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3 years ago

The orbiting of the moon around the Earth can be explained by

1 answer:

3 0

When talking about orbits, it would have to be a mixture of both A. and B. since Newton's first law, gravity plays a huge part in an orbit. However, the universal gravitation law also tells us the relationship between two massive objects in orbit. But to choose only one, it would have to be B. Newton's first law

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We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e

Tju [1.3M]

The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>

So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
Since the Moon and Sun are on opposite sides of Earth during a full moon, the light's distance traveled equals,

$d=(1.5*10^8km)+2(3.8*10^5km)=1.51*10^8km=1.51*10^{11}m$

As we know that light travels at a speed of 300,000 km per second. then, the time taken by the light reflected from sun to reach on earth will be,

$t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min$

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

Learn more about distance here:

brainly.com/question/11495758

#SPJ4

6 0

1 year ago

A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the

harina [27]

Answer:

$R=2.78\ \Omega$

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

$P=I^2R$

Put all the values to find R.

$R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega$

So, the resistance is equal to $2.78\ \Omega$ .

7 0

2 years ago

You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters

olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car $m=2400 kg$

Speed of car $u=14 m/s$

Distance traveled before coming to halt $s=23.2 m$

Let $\mu$ the coefficient of friction

Maximum deceleration road can provide during motion is

$a=\mu g$

using $v^2-u^2=2 as$

$0-14^2=2\cdot (-\mu g)\cdot 23.2$

$\mu =\frac{196}{454.72}$

$\mu =0.431$

7 0

2 years ago

a 280 nm thin film with index of refraction 1.6 floats on waterwhat is the largest wavelength of reflected light for which const

aksik [14]

Answer:

Inside the film the wavelength will be λ/n

For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur

280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed

Note that 700 nm is also the upper wavelength of the visible spectrum

3 0

2 years ago

Describe why drawing a line of best fit is useful.

Amiraneli [1.4K]

Answer:

Cause life

Explanation:

Cause life

8 0

3 years ago