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3 years ago

What would happen to the moon if there was no sun and earth ?

Physics

1 answer:

iragen [17]3 years ago

4 0

Answer:

High tides would be much smaller than they are now, and low tides would be even lower. This is because the sun would be influencing the tides, not the moon; however, the sun has a weaker pull, which would decrease the tides. ... Winds could become much faster and much stronger without the moon.

Explanation:

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A slice of cheese has a mass of 40 g and a volume of 23 cm^3. What is the density of the cheese in units of g/cm^3 and g/mL?

Mila [183]

The mathematical and proportional relationship between mL and $cm ^ 3$ said us that $1cm ^ 3$ is equivalent to 1mL.

If the density is considered as the amount of mass per unit volume we will have to

$\rho = \frac{m}{V}$

here,

m = mass

V = Volume

Replacing we have that

$\rho = \frac{40g}{23cm^3}$

$\rho = 1.739g/cm^3$

As $1mL = 1cm^3$ we have that the density in g/mL is,

$\rho = 1.739g/mL$

6 0

3 years ago

A migrating bird can travel at a speed of 30 m/s. How far will it travel in 25 minutes at this speed? Give your answer in metres

Nady [450]

Answer:

It will travel 45 km (45000 m)

Explanation:

25 minutes is equal to 1500 seconds (25*60=1500)

distance=speed*time

distance=30m/s * 1500s = 45000 m = 45 km

7 0

1 year ago

A missile is launched upward with a speed that is half the escape speed. What height (in radii of Earth) will it reach?

timofeeve [1]

Answer:

The height (h) will be: $\frac{3}{4}R =h$

Explanation:

The scape speed equation is given by:

$v_{scape}=\sqrt{\frac{2GM}{R}}$

Now, the speed of the missile is

$v_{missile}=\frac{1}{2}v_{scape}$

$v_{scape}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Using the conservation of energy, we can find the maximu height of the missile.

$E_{i}=E_{f}$

$\frac{1}{2}mv_{scape}^{2}-mgR =-mgh$

$\frac{1}{2}\frac{2GM}{4R}-gR =-gh$

$\frac{GM}{4R}-gR =gh$

Let's recall that g = GM/R², using the equivalence principle. When R is the radius of the earth and M is the mass of the earth.

$\frac{1}{4}gR-gR =-gh$

$\frac{1}{4}R-R =-h$

Therefore the height (h) will be:

$\frac{3}{4}R =h$

I hope it helps you!

4 0

2 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

Lyrx [107]

Answer:

$\theta=4.64^{\circ}$

Explanation:

It is given that,

The frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Slit separation, $d=2.2\times 10^{-5}\ m$

Let $\theta$ is the angle away from the central bright spot the third bright fringe past the central bright spot occur. The condition for bright fringe is :