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Harman [31]
3 years ago
15

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the

mountain. (b) Compare the mountain’s mass with that of Earth. (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent?
Physics
1 answer:
LiRa [457]3 years ago
8 0

Answer:

a)   M = 2,939  10¹⁷ kg , b)   M_{e} / M  = 2 10⁷

Explanation:

a) The equation for gravitational force is

         F = G m M / r²

Where G is the gravitational constant that is worth 6.67 10⁻¹¹ N m² / kg², m the mass epa person. M the mass of the Mountain and r the distance between them.

The value of this force is 2% of the person's weight

          F = 0.02 W = 0.02 mg

we replace

         0.02 mg = G m M / r²

         M = 0.02 g r² / G

         r = 10 km = 10 10³ m = 1.0 10⁴ m

         M = 0.02  9.8  (10⁴)² / 6.67 10⁻¹¹

         M = 2,939  10¹⁷ kg

b) to compare the masses we find their relationship

       M_{e} / M = 5.98 1024 / 2,939 1017

       M_{e} / M  = 2 10⁷

c) treating the mountain as a point object

d) The mountain is not spherical so the distance changes depending on the height of the mountain

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What happens when polar molecules are between oppositely charged metal plates
KatRina [158]

Answer:

They will become aligned according to the charges on the metal plate.

Explanation:

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Standing waves can be established within a cylindrical volume. However, it is not known if the volume is closed at both ends, op
Evgen [1.6K]

The frequencies are missing in the question. The three successive resonance frequencies within the volume are $f_1, f_2 \text{ and}\ f_3$.

Solution :

Let the volume be : v

The frequency for one end open and one end closed is given by :

So, $ f_n = \frac{(2n-1)v}{4L}$

Therefore,

$f_1 = \frac{v}{4L}$     ,     $f_2 = \frac{3v}{4L}$  ,     $f_3 = \frac{5v}{4L}$

So, $f_2 - 3f_1$  and  $f_3=\frac{5}{3}f_2$

Therefore, the ratio of   $\frac{f_3}{f_2}=\frac{5}{3}$    which is not a whole number.

Now the frequency for open volume and closed at both end

$f_n=\frac{nv}{4L}$

So,    $f_1=\frac{v}{4L}$   ,    $f_2 = 2f_1$ ,  $f_2=3f_1$

From above formulae we can see that  ratio of     is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.

Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.

4 0
3 years ago
A small truck has a mass of 2085 kg. How much work is required to decrease the speed of the vehicle from 22.0 m/s to 13.0 m/s on
Alisiya [41]

Answer:

Work required is 328387.5 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 2085kg

Initial velocity, Vi = 13m/s

Final velocity, Vf =22m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

Where;

W is the work required.

Kf is the final kinetic energy possessed by the object.

Ki is the initial kinetic energy possessed by the object.

But Kinetic energy = ½MV²

W = ½MVf² - ½MVi²

Substituting into the equation, we have;

W = ½(2085)*22² - ½(2085)*13²

Simplifying the equation, we have;

W = 1042.5 * 484 - 1042.5 * 169

W = 504570 - 176182.5

W = 328387.5J

Therefore, the work required to decrease the speed of the vehicle is 328387.5 Joules.

8 0
3 years ago
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