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Lapatulllka [165]
3 years ago
12

If astronauts could travel at v disagree. 0.945c, we on Earth would say it takes 4.200 945 4 44 years to reach pha centaur 42Ф

Çgint ears away. The astronauts (a) How much time passes on the astronauts' clocks? Your response is within 10% of the correct value. This may be due to roundoff error or you could have a mistake in your calculation. results to at least four-digit accuracy to minimize roundoff error. years arn out all men mediate (b) What is the distance to Alpha Centauri as measured by the astronauts?
Physics
1 answer:
kolezko [41]3 years ago
6 0

Answer:

1.3734 years pass on the astronauts' clock.

The distance measured by astronauts is 1.22*10^{16} \ m

Explanation:

The fomula for time dilation is:

t'=t\sqrt{1-\frac{v^2}{c^2} }

where t' is the time observed by the astronauts, t is the time observed form Earth, v is the velocity of the astronaut and c is the velocity of light in a vacuum.

Replacing our values: t=4.2 year, v=0.945c

t'=4.2\sqrt{1-\frac{{(0.945c)}^2}{c^2} } =\\\\=4.2\sqrt{1-0.8930 }=4.2*0.3270=1.3734\ years

The distance measured by astronauts will be their speed multolied by the time is takes them to get to alpha centauri:

D=v*t'=0.945*c*1.3734=0.945*3.00*10^8 m/s*1.3734*3.154*10^7

To acomodate fot units we write the speed of light in m/s and the amount of seconds in a year, so the resulting distance will be in meters.D=1.22*10^{16} \ m

We should compare this to the distance measured from Earth which is D=c*t=3.00*10^8 m/s*4.2*3.154*10^7=3.974*10^{16}\ m

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Mathphys :( im sorry i annoy you
Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

3 0
3 years ago
A 72-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.61 m,
chubhunter [2.5K]

Answer:

2849.98 J

Explanation:

From the question,

Work done by the boy = change Potential energy of the boy + change in kinetic energy of the boy

W = ΔP + ΔK..................... Equation 1

Where W = work done by the boy, ΔP = change in potential energy of the boy, ΔK = Change in kinetic energy of the boy.

But,

ΔP = mgΔh.................... Equation 2

ΔK = 1/2mΔv²................. Equation 3

Where m = mass of the boy, Δh = change in height of the boy, Δv = change in velocity of the boy.

Substitute equation 2 and 3 into equation 1

W = mgΔh+1/2mΔv²................. Equation 4

Given: m = 72 kg, Δh = 1.61 m, Δv = 8.5-1.6 = 6.9 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 72(9.8)(1.61)+1/2(72)(6.9²)

W = 1136.016+1713.96

W = 2849.98 J

8 0
3 years ago
The layer of the Earth that is composed of large plates that interlock and move over time is the _______. asthenosphere lithosph
siniylev [52]
The lithosphere because it includes the outer region of the earth including the crust and outer mantle
4 0
2 years ago
Read 2 more answers
The accompanying map shows the route traveled by a school bus.
Shtirlitz [24]

Answer:

C

Explanation:

6 0
3 years ago
A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

                 v =  \sqrt{ [(k)(x^2)]/m}

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

8 0
2 years ago
Read 2 more answers
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