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Nesterboy [21]
2 years ago
13

A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what

is the primary voltage? (in this case, the voltages and currents are inversely proportional.)
Physics
1 answer:
Lynna [10]2 years ago
7 0

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

V_1 i_1 = V_2 i_2

here we know that

V_2 = 140 volts

i_2 = 3.5 A

i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1

V_1 = 49 volts

So primary coil voltage is 49 Volts

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13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?
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The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

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The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of  mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

Q = mc del T

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

Q = 20*C*80

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

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In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is
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The expression for the initial speed of the fired projectile is:

\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})

And the initial speed ratio for the 9.0mm/44-caliber bullet is 1.773.

Explanation:

For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.

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In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:

P_0=P_f\\v_0m=v_i(m+M)\\v_0=v_i\frac{m+M}{m}  (1)

Second Phase: pendular movement

After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:

Em_i=Em_f\\\frac{1}{2}mv^2_i=mgH\\v_i=[2gH]^\frac{1}{2}  (2)

The height of the pendulum respect L and θ is:

H=L(1-cos(\theta)) (3)

Using equations (1),(2) and (3):

\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}}) (4)

The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):

\displaystyle \frac{v_{9mm}}{v_{44}} =\frac{(M+m_{9mm})m_{44}}{(M+m_{44})m_{9}}(\frac{1-cos(\theta_{9mm})}{1-cos(\theta_{44})} )^{\frac{1}{2}}=1.773

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