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3 years ago

A car moves at 18m/s North. What is the x component of the velocity?

Physics

1 answer:

Marina86 [1]3 years ago

6 0

Velocity, v = 18 m/s, makes an angle 60 degrees with the horizontal (x-axis).

(horizontal) x-component of velocity, v(x) = v cos (60) = 18 m/s x cos(60) = 9 m/s

If you need to determine the y-component (vertical) of velocity,

(vertical) y-component of velocity, v(y) = v sin (60) = 18 m/s x sin(60) = 15.6 m/s

Hope it helps

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A rocket is launched from the surface of Earth with a speed v0 that will allow the rocket to escape the gravitational field of E

weeeeeb [17]

Answer:

option ( a ) is correct .

Explanation:

Escape velocity on the earth = √ ( 2 GM / R )

where G is universal gravitational constant , M is mass of the earth and R is radius .

V₀ = √ ( 2 GM / R )

escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth

Radius of the planet = 4 R

escape velocity of planet = √ ( 2 GM / 4R )

= .5 x √ ( 2 GM / R )

= .5 V₀

option ( a ) is correct .

8 0

2 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

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3 years ago

How does 'g' vary from place to place?

r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

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2 years ago

A 30-kg child rides a 20-kg bicycle together,the child and the bicycle have a momentum of 110 kg•m/s. What is the velocity of th

Korolek [52]

What are the options for this?

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3 years ago

The transfer of thermal energy is a major aspect of which branch of physics?

scZoUnD [109]

Mechanics is dealing with forces that are effecting some body, electrostatics is about electrical fields of not moving bodies, and quantum mechanics is dealing with quantum states of atoms.

Thermodynamics as the word say, is dealing with thermal energy that is moving (transferring from one body to another or even better from one medium to another).

Answer is C <span />

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3 years ago

Read 2 more answers