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Sladkaya [172]
2 years ago
10

What is the energy of a photon of blue light with a frequency of 6.53 × 10¹4 Hz?

Physics
1 answer:
shusha [124]2 years ago
3 0

Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

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The (blank) is something that should be tested according to the comfort level of the golfer. a. GRIP b. PUTTING GREEN c. SAND TR
gulaghasi [49]

Answer:  <em> a). Grip</em>

The grip is the important thing to be consider for golfer. It is the most fundamental important aspects to the game, because it provides the solid contact.

Applying the correct grip, results in better golf shots and an increased confidence to the golfer, to address the ball in the right direction.

Simply the grip is the pressure applied to the club with the hand such that the golfer feels comfortable to make a brilliant shot.



4 0
3 years ago
Please help, it's grade 7 science. :)
xeze [42]

Answer:

The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.

The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.

4 0
3 years ago
Read 2 more answers
For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty
Colt1911 [192]

Answer:

0.53\times 10^{-30}kgms^{-1}

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

\Delta x\Delta p=\frac{h}{4\pi }

Given that the uncertainty in x is 0.1 mm.

Therefore,

\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}

Therefore, uncertainty in the transverse momentum of photon is 0.53\times 10^{-30}kgms^{-1}

6 0
3 years ago
To wait until the oncoming vehicle passes before completing a left turn is known as:
Reika [66]

Answer:

Risk rejection

Explanation:

There are several factors that contribute to the degree of driving risks and they include but not limited to the ability of the driver and the condition of a vehicle. Other factors are condition of the environment and the condition of the highway. When driving, a driver may wait until an oncoming vehicle passes before making a complete left turn as a risk rejection strategy. Left turns are more dangerous when making them because drivers tend to accelerate on to a left turn. The wider radius of a  left turn is know to led to higher speeds and greater pedestrian exposure. A driver is advised to have more mental and physical efforts when making a left turn.

5 0
3 years ago
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
Read 2 more answers
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