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4 years ago

what did classical physics predict about electron flow as a result of light shining on a metal surface?

1 answer:

8 0

This looks like the photo electric effect ... classical physics reckoned that if you shone an intense enough light beam on a metal you could get electrons ejected from the metal (maybe in analogy to thermionic emission - heat). It sort of "forgot" about the frequency and photon/particle nature of light.
Enter the "photo electric" effect experiment, Einstein's explanation, and the Nobel committee having an excuse to award E a Nobel prize, even though said prize was probably more for relativity.

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A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane. You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0

3 years ago

. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin

Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

$T-(m*g)=m*a$

Where:

$T=$ thrust $=3 *10^{4} N$

$m=$ module's mass $=1 *10^{4} N$

$g=$ moon's gravity acceleration $=1.623 \frac{m}{s^2}$

$a=$ module's acceleration during takeoff

Then, we can find the acceleration like this:

$a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}$

$a=1.377 \frac{m}{s^2}$

The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

$weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N$

Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

8 0

4 years ago

Who is yr favorite anime?

anyanavicka [17]

Answer:

Kanna Kamui

Explanation:

From Miss Kobayashi's Dragon Maid

7 0

3 years ago

Read 2 more answers

A rocket explodes into two fragments, one 25 times heavier than the other. The magnitude of the momentum change of the lighter f

V125BC [204]

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment. </u>

So, option B is the answer

4 0

3 years ago

When the force applied to an object increases, the motion of the object

Elza [17]

Answer:

Semen is the most important thing in the life cycle of life

6 0

4 years ago