Ask question

Ask question

All categories

2 years ago

12

What is the intensity of the electromagnetic light waves coming from the Sun just outside of the atmosphere of Venus, Earth and

Mars

1 answer:

Firdavs [7]2 years ago

6 0

The sun emits electromagnetic waves with a power of

4.0 ∗ 10 (26) W.

You might be interested in

A water pump is a positive displacement-type pump true or false

Kay [80]

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

6 0

3 years ago

True or false: Everything in the universe works with harmony because of balanced energy

Eduardwww [97]

Answer:

Explanation:

True

7 0

2 years ago

Read 2 more answers

A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi

S_A_V [24]

Answer:

10.2 Watt

Explanation:

$N$ = number of turns in flat coil = 160

$A$ = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

$B$ = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

$t$ = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

$E = \frac{- N A \Delta B}{t}$

$E = \frac{- (160) (0.20) (- 0.80)}{2}$

$E$ = 12.8 volts

$R$ = Resistance of the coil = 16 Ω

Power is given as

$P = \frac{E^{2}}{R}$

$P = \frac{(12.8)^{2}}{16}$

$P$ = 10.2 Watt

6 0

2 years ago

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

or

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

or

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

or

I = $0.0109[1-2.717^{-1}]$

or

I = 0.00688 A

or

I = 6.88 mA

5 0

2 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi

sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

$v_1$ = Velocity of sound = 342 m/s

$v_2$ = Velocity of sound = 346 m/s

$f_1$ = First frequency = 440 Hz

Frequency is given by

$f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz$

Beat frequency is given by

$|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s$

Beat frequency is 5.15348 Hz

Wavelength is given by

$\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m$

Relation between length of the flute and wavelength is

$\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m$

At v = 346 m/s

$\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m$

$L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m$

Difference in length is

$\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm$

It extends to 4.55 mm

7 0

2 years ago