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jolli1 [7]
1 year 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]1 year 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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|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r
Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

F = ma

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

a = \frac{F}{m}

a = \frac{11*10^3}{80}

a = 137.5m/s

From the cinematic equations of motion we know that

v_f^2-v_i^2 = 2ax

Where,

v_f =Final velocity

v_i =Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

v_f^2-v_i^2 = 2ax

0-v_i^2 = 2(-g)x

v_i =\sqrt{2gx}

v_i = \sqrt{2*9.8*4.8}

v_i = 9.69m/s

From the equation of motion where acceleration is equal to the velocity in function of time we have

a = \frac{v_i}{t}

t = \frac{v_i}{a}

t =\frac{9.69}{137.5}

t = 0.0705s

Therefore the time required is 0.0705s

4 0
2 years ago
Read 2 more answers
The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b
Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0
2 years ago
Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

7 0
1 year ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
3 0
2 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
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