The strength of an electromagnet CANNOT be increased by?

A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/

Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly $y = 0.0204 \ m$

Explanation:

From the question we are told that

The diameter of the glass tube is $d = 1 \ mm = 0.001 \ m$

The density of glycerin is $\rho = 1260 \ kg /m^3$

The surface tension of the glycerin is $\sigma = 6.3 *10^{-2} \ N /m$

The capillary rise of the glycerin is mathematically represented as

$y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}$

substituting value

$y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}$

$y = 0.0204 \ m$

Therefore the height of the glass tube the glycerin was able to cover is

$y = 0.0204 \ m$

4 0

4 years ago

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

A certain lightning bolt moves 50.0 C of charge. How many units of fundamental charge is this?

liubo4ka [24]

Answer: There are number of electrons.
Explanation:
We are given 50 Coulombs of charge and we need to find the number of electrons that can hold this much amount of charge. So, to calculate that we will use the equation:

where,
n = number of electrons
Charge of one electron =
Q = Total charge = 50 C.
Putting values in above equation, we get:

Hence, there are number of electrons.

7 0

4 years ago

Which of the following could be the source of resistance in a household electric circuit?

Tpy6a [65]

D. all of these

all of these use electricity

Hope I helped!

8 0

3 years ago

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