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leva [86]
4 years ago
12

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward

acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 97 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. What is the upward acceleration of the rocket during the burn phase?
8.0 m/s2

7.3 m/s2

7.5 m/s2

8.2 m/s2

7.8 m/s2
Physics
1 answer:
Ksenya-84 [330]4 years ago
6 0
I think th<span>e upward acceleration of the rocket during the burn phase is 7.</span>8 m/s2
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A region of space in which a test charge experiences a force is referred to as
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Answer: The correct answer is- Field.

Explanation -

The modified space around a charged particle is known as electric field.

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Thus, a region of space in which a test charge experiences a force is referred to as Field.

4 0
3 years ago
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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
Vika [28.1K]

Answer:

Explanation:

Let the required velocity of rocket be v .

We shall use the formula of time dilation to find the velocity of rocket .

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2} } }

t = 430

t' = 38

430=\frac{38}{\sqrt{1-\frac{v^2}{c^2} } }

\sqrt{1-\frac{v^2}{c^2} } }=\frac{38}{430}

1-\frac{v^2}{c^2} = .0078

\frac{v^2}{c^2} =.9922

\frac{v}{c} = .996

v  = .996 x 3 x 10⁸ m /s

= 2.988 x 10⁸ m /s

B )

Kinetic energy of rocket

= 1/2 m v²

= .5 x 20000 x (2.988 x 10⁸ )²

= 8.9 x 10²⁰ J .

C )

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7 0
3 years ago
A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from res
Bumek [7]

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

t_1 = \sqrt{\frac{2H}{g}}

now the sound will come back to the observer in the time

t_2 = \frac{H}{v}

so we will have

t_1 + t_2 = 2.42

\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42

so we have

\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42

solve above equation for H

H = 26.8 m

Part b)

If sound reflection part is ignored then in that case

H = \frac{1}{2}gt^2

H = \frac{1}{2}(9.81)(2.42^2)

H = 28.7 m

so here percentage error in height calculation is given as

percentage = \frac{28.7 - 26.8}{26.8} \times 100

percentage = 7.18

5 0
3 years ago
HELP!!!!!!!!!!!!
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Answer:

-4,200,000 J

Explanation:

4 0
3 years ago
A circuit has an AC voltage source in series with a 50 ohm resistor and a 113 mH inductor. The frequency is 100 cycles/sec, and
user100 [1]

Answer:

Explanation:

The rms voltage = 140/√2 = 140/1.414 = 99 V.

Reactance of inductor  = wL = 2 X 3.14 X 100 X 113 X 10⁻³ =70.96 ohm.

Total resistance in terms of vector = 50+70.96j

j is imaginary unit  number

Magnitude of this resistance = √ 50² + 70.96² = 86.80 ohm

current in resistance (rms) ( I ) = 99/86.80 = 1.14 A.

Power dissipated in resistor = I² R = 1.14 X 1.14 X 50 = 65 W( approx)

7 0
4 years ago
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