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2 years ago

(b) Can the speed of a rocket exceed the exhaust speed of the fuel? Explain.

Physics

1 answer:

muminat2 years ago

4 0

<u>Yes. The speed of a rocket can exceed the exhaust speed of the fuel.</u>

How this is explained?

The thrust of the rocket does not depend on the relative speed of the gases or the relative speed of the rocket.
It depends on conservation of momentum.

What is conservation of momentum?

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.

Momentum is equal to the mass of an object multiplied by its velocity and is equivalent to the force required to bring the object to a stop in a unit length of time.
For any array of several objects, the total momentum is the sum of the individual momenta.
There is a peculiarity, however, in that momentum is a vector, involving both the direction and the magnitude of motion, so that the momenta of objects going in opposite directions can cancel to yield an overall sum of zero.

To know more about conservation of momentum, refer:

brainly.com/question/7538238

#SPJ4

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During which lunar phase is the moon s far side entirely dark?

bearhunter [10]

The lunar phase in which the moon's far side (the side that does not get any sunlight) is entirely dark is during a full moon. Since the side that faces the Earth is fully illuminated, the other side would be the exact opposite.

8 0

4 years ago

When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If

masya89 [10]

The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

$Q= ne$

Here Q is the charge, n is the number of electrons and e is the charge on the electron

$n = \frac{Q}{e}$

Replacing,

$n = \frac{4*10^{-6}C}{1.6*10^{-19}}$

$n = 2.5 * 10^{13}77$

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

$m= nm_e$

Here,

m = Mass of the charge

n = Number of electrons

$m_e$ = Mass of the electron

$\text{Percentage change} = \frac{nm_e}{M}*100$

Replacing we have

$\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100$

$\text{Percentage change} = 6.9*10^{-14} \%$

6 0

4 years ago

How much would a 25 kg suitcase weigh on the surface of…?

photoshop1234 [79]

Answer:

A. 95N

B. ?

C.225N

D.?

Explanation:

For Mars and Pluto im not so sure about those too but for A and C I am positive those are correct.

Sorry I could not help you all the way, please dont be mad ;(

8 0

3 years ago

What is the period of a pendulum that takes 3 seconds to move forward, and another 3 seconds to come back?

Serga [27]

The period of the pendulum is 6 seconds.

<h3>What is period?</h3>

Period is the time taken for a simple pendulum to move to(forward) and fro(backward). The s.i unit of period is seconds(s).

Note: When a simple pendulum moves to and fro. it completes one cycle.

From the question, The period of the pendulum is calculated using the formula below.

<h3>Formula</h3>

T = a+b.............. Equation 1

Where:

T = Period of the pendulum
a = Time taken for the pendulum to move forward
b = Time taken for the pendulum to move backward.

From the question,

Given:

a = 3 seconds
b = seconds

Substitute the values above into equation 1

T = 3+3
T = 6 seconds.

Hence, The period of the pendulum is 6 seconds.

Learn more about period here: brainly.com/question/21924087

5 0

3 years ago

A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children

trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

Explanation:

The centripetal acceleration is given by:

$a_{c} = \frac{v^{2}}{r}$

Where:

$v^{2}$ : is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

$a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}$

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0

3 years ago

Read 2 more answers