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jeka57 [31]
2 years ago
11

Pls someone help with number 3

Physics
1 answer:
mars1129 [50]2 years ago
7 0
Rocket thrust equation
= ( mass flow rate of fuel burnt ) X (Velocity of gas ejected ) + ( Exit Pressure - Outdoor Pressure ) X ( Area of exhaust )

In this case, we can assume the exit pressure = outdoor pressure and since area of exhaust is not given, it can be assumed to be negligible.

In this case, by Newton 3rd’s law,

Force exerted by gas on rocket
= Force exerted by rocket on gas
= (10kg/s) X (5 x 10^3 m/s)
= 5 x 10^4 N
You might be interested in
Steam at 4 MPa and 400C enters a nozzle steadily with a velocity of 60 m/s, and it leaves at 2 MPa and 300C. The inlet area of
frez [133]

Answer:

(A) 4.09 kg/s

(B) 589.9 m/s

(C)   0.0008707 m^{3} =  8.71 cm^{2}

Explanation:

inlet pressure of steam (P1) = 4 MPa

inlet temperature of steam (T1) = 400 degree celcius

inlet velocity (V1) = 60 m/s

outlet pressure (P2) = 2 MPa

outlet temperature (T2) = 300 degree celcius

inlet area (A1) = 50 cm^{2} = 0.005 m^{2}

rate of heat loss (Q) = 75 kJ/s

(A) mass flow rate (m) = \frac{A1 x V1}{α1}

where the initial specific volume (α1) for the given temperature and pressure is gotten from tables A-6 = 0.07343 m^3/kg

m = \frac{0.005 x 60}{0.07343}

m = 4.09 kg/s

(B) we can get the outlet velocity using the energy balance equation

  E in = E out

   m(h1 + \frac{(V1)^{2}}{2}) =  m(h2 + \frac{(V2)^{2}}{2})

V2 = \sqrt{2(h1 - h2) +(V1)^{2} - 2\frac{Q}{m}

where h1 and h2 are the enthalpies and are gotten from table A-6

V2 = \sqrt{2 x 1000 x(3214.5 - 3024.2) +(60)^{2} - 2\frac{75 x 1000}{4.09}

V2 = 589.9 m/s  

(C) the outlet area is gotten from mass flow rate (m) = \frac{A2 x V2}{α}

  A2 = (α2 x m) / V2

where the initial specific volume (α2) for the given temperature and pressure is gotten from tables A-6 = 0.12552 m^3/kg

A2 = (0.12552 x 4.09) / 589.5 = 0.0008707 m^{3} =  8.71 cm^{2}

4 0
3 years ago
Please Help Me!
balandron [24]

The image distance when a boy holds a toy soldier in front of a concave mirror, with a focal length of 0.45 m. is -0.56 m.

<h3>What is image distance?</h3>

This is the distance between the image formed and the focus when an object is placed in front of a plane mirror.

To calculate the image distance, we use the formula below.

Formula:

  • 1/f = 1/u+1/v........... Equation 1

Where:

  • f = Focal length of the mirror
  • v = Image distance
  • u = object distance

From the question,

Given:

  • f = 0.45 m
  • u = 0.25 m

Substitute these values into equation 1 and solve for the image distance

  • 1/0.45 = 1/0.25 + 1/v
  • 2.22 = 4+1/v
  • 1/v = 2.22-4
  • 1/v = -1.78
  • v = 1/(-1.78)
  • v = -0.56 m

Hence, The image distance is -0.56 m.

Learn more about image distance here: brainly.com/question/17273444

5 0
2 years ago
A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t
vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be m kg.

Given:

Initial velocity of the car is, u=25\ m/s

As the car stops, final velocity of the car is, v=0\ m/s

Change in momentum is, \Delta p=39400

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg

Therefore, the mass of the car is 1576 kg.

4 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
2 years ago
Um comentarista de futebol certa vez comentou:"A bola bateu na trave e voltou duas vezes mais forte". Sabendo que quando a bola
ryzh [129]

Answer:

Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto

Explanation:

7 0
2 years ago
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