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koban [17]
3 years ago
5

What produces a magnetic field? A. electrical resistance B. increase in Ohms C. transfer of protons D. movement of electrons

Physics
1 answer:
MrRa [10]3 years ago
6 0
The movement of electrons is what produces a magntic field
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A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of
igomit [66]

Answer:t=0.81 s

Explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore F_{net} is zero on Penguin

F_{net}=mg\sin \theta -f_r

f_r=friction Force

f_r=\mu mg

\mu =coefficient of Kinetic friction

mg\sin \theta =\mu mg

\tan \theta =\mu

\mu =\tan 9.85=0.45

after reaching on floor final velocity of penguin will be zero after time t

thus

v=u+at

here a=\mu g

a=0.45\times 9.8=4.41  (deceleration)

0=3.57-4.41\times t

t=\frac{3.57}{4.41}=0.809

t=0.81 s

8 0
3 years ago
Why is it important to consider the experimental error in all the empirical results presented?
irinina [24]

Answer:

It is very important because scientists, especially the ones with empirical experiments and results, are prone to error and the empirical data is in need to be under strict observation done not only by many scientists but also by expermiented ones. This guards everybody to change the parameters suddenly which can affect the real results of an experiment 

Explanation:

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D. What is the net force on the bowling ball rolling lane
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Answer:

Friction

Explanation:

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2 years ago
According to the Hooke’s law formula, the force is proportional to what measurement?
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According to Hooke's Law formula. The force is proportional to the displacement of the spring. I believe
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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
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