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Effectus [21]
2 years ago
8

Please help! you are amazing! brainliest too!

Physics
2 answers:
german2 years ago
6 0
I think it’s B hope it helps:)
Dmitry [639]2 years ago
5 0

Answer:

<em> I think its B and I don't know why my words are like this</em>

Explanation:

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(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg
valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be m_ 1 = 80 kg

The weight of the buddy be m_2 = 120 kg

The weight of  Bubba be m_3 = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2

The length of the boat be x_2 = 4 m

∴

We can find the center of mass of the system by using the formula:

X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}

4 0
3 years ago
Please Help! 30 points! ASAP PLZ :)
Savatey [412]
The best logical answer is A
5 0
2 years ago
Read 2 more answers
A car travels west at 40 km/h
Nesterboy [21]

Answer:

The car is going 0 km/h more than the bike

Explanation:

3 0
3 years ago
A heating element has a resistance of 180 Ω and draws a current of 1.60 A. What is its power?
iragen [17]

its either 288Watts or 112.5 Watts

5 0
3 years ago
Read 2 more answers
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
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