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QveST [7]
3 years ago
9

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

he nearest star is 4.1×1016m 4.1 × 10 16 m away. Suppose you had a spacecraft that could accelerate at 1.5 g g for two thirds of a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.)How long would it take you to reach the nearest star to earth?
Physics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

It would take 8.22037 hrs away. Wouldn't it?

Explanation:

Because

    4.11016

    4.11016

            15

= 8.22037

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An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys
sammy [17]

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

Thus the rope around the pulley must be pulled 4.2 m

7 0
2 years ago
Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s
ozzi

Answer:

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe       ..... (1)

T Cosθ = mg     .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}

tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}

tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}

tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}

As θ is very small, so tanθ and Sinθ is equal to θ.

\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

7 0
2 years ago
Zoning laws establish _______. a. what types of buildings can be built in an area b. the uses an area of land can be put to c. w
fiasKO [112]

Answer:

Zoning laws establish b. the uses an area of land can be put to.

5 0
3 years ago
Read 2 more answers
An object to charge 2.00 c is a pretty from a second object with the same charge by a distance of 1.50 m what is the electric fo
Orlov [11]
F = q₁q₂C / r²

F force
q charge
C Coulomb constant
r separation between charges
5 0
2 years ago
A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p
frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

a(t)=(5-10t)\ m/s^2

(a) Since, a=\dfrac{dv}{dt}

v = velocity

dv=a.dt

v=\int(a.dt)

v=\int(5-10t)(dt)

v=5t-\dfrac{10t^2}{2}=5t-5t^2

(b) v=\dfrac{dx}{dt}

x = position

x=\int v.dt

x=\int (5t-5t^2)dt

x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3

(c) Velocity function is given by :

v=5t-5t^2

5t-5t^2=0

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0
2 years ago
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