3 years ago

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Engineering

2 answers:

34kurt3 years ago

6 0

Answer: yayyyy free sh.it

Explanation:

Scorpion4ik [409]3 years ago

3 0

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The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel = 4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

$\dot{m}=\rho AV$

Putting all the value to get the velocity of the flow

$\frac{\dot{m}}{\rho A} = V$

$V = \frac{4000}{1000*4*2}$

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3 years ago

There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter

weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

u_e = 0.5*e_o * E^2

- Plug in the values given:

u_e = 0.5*8.854 *10^-12 *145^2

u_e = 9.30777 * 10^-8 J/m^3

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Water is contained in a piston-cylinder assembly undergoes four processes separately, all started from an initial state where th

Andrews [41]

Answer:

see pictures

Explanation:

In the pictures you can see the different processes.

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3 years ago

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

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