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3 years ago

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8. The mass of the sun is 1.99 x 1030 kilograms and its distance from Earth is 150 million kilometers (150 x 109 meters). What i

s the gravitational force between the sun and Earth?
please help

1 answer:

solmaris [256]3 years ago

4 0

Answer:

$F=3.53\times 10^{22}\ N$

Explanation:

Given that,

Mass of the Sun, $m_1=1.99\times 10^{30}\ kg$

The distance of Sun from the Earth, $r=150\times 10^9\ m$

We know that,

The mass of the Earth, $m_2=6\times 10^{24}\ kg$

We need to find the gravitational force between the Sun and the Earth.

The formula between two masses is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{1.99\times 10^{30}\times 6\times 10^{24}}{(150\times 10^9)^2}\\\\F=3.53\times 10^{22}\ N$

So, the gravitational force between the Sun and the Earth is $3.53\times 10^{22}\ N$

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In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

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A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.

nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

<em>momentum of the ball, P = 0.9 kgm/s</em>
<em>weight of the ball, W = 0.14 N</em>

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where;

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2 years ago

Please help i'm going to throw up from stress

Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

$t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}$ where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

$16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}$ and

$16.0=\frac{83640x+2928}{4182x+292.8}$ and

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3 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

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The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

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3 years ago

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