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hammer [34]
3 years ago
5

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

s 1.5multiply1011 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on the Earth. N (b) Calculate the magnitude of the gravitational force exerted by the Earth on the Sun.
Physics
1 answer:
spayn [35]3 years ago
7 0

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

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Answer:

The samples specific heat is 14.8 J/kg.K

Explanation:

Given that,

Weight = 28.4 N

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Q=mc\Delta T

c=\dfrac{Q}{m\Delta T}

Where, m = mass

c = specific heat

\Delta T = temperature

Q = heat

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c=\dfrac{1.25\times10^{4}}{\dfrac{28.4}{9.8}\times(18+273)}

c=14.8\ J/kg. K

Hence, The samples specific heat is 14.8 J/kg.K

8 0
3 years ago
A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm
bekas [8.4K]

Answer:

f = 8.89 cm

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

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now the distance of lens from eye is 2 cm

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so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

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3 0
3 years ago
Compare and contrast static, sliding, and rolling friction.
Mama L [17]
Static friction - the friction that occurs when a force is being applied on a still object. 

Sliding friction - the friction that occurs upon a moving object; it is always lower than the static friction of the same object.

Rolling friction - the friction on a rolling object; it is actually a form of static friction with the contact point of the wheel with the surface.
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3 years ago
A manufacturer sells two goods, one at a price of $1000 a unit and the other at a price of $16,500 a unit. A quantity q1 of the
vaieri [72.5K]

Answer:

manufacturing cost = $16.5 q₂ + $1 q₁ - $ 5

Explanation:

given,

one unit price of  first good is = $ 1000

one unit price of second good is = $ 16500

quantity of the first good = q₁

quantity of the second good = q₂                      

total cost to the manufacturer =  $ 5000

revenue from  good one = $1000 q₁                    

revenue from second good = $16500 q₂                      

manufacturing cost = total revenue - total manufacturing cost

manufacturing cost = $16500 q₂ + $1000 q₁ - $ 5000        

but in the question we have to answer in thousands of dollar

so, divide the result by 1000                  

manufacturing cost = $16.5 q₂ + $1 q₁ - $ 5                

4 0
3 years ago
I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.
Phantasy [73]

Answer:

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5 0
3 years ago
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