Scientists might model volcanic eruptions using a computer model. What is

Q4. Consider the skier on a slope shown in the figure below. Her mass including equipment is 55.0 kg.

Shkiper50 [21]

Answer:

Part a)

When there is no friction then acceleration is

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then acceleration is

$a = 3.33 m/s^2$

Explanation:

Part a)

As we know that the skier is on inclined plane

So here if there is no friction then net force along the inclined plane is given as

$F = mg sin\theta$

now acceleration of the skier is given as

$a = \frac{F}{m}$

$a = g sin\theta$

$a = 9.81(sin25)$

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then net force along the inclined plane is given as

$F = mg sin\theta - F_f$

now acceleration of the skier is given as

$a = \frac{F}{m}$

$a = g sin\theta - \frac{F_f}{m}$

$a = 9.81(sin25) - \frac{45}{55}$

$a = 3.33 m/s^2$

5 0

4 years ago

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

Edwin Hubble categorized galaxies according to their

Llana [10]

Your answer would be that Hubble categorized galaxies by their shape.

6 0

4 years ago

As you watch the video, notice that the size of the tidal bulges varies with the Moon's phase, which depends on its orbital posi

Naya [18.7K]

Answer:

Both

A. Low tides are lowest at both full moon and new moon.

B. High tides are highest at both full moon and new moon.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the moon across to the Earth sphere.

Since gravity variate with the distance:

$F = G\frac{m1\cdot m2}{r^{2}}$ (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, see the image below, point A is closer to the moon than point b and at the same time the center of mass of the Earth will feel more attracted to the moon than point B. Therefore, that creates a tidal bulge in point A and point B.

On the other hand, a full moon it gets when Sun, the Earth and the moon are in a line and the moon is reflecting the sunlight.

When the Moon is between the Earth and the Sun it will be illuminated in its back, so it is not possible to see it from the Earth (that is called new moon).

In those two cases mentioned above, the Sun tidal force contributes to the tidal force of the moon over the earth making high tides higher and low tides lower.

8 0

4 years ago

how does the size of objects impact the pull of gravity between Earth and a baseball thrown into the air

Reil [10]

it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.

3 0

3 years ago

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