Please help quick please

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

The weight of an object is measured in air to be 7.0 N. The

Vinil7 [7]

Answer:

Buoyant force = 3.0 N

The object will not float.

Explanation:

Apparent weight of a body immersed in water is the actual weight of object minus buoyant force

Given in the question that;

Weight of object in air = 7.0 N

Apparent weight of object = 4.0 N

4.0 N = 7.0 N - Buoyant force

Buoyant force = 7.0 - 4.0 = 3.0 N

In this case, the buoyant force is less than weight of the object thus the object will sink.

5 0

3 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Troyanec [42]

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=2.00^2+5.50^2$

$AC=\sqrt{(2.00^2+5.50^2)}$

$AC=5.85\ m$

We need to calculate the path difference

Using formula of path difference

$\Delta x=AC-BC$

Put the value into the formula

$\Delta x=5.85-5.50$

$\Delta x=0.35\ m$

We need to calculate the lowest possible frequency of sound

Using formula of frequency

$f=\dfrac{nv}{\Delta x}$

Put the value into the formula

$f=\dfrac{1\times340}{0.35}$

$f=971.4\ Hz$

Hence, The lowest possible frequency of sound is 971.4 Hz.

8 0

2 years ago

I’ll mark brainliest

Flura [38]

Answer:

When the image distance is positive, the image is on the same side of the mirror as the object, and it is real and inverted. When the image distance is negative, the image is behind the mirror, so the image is virtual and upright.

Explanation:

3 0

3 years ago

Neil Armstrong, the first human being to step on the moon, left a foot print that is essentially unchanged nearly 50 years later

Likurg_2 [28]

Since the boot-print was left there nearly 50 years ago, there has been very little wind and very little rain in that area, and plus, there have been very few people or other animals walking around in that spot to disturb it.

6 0

2 years ago