PHYSICS COLLEGE

A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?

Answers

Answer 1

Answer:

Answer:

the lowest frequency is $7.5* 10^(14) Hz$

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν= $(v)/(L)$

frequency ν be lopwest when L will be heighest

ν(lowest)= $(3* 10^8)/(400* 10^-9)$

ν= $7.5* 10^(14) Hz$

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Answers

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

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If a barometer reads 772 mm hg, what is the atmospheric pressure expressed in pounds per square inch?

Answers

15.23.....................

I think it would be 15.23 not so sure but
hope this helps! (:

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm

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I need a good phisics teacher for one day online please123

Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answers

Answer:

a) 343.0 Hz b) 686.0 Hz

Explanation:

a) First, we need to know the distance to both speakers.

If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.

So, if he moves 0.25 m towards one of them, the distance from any speaker will be as follows:

d₁ = 2.0 m-0.25 m= 1.75 m

d₂ = 2.0 m + 0.25 m = 2.25 m

The difference between these distances is the path difference between the sound from both speakers:

d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m

If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:

d = (2*n-1)*(λ/2) = 0.5 m

The minimum distance is for n=1:

⇒ λ = 2* 0.5 m = 1 m

In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:

v = λ*f, where v= 343 m/s and λ=1 m.

Solving for f, we have:

$f =(343.0 m/s)/(1.0 m) = 343 Hz$

b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met the following condition:

d = 0.5 m = 2n*(λ/2) = λ (for n=1)

if the speed remains the same (343 m/s) we can find the new frequency as follows:

$f =(v)/(d) =(343 m/s)/(0.5m) =686.0 Hz$

⇒ f = 686.0 Hz

Final answer:

Two speakers create peaks and troughs of sound intensity due to constructive and destructive interference of waves. Using wave properties, the frequency of the sound when a minimum intensity is experienced 0.25m from the center is 680Hz. Increasing the frequency, the first to produce maximum intensity at the same position is about 2720Hz.

Explanation:

The behavior of sound intensity in this question is due to wave interference, specifically, constructive and destructive interference of sound waves. When you stand halfway between the speakers, the sound waves from each speaker are in phase, which means the pressure variations combine to create an intensified sound, known as constructive interference.

When you move towards one of the speakers and detect a minimum in sound intensity, this is due to destructive interference, which occurs when the crest of one wave overlaps with the trough of another, canceling each other and producing a minimum sound level.

a. The frequency of the sound can be calculated using the formula for wave speed, v = f.lambda, where v is the speed of sound (340 m/s under normal conditions), f is the frequency, and lambda is the wavelength. In this case, a minimum sound intensity indicates one-half wavelength. So, lambda = 0.5 m. Thus, frequency, f = v/lambda = 340/0.5 ~ 680 Hz.

b. When you increase the frequency while remaining 0.25m from the center, the first frequency for which the location will be a maximum of sound intensity will be when you are an integral multiple of the wavelength away from the source. Thus if we let this be 2λ, we can calculate the frequency as f = v / λ = v / (0.25m / 2) = 340 / 0.125 ~ 2720 Hz.

Learn more about Sound Wave Interference here:

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You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

Answers

Answer:

$d'=√(v^2+w^2)$

Explanation:

Rate of Change

When an object moves at constant speed v, the distance traveled at time t is

$x=v.t$

We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

$y=v.t$

and the horizontal distance is

$x=w.t$

The distance between both friends is computed as the hypotenuse of the triangle

$d^2=x^2+y^2$

We need to find d', the rate of change of the distance between both friends.

Plugging in the above relations

$d^2=(v.t)^2+(w.t)^2$

$d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2$

Solving for d

$d=√((v^2+w^2)t^2)$

$d=√((v^2+w^2)).t$

Differentiating with respect to t

$\boxed{d'=√(v^2+w^2)}$

Final answer:

The problem is solved using Pythagoras' Theorem, representing the two travel paths forming a right triangle. The rate at which the distance increases between two points moving perpendicularly can be found by differentiating the resulting equation, which yields the expression sqrt[(v^2)+(u^2)].

Explanation:

The question is about the rate at which the distance between you and your friend is increasing at time t. It's a typical problem in kinematics. Because the roads are perpendicular to each other, we can solve the problem using Pythagoras' Theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's denote the distance you've traveled as D1 = v*t (because distance = speed * time) and the distance your friend has travelled D2 = u*t. The distance between you can be computed using Pythagoras' Theorem as D = sqrt(D1^2 + D2^2). Hence, D = sqrt[(v*t)^2 + (u*t)^2]. Differentiating D with respect to t using the chain rule will give us the rate at which the distance between you is increasing, which is sqrt[(v^2)+(u^2)].

Learn more about Relative Velocity here:

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