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3 years ago

A transformer steps up an alternating current from

Physics

1 answer:

Deffense [45]3 years ago

4 0

Answer:

See the answers below.

Explanation:

A transformer is a static electrical machine, it should be borne in mind that although the voltage can be increased or decreased, the power must always be equal.

For this case:

The power in the first coil can be calculated by means of the following formula.

$P=V*I\\$

where:

V = voltage = 12 [V]

I = current =2 [A]

Therefore the power:

$P=12*2\\P=24[W]$

Now in the second coil, the power must be the same:

$P_{secondcoil}=V*I\\24=144*I_{secondcoil}\\I_{secondcoil}=24/144\\I_{secondcoil}=0.166[A]$

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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

m_a_m_a [10]

Answer:

a The kinetic energy is $KE = 0.0543 J$

b The height of the center of mass above that position is $h = 1.372 \ m$

Explanation:

From the question we are told that

The length of the rod is $L = 1.4m$

The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$

The angular speed at the lowest point is $w = 1.09 \ rad/s$

Generally moment of inertia of the rod about an axis that passes through its one end is

$I = \frac{mL^2}{3}$

Substituting values

$I = \frac{(0.140) (1.4)^2}{3}$

$I = 0.0915 \ kg \cdot m^2$

Generally the kinetic energy rod is mathematically represented as

$KE = \frac{1}{2} Iw^2$

$KE = \frac{1}{2} (0.0915) (1.09)^2$

$KE = 0.0543 J$

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

$KE = PE = mgh$

$0.0543 = mgh$

$h = \frac{0.0543}{9.8 * 0.140}$

$h = 1.372 \ m$

5 0

3 years ago

Read 2 more answers

The IMA of the pulley show is

Lera25 [3.4K]

The IMA of the pulley shown is 2.

4 0

3 years ago

Read 2 more answers

Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge b

Elena L [17]

Answer:

a) 0.167 μC/m^2

b) 1.887 * 10^4 V/m

Explanation:

Hello!

First let's find the surface charge density:

Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:

σ = Q/A

Where A is the area of the satellite, which is:

A=4πr^2 = πd^2 = π(1.9m)^2

Therefore:

σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2

Now let's calculate the electric field

Just outside the surface of the satellite the elctric field will be:

E = σ/ε0

Where ε0=8.85×10^−12 C/Vm

Therefore:

E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m

E = 1.887 * 10^4 V/m

5 0

3 years ago

Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less thanthat o

snow_tiger [21]

Answer:

41.4 g

41.4 cm³

1.08695 g/cm³

Explanation:

$\rho$ = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

$m=45-3.6\\\Rightarrow m=41.4\ g$

Mass of water displaced is 41.4 g

Density is given by

$\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{41.4}{1}\\\Rightarrow v=41.4\ cm^3$

So, volume of bone is 41.4 cm³

Average density of the bird is given by

$\rho=\dfrac{45}{41.4}\\\Rightarrow \rho=1.08695\ g/cm^3$

The average density is 1.08695 g/cm³

3 0

3 years ago

The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

$=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m$

Δh_air=714m

7 0

3 years ago