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11111nata11111 [884]
3 years ago
11

A student is investigating the best method for heating marshmallows using an open flame.

Physics
1 answer:
Llana [10]3 years ago
7 0

Answer:

The answer is Conduction.

Explanation:

In the first scenario given in the question where the student is holding the marshmallows on the side of the flame, she is subjecting the marshmallow the a heat transfer by radiation since the marshmallow is not in direct contact with the fire.

Heat transfer by radiation is done so with the heating of the air between the object and the heat source and does not need any contact between the source and the heated material.

And later when she places the marshmallow over the flames, the heat transfer type she is observing is heat transfer by conduction.

I hope this answer helps.

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An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

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3 years ago
One of the light bulbs in this series circuit burns out, causing a break in the circuit.
ivanzaharov [21]

Answer:

It doesn't give light

Explanation:

No Flowing of electricity

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Answer:

a. Near both the equator and the prime meridian.

Explanation:

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Answer:

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