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2 years ago

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Consider (a) an electron (b) a photon, and (c) a proton, all moving in vacuum. Choose all correct answers for each question. (iv

) Which carry momentum?

1 answer:

DiKsa [7]2 years ago

4 0

Electron, photon and proton carry momentum .

De Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.
De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ ...(1)
Momentum is the product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.

Equation (1) can be written as $\lambda=\frac{h}{p}$ where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant

For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

From equation (1) and (2) we can conclude that electron, photon and proton carry momentum .

Learn about more photon momentum here :

brainly.com/question/13112527

#SPJ4

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A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway

Andru [333]

Answer:

<em>63.44 rad/s</em>

<em></em>

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>

5 0

3 years ago

It is easy..plz help!

Aleks04 [339]

No because point A has to be higher for more KE to push the object and make the look

3 0

4 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

6 0

3 years ago

Cells store energy in the form of what

lukranit [14]

Cells store energy either as carbohydrates or lipids (fats and oils).

Carbohydrates provide for fast, short term energy. They are a key part of cellular respiration.

Lipids provide for long term energy storage because storing lipids is much more effective than storing carbs. One gram of lipid can store twice as much as one gram of carbs. So to access this energy storage, cells must conduct hydrolysis (inserting a hydrogen and a hydroxide group to split off the energy).

Hope this helps!

3 0

3 years ago

A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long

KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

$h=Lsin30$

$h=10sin30$

$h=5m$

In the case of Inertia would be given by

$I = \frac{mR^2}{2}$

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

$I = mk^2$

$\frac{mR^2}{2}= mk^2$

$\frac{k^2}{R^2}=\frac{1}{2}$

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

$mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})$

$9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})$

$v^2 (1.5) = 98$

$v=8.0829m/s$

Therefore the correct answer is C.

3 0

3 years ago