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3 years ago

An object possesses both potential (PE) and kinetic (KE) energy. The object’s total mechanical energy is

Physics

2 answers:

tia_tia [17]3 years ago

7 0

Answer:

a- ME=KE+PE

Explanation:

edg2020

Nina [5.8K]3 years ago

5 0

Answer: PE+KE

Explanation:

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4 years ago

An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s2

Stolb23 [73]

Answer:

uk = 0.25

Explanation:

Given:-

- An object comes to stop with acceleration, a = -2.45 m/s^2

Find:-

What is the coefficient of kinetic friction between the object and the floor?

Solution:-

- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).

- We will apply equilibrium equation on the object in vertical direction.

N - m*g = 0

N = m*g

Where, N : Contact force exerted by the surface on the floor

g : Gravitational acceleration constant = 9.81 m/s^2

- Now apply Newton's second law of motion in the horizontal ( x-direction ):

- Ff = m*a

- The frictional force is related to contact force (N) by the following expression:

Ff = uk*N

- Substitute the 1st and 3rd expressions in the 2nd equation:

uk*m*g = -m*a

uk = a / g

- Plug in the values and solve for uk:

uk = - (-2.45) / 9.81

uk = 0.25

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3 years ago

A dog is running at an initial speed of 10 m/s. He covers 50 m in 4 seconds. What is the acceleration of the dog?

nikdorinn [45]

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3 years ago

Which property increases as an electromagnetic waves energy decreases

Rufina [12.5K]

Answer:

Wavelength

Explanation:

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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:

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.

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:

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

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4 years ago