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3 years ago

7

A visible light emits light with a wavelength of 4.00x10^-7m.

1 answer:

velikii [3]3 years ago

8 0

answer

a) 7.50x10^4Hz

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Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)

ipn [44]

Answer:

4 x 10¹⁵

Explanation:

5 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

Does a star's apparent brightness depend only on its distance from earth?

Arisa [49]

I dont think so. It also probably matters what type of str it is

4 0

3 years ago

Read 2 more answers

How can density be determined in a lab of a rectangular solid?

True [87]

We will first record its mass and then its volume by measuring its dimensions
then divide mass by volume and will get density of regular solid

6 0

4 years ago

If this did not happen, what would be the approximate force on an eardrum of area 0.22 cm2 if a change in altitude of 1500 m tak

jeyben [28]

Complete Question

When you ascend or descend a great deal when driving in a car yours ears "pop," which means that the pressure behind the eardrum is being equalized to that outside. If this did not happen, what would be the approximate force on an eardrum of area .50 cm2 if a change in altitude of 950 m takes place?

Answer:

The value is $F = 0.60 \ N$

Explanation:

From the question we are told that

The area of the ear drum is $A = 0.5 \ cm^2 = 0.50 *10^{-4} \ m^2$

The change in altitude is $\Delta d = 950 \ m$

Generally the change in pressure is mathematically represented as

$\Delta P = \frac{F}{A}$

This can also be mathematically represented as

$\Delta P = \rho * g * \Delta d$

So

$\frac{F}{A} = \rho * g * \Delta d$

=> $F = \rho * g * \Delta d * A$

=> $F = \rho * g * \Delta d * A$

Here $\rho$ is the density of dry air with value $\rho = 1.29 \ kg /m^3$

So

$F = 1.29 * 9.8 * 950 * 0.50 *10^{-4}$

=> $F = 0.60 \ N$

3 0

3 years ago