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3 years ago

13

Consider two spaceships, each traveling at 0.50c in a straight line. Ship A is moving directly away from the Sun and ship B is a

pproaching the Sun. The science officers on each ship measure the velocity of light coming from the Sun. What do they measure for this velocity?

1 answer:

attashe74 [19]3 years ago

3 0

Answer:

The velocity of the light will be 1.0c only

Explanation:

The velocity of the light measured in the case given in question will be 1.0c only.

This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum

The velocity of the light cannot be scaled down in no case

Thus, the velocity of the light remains as constant.

Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.

You might be interested in

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the

belka [17]

We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values,
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶) / (3 * 10⁻²)²

F = 100 N [ East of positive charge ]

Hope this helps!

5 0

3 years ago

Read 2 more answers

A baseball with a mass of 0.8 kg is given an acceleration of 20m/s.how much force was applied to the ball

Delicious77 [7]

F=mass x acceleration = ma= 0.8*20 = 16N

5 0

3 years ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

leonid [27]

Answer:

The magnitude of the force is $F_{net}= 1.837 *10^4N$

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

At t = 0 , $\theta = 20^o$

The rate at which the angle increases is $w = 2 \ ^o/s$

Converting this to revolution per second $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The length of the rope is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , The tension on the rope T = 18 kN

Mass of the para-sailor is $M_p = 75kg$

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

Now the derivative of displacement is velocity

So

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

represents the velocity, again the derivative of velocity gives us acceleration

So

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

Now considering the interval between $20^o \ to \ 30^o$ we have

$2 = \frac{30 -20 }{t -0}$

making t the subject

$t = \frac{10}{2}$

$= 5s$

Now at this time the displacement is

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity is

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

The linear acceleration is

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally radial acceleration is mathematically represented by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Generally angular acceleration is mathematically represented by

$\alpha_t = r \theta'' + 2 r' \theta '$

Now $\theta '' = \frac{d (0.0349)}{dt} = 0$

So

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The force on the para-sailor along y-axis is mathematically represented as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The force on the para-sailor along x-axis is mathematically represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The net resultant force is mathematically evaluated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

3 years ago

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on

Darya [45]

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²) (since R' = 0)

= √R²

= R

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these two forces on the barge is 1.621 kN

4 0

3 years ago

A closed system consisting of 10 lb of air undergoes a polytropic process from p1 = 75 lbf/in2, v1 = 4 ft3/lb to a final state w

Vitek1552 [10]

Answer:

The polytropic exponent (n) is 1.2

The amount of energy transfer by work is 552.45 Btu

Explanation:

P1V1^n = P2V2^n

P1 = 75 lbf/in^2 = 75 lbf/in^2 × 143.3 in^2/ft^2 = 10,747.5 lbf/ft^2

V1 = 4 ft^3/lb × 10 lb = 40 ft^3

P2 = 20 lbf/in^2 = 20 × 143.3 = 2866 lbf/ft^2

V2 = 12 ft^3/lb × 10 lb = 120 ft^3

(V2/V1)^n = P1/P2

(120/40)^n = 10747.5/2866

3^n = 3.75

ln 3^n = ln 3.75

n ln 3 = ln 3.75

n = ln 3.75/ln 3 = 1.2

W = (P2V2 - P1V1)/1 - n = (2866×120 - 10747.5×40)/1-1.2 = (343,920 - 429,900)/-0.2 = -85,980/-0.2 = 429,900 lbf.ft = 429,900/778.17 Btu = 552.45 Btu

4 0

3 years ago