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Anon25 [30]
4 years ago
6

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is

6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.
Physics
1 answer:
mezya [45]4 years ago
4 0

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= \frac{1}{2}at^2

h= \frac{1}{2}4.71\times4.96^2

h= 58.012 m

h≅ 58 m

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Mamont248 [21]

Answer:

Option (b)

Explanation:

let the weight of boat is W.  In equilibrium condition, the weight of boat is equal to the buoyant force acting on the boat.

The buoyant force acting on the boat is equal to the weight of water displaced by the boat.

In case of fresh water:

Weight of the boat = weight of fresh water displaced by the boat

W = Volume of fresh water displaced x density of fresh water x g

W = V x 1 x g

W = V x g ....... (1)

In case of salt water:

Let the volume of salt water displaced is V'.

Weight of the boat = weight of salt water displaced by the boat

W = Volume of salt water displaced x density of salt water x g

W = V' x 1.02 x g    ..... (2)

Equate equation (1) and equation (2), we get

V x g = V' x 1.02 x g

V' = 0.98 V

Thus, option (b) is true.

7 0
4 years ago
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4 0
3 years ago
Which example shows that electromagnetic waves can travel without moving through a specific material (medium)? . . . A.. visible
MrMuchimi
<span>example shows that electromagnetic waves can travel without moving through a specific material  is
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4 0
3 years ago
Which is heavier, the earth or the moon
Aloiza [94]

When you say "heavy", you're talking about the gravitational force
between that object and another object, so it depends on what the
"other object" is.

If the "other object" is, let's say, the sun, then the gravitational attraction
between the Earth and sun is about 80 times as much as the gravitational
attraction between the Moon and sun, because the Earth has about 80 times
the mass of the Moon.

But if, somehow, the weight you have in mind is the gravitational attraction
between the Earth and the Moon, then those forces are equal.  The force
of gravity between two objects depends on the product of both masses,
and it's equal in both directions.

If that isn't clear to you, let me give you this additional fact that's guaranteed
to knock you even further off-balance:

Your weight on the Earth is determined by the product of

                           (your mass) times (the Earth's mass).

The Earth's weight on you is determined by the product of

                           (your mass) times (the Earth's mass).

Your weight on Earth is the same as the Earth's weight on you.

Would you like to prove it ?

-- Turn the bathroom scale upside-down, so that the step-pad
    is on the floor.

-- Then step on it, so that you're standing on the bottom, which
    is facing up.

-- If you placed a little mirror on the floor, so that you can read
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5 0
3 years ago
The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

5 0
3 years ago
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