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WITCHER [35]
3 years ago
5

A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times

the mass of the empty rocket. Part A What is the rocket's speed when half the fuel has been burned?
Physics
1 answer:
notka56 [123]3 years ago
3 0

Answer:

 v_{f} = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

       v_{f} - v₀ =  v_{r} ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is    

       M = 2.5 Mf

       v_{f} - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

       v_{f} = 1,386 m / s

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a fixed amount of ideal gas is held in a rigid container that expands negligibly when heated. at 20 the gas pressure is p. if we
pantera1 [17]

Answer:

When the temperature of the gas is increased from 20 to 40, the pressure will be 2p

Explanation:

Given;

initial temperature of the gas, T₁ = 20 K

final temperature of the gas, T₂ = 40 k

initial pressure of the gas, P₁ = P

final pressure of the gas, P₂ = ?

Apply pressure law of gases;

\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} \\\\P_2 = \frac{40P}{20} \\\\P_2 = 2P

Therefore, when the temperature of the gas is increased from 20 to 40, the pressure will be 2p

5 0
3 years ago
a. A beam of light is incident from air on the surface of a liquid. If the angle of incidence is 26.7° and the angle of refracti
Diano4ka-milaya [45]

Answer:

(a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Explanation:

Given that,

Incidence angle = 26.7°

Refraction angle = 18.3°

(a). We need to calculate the refraction of liquid

Using Snell's law

n=\dfrac{\sin i}{\sin r}

Put the value into the formula

n=\dfrac{\sin 26.7}{\sin 18.3}

n=1.43

We need to critical angle for the liquid when surrounded by air

Using formula of critical angle

C=\sin^{-1}(\dfrac{1}{n})

Put the value into the formula

C=\sin^{-1}(\dfrac{1}{1.43})

C=44.37^{\circ}

(b). Given that,

Incidence angle = 37.5°

Speed of light in mineral v=2.17\times10^{8}\ m/s

We need to calculate the index of refraction

Using formula of index of refraction

n=\dfrac{c}{v}

Put the value into the formula

n=\dfrac{3\times10^{8}}{2.17\times10^{8}}

n=1.38

We need to calculate the angle of refraction

Using Snell's law

n=\dfrac{\sin i}{\sin r}

\sin r=\dfrac{\sin i}{n}

Put the value into the formula

\sin r=\dfrac{\sin 37.5}{1.38}

r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})

r=26.17^{\circ}

Hence, (a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

3 0
3 years ago
Read 2 more answers
What is the speed of a 200-kilogram car that is driving with 2000 joules of kinetic energy? (SHOW ALL WORK)
Katyanochek1 [597]

Answer:

v ≈ 4.47

Explanation:

The Formula needed = <u>KE = </u>\frac{1}{2}<u> m v²</u>

<u></u>

Substitute with numbers known:

2000J = \frac{1}{2} × 200kg × v²

Simplify:

÷100       ÷100      (Divide by 100 on both sides)

2000J = 100 × v²

\frac{2000J}{100} =  v²

20 = v²

√         √             (Square root on both sides)

√20 = √v²

4.472135955 = v (Round to whatever the question asks)

v ≈ 4.47       (I rounded to 2 decimal places or 3 significant figures, as that is what it usually is)

3 0
3 years ago
A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave
bonufazy [111]

Answer:

(a) P=33000W

(b) P=51000W

Explanation:

The average power is defined as the amount of work done during a time interval:

P=\frac{W}{t}(1)

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)

Recall that the weight is given by:

w=mg\\m=\frac{w}{g}(3)

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W=\frac{wv_f^2}{2g}

(a) Finally, we replace this in (1):

P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W

(b)

P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W

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4 years ago
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Bezzdna [24]
Trips/slips/falls are among the most common types of work related injuries and/or deaths.
7 0
4 years ago
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