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3 years ago

A closed path that allows electrical energy to flow is called a(n)

Physics

2 answers:

Bumek [7]3 years ago

4 0

Answer:

Circuit

Explanation:

Tresset [83]3 years ago

4 0

Circuit jwodnsdnsokdoxx

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6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440

MrRissso [65]

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

7 0

4 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.

dem82 [27]

Angle (θ) = 60°
Force (F) = 20 N
Distance (s) = 200 m
Therefore, work done
= Fs Cos θ
= (20 × 200 × Cos 60°) J
= (20 × 200 × 1/2) J
= (20 × 100) J
= 2000 J

Answer:

2000 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

Read 2 more answers

Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by

HACTEHA [7]

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408*10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M = M_1 = M_2$ is the masses of the 2 objects. and R = 2.6m is the distance between them.

$F_G = 5.3*10^{-4}N$ is the attraction force.

$5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}$

$7941265 = \frac{M^2}{2.6^2}$

$M^2 = 53682948.76$

$M = \sqrt{53682948.76} \approx 7327 kg$ or 7.3 tons

3 0

4 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

Which of the following amounts to a scientific research process that can be replicated?

STALIN [3.7K]

Answer:

D.all of the above

Explanation:

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3 0

2 years ago