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Xelga [282]
2 years ago
7

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

ose sides are 3.84 x 10^5 km in length.-What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.
Physics
1 answer:
Afina-wow [57]2 years ago
6 0

Answer:

W = 1.22 \times 10^9 J

Explanation:

Initial potential energy of the given spacecraft is given as

U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}

so we have

U = - \frac{Gm}{r}(M_e + M_m)

so we have

M_e = 5.98 \times 10^{24} kg

M_m = 7.35 \times 10^{22} kg

m = 1160 kg

r = 3.84 \times 10^8 m

U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})

U = -1.22 \times 10^9 J

now total work done to move it to infinite is given

W = 0 - U

W = 1.22 \times 10^9 J

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A person can run 280 m in 68 s. At what speed are they running?
ella [17]

Answer:

Explanation:

Hi there,

280 miles in 68 seconds

280/68 in 1 second

= 4.12 miles per second

 

or 4.12 x 60

247.2 miles per minute

or 247.2 X 60

= 14832 miles per hour

Hope this helps :-)

6 0
3 years ago
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Consider this statement: Air is matter. Which facts best support the statement?
lyudmila [28]
A. Balloons can be filled with air.
C. Air has mass.
7 0
2 years ago
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Two long, straight wires, one above the other, are seperated by a distance 2a2a and are parallel to the x−axisx−axis. Let the +y
kompoz [17]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
3 0
3 years ago
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0
3 years ago
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