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Hatshy [7]
4 years ago
12

I am currently studying about spectrums, difraction Gratings, and many other spectrum. So my question is: What should the Spectr

um of the Moon look like? Justify your answer.
This is the only question I am struggling with...so could you please help me? Many Thanks in the future
Physics
1 answer:
saveliy_v [14]4 years ago
3 0
The spectrum of light from the moon should very strongly resemble the spectrum of sunlight. The reason is that any light from the moon started out from the sun. Any difference in their spectra is only due to the moon absorbing more of some wavelengths and less of others. But since the moon appears colorless gray, we don't expect any particular colors to be strongly absorbed, otherwise the moon would look to be the colors of the light that's left.
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A 762 kg car experiences a braking force of 9045 N and skids to a stop in 4.3 seconds. What is the speed of the car just before
PilotLPTM [1.2K]

Answer:

The solution for this problem is:

We will be using the formula for force which is F = ma 

=>10,000 = 2000 * a 

but we need to solve for acceleration so divide both sides by 2000, we will get:

=>a = 5 m/s^2 

Let the initial velocity was u m/s 

=>By v = u - at 

=>0 = u - 5 x 6 

Since acceleration is constant the velocity can be computed by multiplying the acceleration by 6 seconds. 

=>u = 30 m/s

Explanation:

6 0
3 years ago
Which quote best reveals why sowers was successful in achieving her goal?
BabaBlast [244]
Sowers has spent her career fighting for gender equality and it passionate about it .
5 0
3 years ago
How are stars formed and classified?
Romashka-Z-Leto [24]
Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. Stars are typically classified by their spectrum in what is known as the Morgan-Keenan or MK system.
6 0
3 years ago
The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.
lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

(c) v = 7.12 x 10⁵ m/s

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume  of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

<u>m = 1.6 x 10²¹ kg</u>

<u></u>

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

<u>K.E = 2.536 x 10¹¹ J</u>

<u></u>

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

<u>v = 7.12 x 10⁵ m/s</u>

7 0
3 years ago
What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable
VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: 1.92\cdot 10^{-18}J

Explanation:

When a particle with charge q travels across a potential difference \Delta V, then its change in electric potential energy is

\Delta U = q \Delta V

In this problem, we know that:

The particle is an electron, so its charge is

q=-1.60\cdot 10^{-19}C

We also know that the positive terminal is at potential

V_+=0V

While the negative terminal is at potential

V_-=-12 V

Therefore, the potential difference (final minus initial) is

\Delta V = -12-0 = -12 V

So, the change in potential energy of the electron is

\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J

This means that the electron when it is at the negative terminal has 1.92\cdot 10^{-18}J of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

1.92\cdot 10^{-18}J

3 0
4 years ago
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