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jolli1 [7]
2 years ago
10

A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e

lapsed time of 8 years for the trip. An identical clock on Earth shows that the total elapsed time for the trip was 10 years. What was the speed of the spaceship relative to the Earth?
Physics
1 answer:
m_a_m_a [10]2 years ago
4 0

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

10 years = \frac{2d}{v}

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }

0.64 = 1-\frac{v^{2} }{186282^{2} }

solving for v, gives = 35 288 miles/s

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A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
The two major problems with most motor vehicles are that they burn fossil fuels and _____________.
Citrus2011 [14]

Answer:

A. Create radioactive waste i believe

Explanation:

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2 years ago
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Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den
Ymorist [56]

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

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2 years ago
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A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a
Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

y = 0.049 \cos(7t) .......(1)

The general equation of a wave is given by :

y=A\cos(\omega t) .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0
2 years ago
Fields of Point Charges Two point charges are fixed in the x-y plane. At the origin is q1 = -6.00 nC . and at a point on the x-a
My name is Ann [436]

Answer:

Part A) Electric fields at the point due to q₁ and q₂:

E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Net electric field at P (Ep)

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ = -6.00 nC = -6 *10⁻⁹C

q₂ = +3.00 nC = +3*10⁻⁹C

d₁ = 4cm = 4 *10⁻²m

d_{2} =\sqrt{(4*10^{-2})^{2}+((3*10^{-2})^{2} }

d₂ = 5 *10⁻²m

Part A) Calculation of the electric fields at the point due to q₁ and q₂

Look at the attached graphic:

E₁: Electric Field at point  P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge

E₂: Electric Field at point  P(0,4) cm  due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² =  10.8*10³ N/C

E₁ = 33.75*10³ N/C (-j)

E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C

E₂y=E₂sinβ = 10.8*(4/5) =  8.64*10³ N/C

E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Calculation of the net electric field at P (Ep)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep=Epx (i) + Epy (j)

Epx= E₂x= 6.48*10³ N/C (-i)

Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Ep=   (6.48*10³ (-i)+25.11 10³ (-j) )N/C

3 0
2 years ago
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