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mr Goodwill [35]

3 years ago

6

A rocket of mass 1200kg is travelling at 2000m/s .It fires its engine for 1 min .The forwarded thrust provided by the rocket eng

ines is 10KN (10000 N) .
a)Use the equation of momentum x t to calculate the increase in the momentum of the rocket .
b)use your answer a to calculate the increase in velocity of the rocket and its new velocity after firing the engines .

1 answer:

hodyreva [135]3 years ago

5 0

Answer:it fires its engine for 1 minute. If the forward thrust provided by the rocket engines is 10000 N, what is the increase in ... From this, work out the increase in velocity of the rocket and its new velocity ... You can put this solution on YOUR website! . From Physics (Mechanics), there is this impulse-momentum change equation: ...

Explanation:

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An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi

sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

now we know that

object distance from mirror is

$d_o = -50 cm$

Focal length of the mirror is given as

$f = 40 cm$

now we have

$\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}$

$\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}$

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6 0

3 years ago

Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i

ELEN [110]

Answer:

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Explanation:

5 0

3 years ago

A hockey puck slides 55.0 m along the length of the rink in just 1.25 s. The slight friction between the puck and the ice provid

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Answer:

44.8 m/s

Explanation:

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InS = 2(d/t) - Final Speed

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2 years ago

Which is the result of mitosis and cytokinesis?

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3 0

3 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago