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vodomira [7]
4 years ago
6

A rocket has initail mass M begins to move from space , with an exhaust constant speed . Find the mass of the rocket while has t

he maxmum momentum.
Physics
1 answer:
irina1246 [14]4 years ago
3 0

Answer:

m=\frac{m_{0}}{e}

Explanation:

Equation of the rocket is,

m\frac{dv}{dt} =F-v'\frac{dm}{dt}

Here, v' is the relative velocity of rocket.

In space F is zero.

So,

m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }

Now the momentum can be obtained by multiply by m on both sides.

P=-v'mln\frac{m}{m_{0} }

Now for maxima, \frac{dP}{dm}=0

-v'ln\frac{m}{m_{0} }-v'm\frac{m_{0}}{m }m_{0=0

Now,

ln(\frac{m}{m_{0} } )=-1\\\frac{m}{m_{0} }=\frac{1}{e} \\m=\frac{m_{0}}{e}

Therefore, the mass of the rocket while having maximum momentum is \frac{m_{0}}{e}

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A pendulum is observed to complete 23 full cycles in 58 seconds. Determine the period and
Fynjy0 [20]

Answer:

Frequency=0.39Hz

Explanation:

A pendulum completes 23 full cycles in 58 seconds

A pendulum completes 1 full cycle in 58/23=2.52 seconds

Time taken(t)=2.52 seconds

Now,

Frequency=1/time taken

Frequency=1/2.25

Frequency=0.39Hz

6 0
2 years ago
The amount of work done by two boys who apply 200 N of force in an
Aleksandr [31]

The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

<u>Explanation: </u>

Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.  

           \text { Work done }=\text { Force } \times \text { displacement }

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

           \text { Work done }=200 \mathrm{N} \times 0=0

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.

8 0
3 years ago
A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretc
Anvisha [2.4K]

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz

5 0
4 years ago
Due to design changes, the efficiency of an engine increases from 0.28 to 0.51. For the same input heat |QH|, these changes incr
Volgvan

Answer:

the ratio of the heat rejected to the cold reservoir for the improved engine to that for the original engine is 0.68

Explanation:

Given information

initial efficiency, η_{1} = 0.28

final efficiency, η_{2} = 0.51

ratio of the heat rejected = (1 - η_{2})/(1 - η_{1})

                                        = (1 - 0.51)/(1 - 0.28)

                                        = 0.68

4 0
4 years ago
An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat
KATRIN_1 [288]

Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

S= distance travelled

a = acceleration due gravity

t= time.

1 foot = 0.305m so,

S= 2860 feet =872.3m

S= ut+1/2 at²

872.3 = 0×t + 1/2×10 × t²

872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

t = 13.2 seconds to one decimal place.

8 0
3 years ago
Read 2 more answers
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