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ivolga24 [154]
2 years ago
8

Four ways to increase magnitude of current in dynamo​

Physics
1 answer:
8090 [49]2 years ago
5 0

Answer:

hmm

Explanation:

By increasing the number of turns in the coil, strength of magnetic field, speed of rotation of the coil in the magnetic field and by decreasing the distance between the coil and the magnet the magnitude of the induced e.m.f. can be increased in generator/dynamo.

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What does energy and Newton's Laws (all three) have to do with roller coasters?
Kipish [7]

Answer:

Potential Energy is the energy that is waiting to be released like at the top of the roller coaster, Kinetic energy is the energy that is moving the rollercoaster downhill fast.

Three Laws:

The rollercoaster will rest at the top, and then the gears will make it move downhill, and make it stop.

The gears make the rollercoaster move, and in reaction, it pushes back on the gears.

7 0
2 years ago
Read 2 more answers
Olivia is on a swing at the playground. if the swing is moving from W to Z, at which point is her kinetic energy increasing and
s2008m [1.1K]

Answer:

from W-Z.. i think on a swing you get your most potential energy at W and Z is where you go up so Z would be where the kinetic energy increased and W is where potential energy decrease

Explanation:

hope this helps

4 0
2 years ago
A soccer player is running upfield at 10 m/s and comes to a stop in 3 seconds facing the same direction. What is his acceleratio
I am Lyosha [343]
His acceleration would be 3.34 m/s 
5 0
3 years ago
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
A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb
marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

n_1 =\dfrac{46 \times 1}{0.0821\times 298}

  n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

n_2 =\dfrac{12 \times 1}{0.0821\times 298}

  n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

P_1=\dfrac{n_1 R T}{V}

P_1=\dfrac{1.89\times 0.0821\times 298}{5}

     P₁ = 9.25 atm

Partial pressure of oxygen

P_2=\dfrac{n_2 R T}{V}

P_2=\dfrac{0.49\times 0.0821\times 298}{5}

    P₂ = 2.39 atm

total pressure

    P = P₁ + P₂

    P = 9.25 + 2.39

    P = 11.64 atm

8 0
2 years ago
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