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stira [4]
3 years ago
11

Visible light falls into wavelength ranges of 400-700 nm, for which 1 m = 1 × 10 9 nm . The energy and wavelength of light are r

elated by the equation E = h c λ where E is energy in Joules, h is Planck's constant ( 6.626 × 10 − 34 J-s ), c is the speed of light ( 2.998 × 10 8 m/s ), and λ is the wavelength in m. If a visible light photon has a wavelength of 560.6 nm, what is the energy of the photon (in J)?
Physics
1 answer:
Paul [167]3 years ago
7 0

Answer:

E = 3.54 x 10⁻¹⁹ J

Explanation:

The energy of the photon can be given in terms of its wavelength by the use of the following formula:

E = \frac{hc}{\lambda}

where,

E = energy = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light = 2.998 x 10⁸ m/s

λ =  wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m

Therefore,

E = \frac{(6.626\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{5.606\ x\ 10^{-7}\ m}\\

<u>E = 3.54 x 10⁻¹⁹ J</u>

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As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave
DedPeter [7]

Answer:

-29.2\times 10^{3} N

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time=\Delta t=1.3 s

We have to find the average force exerted on the car.

Average force=\frac{change\;in\;momentum}{\Delta t}

F_{avg}=\frac{mv-mu}{1.3}

F_{avg}=\frac{1900(0)-1900(20)}{1.3}

F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N

Hence, the average force exerted on the car that hits a line of water barrels=-29.2\times 10^{3} N

8 0
3 years ago
Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, t
makvit [3.9K]

Answer:

A) 30 s, 792 m

B) 10.28 s, 4108.2 m = 4.11 km

Explanation:

A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

Using the equations of motion.

v = u + at

v = final velocity = 120 km/h

u = initial velocity = 70 km/h

a = acceleration = 6000 km/h²

t = ?

120 = 70 + 6000t

6000t = 50

t = (50/6000) = 0.0083333333 hours = 30 seconds.

Using the equations of motion further,

v² = u² + 2ax

where x = horizontal distance covered by the car during this time

120² = 70² + 2×6000×x

12000x = 120² - 70² = 9500

x = (9500/12000) = 0.79167 km = 791.67 m = 792 m

B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

For the bullet A

u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

t = T

y = 600T - 4.9T²

At the point where the two bullets pass each other, the vertical heights covered are equal

y = y

450t - 4.9t² = 600T - 4.9T²

But, note that, since T starts reading, 3 seconds after t started reading,

T = (t - 3) s

450t - 4.9t² = 600T - 4.9T²

450t - 4.9t² = 600(t-3) - 4.9(t-3)²

450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)

450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1

600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0

179.4t - 1844.1 = 0

t = (1844.1/179.4) = 10.28 s

Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.

y = 450t - 4.9t²

= (450×10.28) - (4.9×10.28×10.28)

= 4,108.2 m = 4.11 km

Hope this Helps!!!!

6 0
2 years ago
Timbre describes the sound’s _____.<br><br> pitch<br><br> loudness<br><br> tone<br><br> quality
Bumek [7]

Answer:

A) pitch

Explanation:

I did band and also g*ogled it to check :)

If it isn't A, however, it is C

Hope this helps :D

7 0
2 years ago
Read 2 more answers
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
2 years ago
If Dwight races his Chevy towards Chatham and travels 2460 meters in 60 seconds. what is his velocity?
labwork [276]

Answer:

divide

Explanation:

whenever looking for velocity.just devide

8 0
2 years ago
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