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1 month ago

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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.)

1 answer:

kicyunya [3.2K]1 month ago

7 0

In the case of an ideal transformer, power loss is regarded as nonexistent

meaning that the power in the primary coil resulting from the input voltage must equal the power in the secondary coil resulting from the output voltage

This relationship can be expressed as an equation

$V_1 i_1 = V_2 i_2$

Here, we know that

$V_2 = 140 volts$

$i_2 = 3.5 A$

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

$V_1 = 49 volts$

Thus, the voltage across the primary coil is 49 Volts

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Add a third force that will cause the object to remain at rest. Label the new force F⃗ 3. Draw the vector starting at the black

kicyunya [3294]

The new force F3 is added in the same direction as F2. To analyze the forces acting on an object in this scenario, we observe that they operate along the vertical axis, with F1 acting upward and F2 downward. To determine the necessary vector F3 to counteract the net force, it's important to calculate the length difference between F1 and F2. The direction of F3 will match that of the smaller force. If F2 is less than F1, F3 will align with F2.

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1 month ago

A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

Ostrovityanka [3204]

Answer

Given:

Wavelength = λ = 18.7 cm

= 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A) Calculate the angular frequency.

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

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3 months ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Softa [3030]

Answer:

Baseball mass: $m_b=3.992*10^{14}kg$

Explanation:

Circumference of a baseball is calculated using 2πr = 23 cm

Thus, the radius comes out to be 3.66 cm, which equals $3.66*10^{-2}$ m

The mass density of the baseball matches that of a neutron or proton.

Proton mass = $10^{-27}$ kg

Proton diameter = $10^{-15}$ m

Proton radius = $5*10^{-16}$ m

Volume of the baseball is $\frac{4}{3} \pi r^3$

Now by substituting all values into the mass per unit volume equation for the baseball, we get:

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

Therefore, the baseball mass amounts to $m_b=3.992*10^{14}kg$

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2 months ago

Read 2 more answers

A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag

Softa [3030]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

$\Phi =E\times S\times COS\Theta$

where Ф represents flux

E indicates electric field

S denotes surface area

θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 $\frac{Nm^{2}}{sec}$

E= $1.44\times 10^{4}\frac{Nm}{C}$

S= $\pi \times 0.057^{2}$

$COS\Theta =\frac{\Phi }{S\times E}$

= $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

=0.5306

θ=57.94°

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2 months ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m

Maru [3345]

Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by $\frac { m{ v }^{ 2 } }{ r }$ .

Here, m denotes the mass of the object

v is the velocity or speed of the object

r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F = $\frac { m{ v }^{ 2 } }{ r }$

where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F= $\frac { m{ (2v) }^{ 2 } }{ r } =\frac { 4m{ v }^{ 2 } }{ r }$

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.

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1 month ago