The velocity of a wave with a wavelength of 4.7000 m and frequency of 34.00 hz

At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either

GrogVix [38]

Answer:

The direction of the force at A and B is perpendicular to the walls of the container.

The direction of the force at C is down.

The direction of the force in D is up

The direction of the force at E is to the left.

The attached figure shows the forces exerted by the water at points A, B, C, D and E.

Explanation:

The water is in contact with the bowl and with the fish. It exercises at points A, B, C, D and E, but the direction is different from the force.

The fish has a buoyant force on the water and that direction is up. The direction of at point D is up.

The column of water on the fish has a downward force, therefore the direction of the force at point C is down. The water column to the right of the fish has a force to the left, and the direction at point E is to the left.

The water will exert a force on the walls of the container and this force at points A and B is a on the walls of the container.

4 0

3 years ago

In which of the following situation is work being done?

Afina-wow [57]

A wagon is pulled at an angle of 30 degrees to the horizontal.

6 0

3 years ago

Why is it expensive to bring electricity into urban areas?

motikmotik

Because of the pole and the generator you would have to biuld

8 0

3 years ago

Read 2 more answers

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago

A proton is moving toward a second, stationary proton. What happens as the protons get closer?

MatroZZZ [7]

Answer:

A. Kinetic energy is converted to electric potential energy, and the proton moves more slowly.

Explanation:

When a moving proton is brought close to a stationary one, the kinetic energy of the moving one is converted to electric potential and the proton moves more slowly.

Kinetic energy is the energy due to the motion of a body. A moving proton will possess this form of energy.

Two protons according to coulombs law will repel each other with an electrostatic force because they both have similar charges. This will increase their electric potential energy of both of them.

Potential energy is the energy at rest of a body. As it increases, the motion of a body will be slower and it will tend towards being stationary.

5 0

3 years ago