Light wave a has a greater frequency than light wave b. Which has a greater wavelength?.

QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of

kvv77 [185]

The volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

According to bernaulli's theorem theorem

P+1/2pV²+pgy = constant

where p fluid density

g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.

As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is A1V1=A2V2 .

rest is attached in image

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4 0

2 years ago

The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

We simply replace the values in that equation :

$\vartriangle r= m \lambda =2.\ 460\ nm$

$\vartriangle r= 920\ x\ 10^{-9} m$

The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.

3 0

3 years ago

A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it

mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0

3 years ago

Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnect

Iteru [2.4K]

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

$V = V_{o} e^{\frac{-t}{T}}$

$2.2 = 6 e^{\frac{-t}{66}}$

t = 66.2 sec

5 0

3 years ago

The diagram shows a screwdriver being used as a lever to open a tin

Musya8 [376]

Answer:

(a) Most reactive ${}$ Metal B

${}$ Metal D

${}$ Metal A

Least reactive ${}$ Metal C

(b) (i) Bubbles should form very slowly

(ii) No reaction takes place

Explanation:

(a) The given metals arranged in their order of reactivity are;

Most reactive ${}$ Metal B

${}$ Metal D

${}$ Metal A

Least reactive ${}$ Metal C

The other of reactivity is based on the nature of their reactivity of the metals in air

(b) (i) Based on the reactivity of the metals in air, whereby metal A reacts very slowly and an oxide is formed, we have that, based on the reactivity of the metal A, when mixed with dilute hydrochloric acid, bubbles should form very slowly

(ii) Similarly, given that metal C is unreactive, we have that when small pieces of metal C are added to dilute hydrochloric acid, no reaction takes place.

7 0

3 years ago