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goldfiish [28.3K]
3 years ago
15

This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the

gravitational pull from the stars is negligible, because the stars are extremely far. The spacecraft turns its main rocket engine on for a time period of 6.5 hours. The rocket engine provides a thrust of 896 N of force. What is the new speed of the craft after the engine is turned off? The mass of the spacecraft is 1380 kg. The rocket engine is very efficient, it uses only a very small amount of rocket fuel during this whole acceleration process without changing the mass of the craft.
Physics
1 answer:
valentinak56 [21]3 years ago
5 0

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2

Equation of motion

v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

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Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
RideAnS [48]

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

A_{1}\times V_{1} = A_{2}\times V_{2}

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

A_{1} = 0.070 m^{2}

V_{1} = 3.50 m/s

a) A_{2} = 0.105 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.105\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.150}m/s

or

V_{2} = 2.33m/s

b) A_{2} = 0.047 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.047\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.047}m/s

or

V_{2} = 5.21m/s

c) we have,

DischargeQ = Area (A)\times Velocity(V)

thus from the given value, we get

Q = 0.070m^{2}\times 3.5m/s\

Q = 0.245 m^{3}/s

Also,

DischargeQ = \frac{volume}{time}

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

0.245m^{3}/s = \frac{volume}{3600s}

or

0.245m^{3}/s\times 3600 = Volume

volume of flow = 882 m^{3}

8 0
3 years ago
Mountains are _____.
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<h3><u>Answer;</u></h3>

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5 0
3 years ago
A 1.005 m chain consists of small spherical beads, each with a mass of 1.00 g and a diameter of 5.00 mm, threaded on an elastic
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Answer:

1) μ = 1.33 10⁻³ kg / m , F = - 14,256 ,  2) v= 103.53 m/s, 3)  f = 138.04 Hz , 4)  1, 25, 50, 76, 101   , 5) A = 0.00869 m , 6)  # _position = (# _account-1) (1.5m / 100 accounts)

Explanation:

1) Linear density is the mass per unit length

     μ = m / L

     μ = 2 1 10⁻³ / 1,5

     μ = 1.33 10⁻³ kg / m

this is the density when the chain is stretched, which is when the pulse occurs

we can find the tension with

     F = - k (x₁-x₀)

where k is the spring constant

     F = - 28.8 (1.5 -1.005)

     F = - 14.256 N

the negative sign indicates that the force is restorative

2) the pulse speed is

      v = √ T /μ

      v = √ 14,256 / 1,33 10⁻³

      v = 103.53 m / s

3) If standing waves are formed with fixed points at the ends and 4 antinodes, the wavelength is

          2 λ = L

            λ = L / 2

wave speed is related to frequency and wavelength

           v = λ f

            f = v / λ

            f = v 2 / L

            f = 103.53 2 / 1.5

            f = 138.04 Hz

4) The marbles are numbered, the marbles that remain motionless are

   the first (1) and the last (101)

Let's look for the distance to each node, for this we must observe that in each wavelength there is a node at the beginning, one in the center and one at the end, therefore the nodes are in

         #_node = m λ / 2 = m L / 4

        #_node     position (m)

         1                  1.5 / 4 = 0.375

         2             2 1.5 / 4 = 0.75

         3             3 1.5 / 4 = 1,125

         

Since there are 101 marbles in the initial length, this number does not change with increasing length, so there is 101 marble in 1.5 m. Let's find with a direct proportion rule the number of marbles at these points with nodes

        #_canica = 0.375 m (101 marble / 1.5 m) 0.375 67.33

        # _canica = 25

        #_canica = 0.75 67.33

        #_canica = 50

        # _canica = 1,125 67.33

        #canica = 75.7 = 76

in short the number of the fixed marbles is

      1, 25, 50, 76, 101 canic

5) The movement of the account is oscillatory at this point, which is why it is described by

          y = A cos wt

          v_{y}= -A w sin wt

the speed is maximum for when the breast is worth ±1

          v_{y} = Aw

           A = v_{y} / w

angular velocity related to frequency

         w = 2π f

          A = v_{y} / 2πf

          A = 7.54 / (2π 138.04)

          A = 0.00869 m

6) for the position of each account we can use a direct proportion rule

      in total there are 100 accounts distributed in the 1.50 m distance, the #_account is in the # _position. Note that it starts to be numbered 1, so this number must be subtracted from the index of the amount

       # _position = (# _account-1) (1.5m / 100 accounts)

#_canic position(m)

   1          0

   2         0.015

   3         0.045

   4         0.06

7) the wave has a constant velocity, but every wave is oscillated perpendicular to this velocity, with an oscillatory movement described by the expression

         y = Acos wt

the maximum speed is

         v_{y} = -Aw sin wt

speed is maximum when the sine is ±1

         v_{y} = A w

to calculate the amplitude of the count we use that for a standing wave

         y = 2Asin kx

          y / A = 2 sin (2π /λ x)

the wavelength is

 λ = 0.75 m

the position is

x (30) = 29 1.5 / 100 = 0.435  m

          y (30) A = 2 sin (2pi 0.435 / 0.75)

          y (30) / A = 0.96 m

8 0
4 years ago
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