago
8 0

Answer + Explanation : Mass is proportional to gravity and gravity is necessary to hold an atmosphere, therefore mass would seem to be very important for the ability for a body to have an atmosphere.

forsale [732]3 years ago
6 0

Answer:

Mass is proportional to gravity and gravity is necessary to hold an atmosphere, therefore mass would seem to be very important for the ability for a body to have an atmosphere. However there are quite a few examples that show that it doesn't require that much mass to get an atmosphere.

Explanation:

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If the electron e − and the positron e + both have rest energy 0.51 MeV, nd the minimum energy a particle of light γ (a photon)
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Answer:

For Total energy and momentum to be conserved, the minimum energy of the photons released is equal to twice the rest mass energy of an electron that is  2 \times 0.51 MeV = 1.02 MeV

The annihilation of electron -positron cannot produce a single photon. It is prohibited by the law of conservation of energy and momentum.

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A T-Bar is similar to a_______
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Donkey

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3 years ago
A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.
Daniel [21]

Answer:

a) a_{center} = 7.38 ~m/s^2

b) a_{end} = 14.77 ~m/s^2

c) v_{center} = 2.43~m/s

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

\tau =I\alpha

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2

So, the Newton's Second Law yields

\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

a = \alpha r = 7.38~m/s^2

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

a = \alpha L = 14.77~m/s^2

c) The conservation of energy can be used to find the velocity when the rod is vertical.

K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s

The linear velocity is v = ωr, so

v = 2.43 m/s.

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3 years ago
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