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jeyben [28]
3 years ago
14

When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth. Which type of pull are the

y experiencing?
A. tension
B. external
C. internal
D. compression
Physics
2 answers:
torisob [31]3 years ago
7 0
When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth, the type of force they are experiencing is <span>A. tension. Tension is the opposite of compression which is pulling of the astronaut from the ground or Earth</span>
Anvisha [2.4K]3 years ago
3 0

Answer:

A. tension

Explanation:

The reason that astronauts float is due to their free fall, that is, when astronauts are in Earth orbit, Earth gravity attracts them. when astronauts travel to the moon, this attraction that the earth imposes on them is much smaller, called the outer attraction. This external attraction exerts on the astronauts a tention, since the gravity of the moon is insignificant.

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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

5 0
3 years ago
It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision (\Delta t):

J=F \Delta t (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

F=ma

So, (1) becomes

J=ma \Delta t

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: a=\frac{\Delta v}{\Delta t}. If we substitute into the previous equation, we find

J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v

And the quantity m\Delta v is equivalent to the change in momentum, \Delta p.

6 0
3 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
Sara walks part way around a swimming pool. She walks 50 yards north, then
Mekhanik [1.2K]

Answer:

20 Yards

Explanation:

|---20----|

|            |

| 50       |50

|---D--->|

Start      End

Total displacement(D)  20 yards (East).

7 0
3 years ago
His the bottom, or lowest point, of a wave.
Gnom [1K]

A trough is the lowest point in a wave.

5 0
3 years ago
Read 2 more answers
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