All categories

1 year ago

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

Physics

1 answer:

muminat1 year ago

4 0

Yes. The speed of a rocket can exceed the exhaust speed of the fuel.

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

You might be interested in

Sara is riding her bike down the road at 35 km/hr. She starts to peddle faster as she

steposvetlana [31]

Answer:

Distance covered to top of the hill was : 1.755 km

Explanation:

Initial velocity = 35 km/hr

Acceleration = 2.0 km/hr²

Time taken to accelerate = 3 minutes = 3/60 hours = 1/20 hours

Formula for acceleration : a = Δv /t

v-u/t ---where u is initial velocity , v is final velocity and t is time taken for acceleration

v- 35 / 0.05 = 2

v = 35.10 km/h

Formula for distance is product of speed and time

Distance covered = 35.10 * 0.05 = 1.755 km

6 0

3 years ago

Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th

Vladimir79 [104]

Answer:

$C = 1.77 \times 10^{-4} F$

Explanation:

As we know that in AC circuit we have

$V = i x_c$

here we have

V = 59 V

i = 5.05 A

so we will have

$x_c = \frac{59}{5.05}$

$x_c = 11.68 ohm$

also we know that

$x_c = \frac{1}{\omega C}$

here we will have

$11.68 = \frac{1}{(2\pi 77)C}$

$C = 1.77 \times 10^{-4} F$

7 0

2 years ago

Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

3 years ago

Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien

otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s

The height at which the block stops rising is approximately 1.1415 m

The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

$Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}$

Coefficient of kinetic friction, $\mu_k$ = 0.39

Therefore, we have;

Friction force = $\mathbf{\mu_k}$ ·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, W ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

$\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d$

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, W ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The velocity of the block after the Spring extends 7 cm, v₂ ≈ 1.73 m/s

The height at which the block will stop moving, h, is given as follows;

$At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x$

Which gives;

$Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536$

The distance up the inclined, the block rises, at maximum height is therefore;

$x_{max}$ ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

The height at which the block stops rising, h ≈ 1.1415 m

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

Length of the incline, $x_{max}$ = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0

2 years ago

2. The ability to keep upright posture while standing or moving

Masteriza [31]

Is balancing .
Hope this is helpful

4 0

3 years ago