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mezya [45]
2 years ago
5

a 1300-kilogram space vehicle travels at 4.8 meters per second along the level surface of mars. if the magnitude of the gravitat

ional field strength on the surface of mars is 3.9 newtons per kilogram, the magnitude of the normal force acting on the vehicle is (1) 320 n (2) 4400 n (3) 930 n (4) 5070 n
Physics
1 answer:
Helga [31]2 years ago
5 0

The magnitude of the normal force acting on the space vehicle will be 5070 newtons. So option (4) is correct.

Given that,

The mass of the space vehicle is 1300 kilograms

It travels at a speed of 4.8 meters per second along the level surface of mars.

The g' (gravitational field strength) on the surface of mars is 3.9 newtons per kilogram

We need to find the normal force acting on the space vehicle

We know, N = mg' (where N is normal force)

N = (1300 × 3.9) Newton

N = 5070 Newton

So, the magnitude of the normal force acting on the space vehicle that is traveling at 4.8 meters per second along the level surface of mars will be 5070 newtons.

The normal force is the force that surfaces exert to prevent solid objects from passing through each other. The normal force is one type of ground reaction force.

Learn more about the normal force here:

brainly.com/question/6922206

#SPJ4

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
4 years ago
What is the velocity of an object with a mass of 4 kg and a momentum of 24 kg.m/s
Eva8 [605]

Answer:

6m/s

Explanation:

momentum = mass × change in velocity

∆p =m(v)

24 = 4(v)

V =>24/4 = 6m/s

5 0
3 years ago
The self inductance relates the magnetic flux linkage to the current through the coil. Calculate the self inductance L in units
Katen [24]

Answer:

Self inductance, L=127\ \mu H

Explanation:

It is given that,

Length of the coil, l = 5 cm = 0.05 m

Area of cross section of the coil, A=3\ cm^2=0.0003\ m^2

Number of turns in the coil, N = 130

The self inductance relates the magnetic flux linkage to the current through the coil and it is given by :

L=\dfrac{\mu_oN^2A}{l}

L=\dfrac{4\pi \times 10^{-7}\times (130)^2\times 0.0003}{0.05}

L = 0.000127 Henry

or

L=127\ \mu H

So, the self inductance of the coil is 127 microhenry. Hence, this is the required solution.

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How do you convert 1.3*10^6cal into joules
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Answer:

5.4×10⁶J

Explanation:

1 cal = 4.184 J

1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J

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There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in
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Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

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