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3 years ago

When playing tug of war and neither side moves, forces are...

Physics

2 answers:

Fantom [35]3 years ago

8 0

Answer:

C.Balanced... The force is balanced so neither one is being pulled harder...

Explanation:

Hope this helps!

natulia [17]3 years ago

8 0

Answer:

Balanced.

Explanation:

Both sides have equal strength. therefore the rope will not move. The Forces are Balanced! :)

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A car has an initial velocity of +30.0 m/s and undergoes an acceleration of -5.00 m/s squared for 5.00 seconds what is the displ

Leni [432]

Ok so the formula is d=vi(t)+½at² and when you substitute it you should get 172.5meters

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3 years ago

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per

Brilliant_brown [7]

Answer:

$a_{cp}=7.77m/s^2$

Explanation:

The equation for centripetal acceleration is $a_{cp}=\frac{v^2}{r}$ .

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference $C=2\pi r$ and the period T, then we have:

$v=\frac{C}{T}=2\pi r f$

Putting all together:

$a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2$

4 0

3 years ago

¿Qué es la velocidad?

egoroff_w [7]

Velocidad-es una cantidad vectorial física; Tanto la magnitud como la dirección son necesarias para definirlo.

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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

3 years ago

Can anyone pls help me with this I’ll appreciate it. I’m trying to study for a test but I don’t know the answer to this.

forsale [732]

Answer:

okay so what you will do is where is says red giant you will write all about what it talks about red giants only, and the box plantary nebulas you will write about what is says about only planetary nebulas. x- hope this helps :)

Explanation:

7 0

3 years ago