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natka813 [3]
3 years ago
7

A tracking station on Earth observes a rocket move away at 0.370c. This rocket is designed to launch a projectile at 0.505c rela

tive to the rocket. What speed (in terms of the speed of light) does the tracking station measure for the projectile if it is fired (a) straight ahead of the rocket or (b) backward from the rocket?
Physics
1 answer:
Orlov [11]3 years ago
7 0

Answer

given,

speed of rocket moving away = 0.370 c

speed of projectile = 0.505 c

a) Speed of the tracking station

  u = 0.505 c + 0.370 c = 0.875 c

 u'= \dfrac{u-v}{1-\dfrac{uv}{c^2}}

 u'= \dfrac{(0.875 c)-0.370 c}{1-0.875 \times 0.370}

        u' = 0.747 c

b)  when the rocket is moving backward

    u = -0.505 c + 0.37 c = - 0.135 c

 u'= \dfrac{-v}{1-\dfrac{uv}{c^2}}

 u'= \dfrac{(-0.135 c-0.370 c}{1-(-0.135) \times 0.370}

        u' = -0.481 c

negative sign represent direction

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Two different liquids are poured into a jar until it is half full. The jar is then sealed shut and shaken. The liquids undergo a
Lilit [14]

Answer:

A closed system.

Explanation:

The three major types of system are: open, closed and isolated. Open system interacts with its surroundings with respect to its particles and energy. A closed system interacts with its surroundings with respect to energy but not its particles. While an isolated system does not interact with its surroundings in any way.

Therefore, after the jar is sealed, it is an example of a closed system. This is because the emitted gas could not escape into the surroundings, but thermal energy was emitted into its surroundings after the chemical reaction has taken place.

7 0
3 years ago
Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th
Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0
3 years ago
A bullet with a mass of 0.3 kg is fired out of a gun with a mass of 4 kg at 600 m/s. What is the recoil velocity on the gun?
slavikrds [6]

Answer:

According to the Conservation of Momentum,

Momentum of the gun = momentum of the bullet

M(gun)×V(gun)=m(bullet)×v(bullet)

4kg × V = 0.3kg × 600m/s²

V = (0.3 × 600)/4 = 45 m/s

The recoil velocity on the gun is <em><u>45 m/s</u></em>

<h3><u>45 m/s</u> is the right answer.</h3>
4 0
2 years ago
An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.
Rudik [331]
The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ\frac{l}{A}
where l = length of the wire
A = area of the wire
A = \pi r^{2} where, r = \frac{diameter of wire}{2}

Thus, on finding the ratio of resistance of the two wires, we get,

\frac{R1}{R2} =  \frac{l1A2}{l2A1}

here, R1 = R
l1 = 8m
l2 = 2m
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7 0
3 years ago
) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th
Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

    m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w

Now, we will substitute 0 for both z_{1} and z_{2}, 0 for w, 334.9 kJ/kg for h_{1}, 2726.5 kJ/kg for h_{2}, 5 m/s for V_{1} and 220 m/s for V_{2}.

Putting the given values into the above formula as follows.

     m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w  

     1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0

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Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

6 0
3 years ago
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