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3 years ago

14

A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

the car

2 answers:

9966 [12]3 years ago

4 0

F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N

olga_2 [115]3 years ago

3 0

Answer:

The magnitude is $2000N$

Explanation:

To solve this question we are going to use the second Newton law.

The second Newton law states that the acceleration of an object is directly proportional to the net force which acts on the object and inverse proportional to the mass of the object ⇒

$a=\frac{F}{m}$

Which can also be written as :

$F=m.a$ (I)

Where F is the force

Where m is the mass of the object and where ''a'' is the acceleration.

In the problem, we know that

$m=1000kg$

$a=2\frac{m}{s^{2}}$

If we replace in the equation (I) :

$F=(1000kg).(2\frac{m}{s^{2}})=2000(kg.\frac{m}{s^{2}})$

The unit $kg.\frac{m}{s^{2}}$ is defined as N which is called Newton. N is an unit of force.

Finally, the net force exerted on the car is $2000N$

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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

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Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

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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:

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

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$m=p*vel*A$ 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:

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

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