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3 years ago

Find the location of the center of mass of the earth-moon system relative to the center of earth.

Physics

1 answer:

mamaluj [8]3 years ago

8 0

Answer:

4765 Km

Explanation:

The center of mass for a two point system is given by the formula:

CM= (mass earth x distance of cm to centre of earth + mass moon x distance of cm to moon ) / mass earth + mass moon

mass earth: 5.97 x 10^24 Kg

mass moon: 7.35 x 10^22 Kg

distance earth-moon = 384,400 Km : 3.84 x 10^6 m

The center of mass will be located between the line distance joining the earth moon but since we are told to calculate relative to the center of earth ,the distance from earth to the center of mass is cero making our calculations easier.

CM = (5.97 x 10 ^24 x 0 + 7.35 x 10 ^22 x 3.84 x 10 ^6 )/ (5.97 x 10^24 + 7.35 x 10^22)

CM= 7.35 x 10 ^22 x 3.84 x 10 ^6/ ( 6.04 x 10 ^24) m = 4.765 x 10 ^6 m = 4765 Km

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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi

algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0

2 years ago

2. A car accelerates down the road. What is the "reaction" to the tires pushing on the road?

Artist 52 [7]

The answer would be friction

7 0

2 years ago

An object accelerating at 16 m/s/s doubles its mass and triples its net force acting on it. What will the new acceleration be? (

nataly862011 [7]

Answer:

24 m/s²

Explanation:

The given parameters are;

The initial acceleration of the object, a = 16 m/s²

Let 'm' represent the initial mass of the object

The initial force acting on the object, F = m × a

∴ F = 16 × m = 16·m

When the mass is doubled, we have;

The new mass of the object, m₂ = 2 × m = 2·m

When the net force acting on the object triples, we have;

The new net force acting on the object, F₂ = 3 × F = 3 × 16·m = 48·m

From F = m × a, we have;

a = F/m

∴ The new acceleration of the object, a₂ = F₂/m₂

From which, by plugging in the values, we have;

a₂ = 48·m/(2·m) = 24

The new acceleration of the object, a₂ = 24 m/s².

6 0

3 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

lukranit [14]

Answer:(a) 2.40 (b) horizontal distance. (c) 0.630. (d) 6.50

Explanation:that's all is talking about a speed and distance and time right

3 0

2 years ago