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zhuklara [117]
2 years ago
5

a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the

moon, where the acceleration due to gravity is determine to three significant figures its weight in si units and its mass in si units
Physics
1 answer:
Katena32 [7]2 years ago
7 0

Answer:

W_{earth} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 5.91 * 10^6 N : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg

g_{moon}= 1.62 \frac{m}{s^2}

g_{earth}= 9.8 \frac{m}{s^2}

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N : Rocket weight on moon

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hence greater the distance from center , greater will be the centripetal force to remain in place.  

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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
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Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

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           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

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