In terms of matter and resources, Earth is essentially a(n) ________ system ; in terms of energy, Earth is a(n) _______

If C is 1kg and D is 100kg, and the initial velocities of both balls are 5m/s, how would the magnitude of the forces exerted by

kupik [55]

Answer:

Explanation:

The forces exerted by each mass is best understood in terms of their momentum.

Momentum is a sort of compelling force or impulse. It is given as:

Momentum = mass x velocity

Let us consider the momentum of the balls;

Substance C;

Mass = 1kg

Velocity = 5m/s

Momentum of C = 1 x 5 = 5kgm/s

Substance D:

Mass = 100kg

Velocity = 5m/s

Momentum of D = 100kg x 5m/s = 500kgm/s

Body D has a higher momentum compared to Body C. This suggests that body D will exert a higher force than C when they collide.

The higher the momentum, the more the force of impact it has.

3 0

3 years ago

Action and reaction force always cancel each other. True or False.

polet [3.4K]

False. They have same magnitude and opposite direction but they never cancel as each of them does the action on the other body, and for the forces to cancel out they need to act ob the same body.

Hope this helps!

5 0

3 years ago

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

3 years ago

A factory line moves parts for an automobile at a constant speed of 0.22 m/s. How long does it take the parts to move 8.5 m alon

OleMash [197]

Answer: C. 39 s

Explanation:

We know the constant speed is 0.22 m/s. We have to get to 8.5 m. We divide 8.5 m by 0.22 = 38.6. After we estimate, 6 is greater than 5, so 39 s.

5 0

3 years ago

How to calculate F2? m=16.4kg f1= 2.7n angle=34.4

V125BC [204]

Option 2 is your answer :)

4 0

3 years ago

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In terms of matter and resources, Earth is essentially a(n) ________ system ; in terms of energy, Earth is a(n) ________ system.

In terms of matter and resources, Earth is essentially a(n) system ; in terms of energy, Earth is a(n) system.