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julsineya [31]
3 years ago
7

The data table shows some data related to the Sun and the planets in our solar system.

Physics
2 answers:
AlexFokin [52]3 years ago
8 0
You have selected the correct answer and blobbed over it with your pencil.

I assume you must have looked at Saturn's average distance, found 1427,
divided that number by 6, got 237 and change, then looked at the others,
and found that 228 was the only one that's anywhere close.
DerKrebs [107]3 years ago
7 0

Answer:

Mars

Explanation:

The distance between Sun and Saturn = 1427 Million km

one sixth of the distance between Sun and Saturn will be

= \frac{1427\times 10^{6} }{6} km

= 237.83 Million km

From the given data it is evident that the right answer will be Mars.

As the distance between Mars and Sun is 228 million km which is the closest value to the one sixth of the distance between Sun and Saturn.

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A 86 kg human stands on the surface of Venus. The mass of Venus is 4.9 × 1024 kg and its radius is 6.1 × 106 m. Collapse questio
iris [78.8K]

Answer:

755.37 N

Explanation:

We are given that

Mass of venus=m_1=4.9\times 10^{24}kg

Radius=r=6.1\times 10^6m

Mass of human=m_2=86 kg

We know that the gravitational force between two bodies

F=G\frac{m_1m_2}{r^2}

Where G=Gravitational constant=6.67\times 10^{-11}Nm^2/kg^2

Using the formula

The magnitude of the gravitational force exerted by Venus on the human=F=\frac{6.67\times 10^{-11}\times 86\times 4.9\times 10^{24}}{(6.1\times 10^6)^2}

The magnitude of the gravitational force exerted by Venus on the human=F=755.37N

7 0
3 years ago
The biggest reason that some fuels are considered nonrenewable
zhuklara [117]
Because many fuels are fossil fuels they take millions of years to form and the known reserves are being used much faster than the new ones being made
4 0
3 years ago
When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

#SPJ4

4 0
1 year ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

8 0
4 years ago
On the surface of the earth the weight of an object is 200 lb. Determine the height of the
siniylev [52]

Answer:

The height of the  object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of Gmm_{e}

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

200=\dfrac{Gmm_{e}}{(3958.756)^2}

200\times(3958.756)^2=Gmm_{e}

Gmm_{e}=3.134\times10^{9}

We need to calculate the height of the  object

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

125=\dfrac{200\times(3958.756)^2}{r^2}

r^2=\dfrac{200\times(3958.756)^2}{125}

r^2=25074798.5

r=\sqrt{25074798.5}

r=5007.4\ miles

Hence. The height of the  object is 5007.4 miles.

7 0
3 years ago
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