What are 3 wether instuerments i know anemometer and barometer but the third one and what there job!!

The linear expansion of a material depends on which of the following?

nordsb [41]

Answer:

1. The change in temperature of the material.

2. The length of the material.

3. The type of material.

Explanation:

The term linear expansion is defined as the increase in length of any conductor when there is a change in temperature. The new length of any wire or rod is given by :

$L=L_o(1+\alpha \Delta T)$

$L_o$ is the initial length of the rod

$\alpha$ is the coefficient of linear expansion

$\Delta T$ is the change in temperature

It is clear that the linear expansion of a material depends on :

1. The change in temperature of the material.

2. The length of the material.

3. The type of material.

So, all options are correct.

7 0

3 years ago

A tight knot can be easily opens by using a longer spanner. Give reasons

kobusy [5.1K]

It is because the effort distance is greater than the load distance

Explanation:

As we know, Effort×effort distance = load × load distance

So when effort distance is increases,

The effort decreases

So when the spanner’s handle is long

A tight knot can easily be opened by less effrot

I hope it helped

6 0

2 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

3 years ago

Given the equation E = P/N , solve for P

Mars2501 [29]

Answer:

E=P/N

multiply both sides by N

P=EN

6 0

3 years ago

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

4 years ago