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2 years ago

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An ideal step-down transformer has a turns ratio of 1/106. An ac voltage of amplitude 170 V is applied to the primary. If the pr

imary current amplitude is 8.10 mA, what is the secondary current amplitude

1 answer:

Leya [2.2K]2 years ago

7 0

Answer:

$I_2=0.8586A$

Explanation:

From the question we are told that:

Turns Ratio i.e $\frac{N_2}{N_1}=\frac{1}{106} (step down)$

Voltage $v=170v$

Primary current amplitude $I_1= 8.10 mA =>8.10*10^{-3}A$

Generally the equation for a Transformer is mathematically given by

$\frac{E_1}{E_2}=\frac{I_2}{I_1}=\frac{N_1}{N_2}$

Therefore

$I_2=\frac{N_1}{N_2}*I_1$

$I_2=106*8.10*10^{-3}$

$I_2=0.8586A$

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A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

A family drives north for 30km then turns east for 20km. The family then decided to turn west for 5km before finally stopping to

salantis [7]

Answer:

A

Explanation:

They drove 30km north. The displacement adds up to 25km therefore making the distance greater

Hope this helps!

6 0

3 years ago

Find the weight of a 25 kg table. (Use g= 10 m/s?)

PtichkaEL [24]

Answer:

weight = 25*10 =250 N

Explanation:

g must be given in units of m/s^2

The weight of any type of body will be the product of his mass by the gravity

where:

m =mass [kg]

F = force [N] or [kg*m/s^2]

g = acceleration [m/s^2]

3 0

3 years ago

Read 2 more answers

What does every magnet possess?

IRINA_888 [86]

Answer:

a North and South Pole :)

Explanation:

5 0

3 years ago