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Lostsunrise [7]
3 years ago
7

Which describes the relationship between photon energy and frequency?

Physics
2 answers:
nasty-shy [4]3 years ago
8 0
E = hf
E : photon energy
h : Plank's constant 6.63×10^-34
f : frequency

Hope it helped!
vovangra [49]3 years ago
8 0

high energy photons have high frequencies hope this helps brother

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A 10,000 N net force is accelerating a car at a rate of 5.5m/s^2. What is the cars mass
ycow [4]

Answer:

If a net horizontal force of 175 N is applied to a bike whose mass is 43 kg what acceleration is produced? What average net force is ... A 10,000 N net force is accelerating a car at a rate of 5.5 m/s2. What is the car's mass? A boy pedals his ...

Explanation:

6 0
3 years ago
Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1
max2010maxim [7]

Answer:

2.5 m/s

Explanation:

Mechanical energy is the sum of the potential and kinetic energy.

E = PE + KE

E = mgh + ½mv²

172.1 J = (7.26 kg) (9.8 m/s²) (2.1 m) + ½ (7.26 kg) v²

v = 2.5 m/s

7 0
3 years ago
Read 2 more answers
A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in ai
Svetach [21]

Answer:

The  values is  f_b  =14.9 \  beats/s

Explanation:

From the question we are told that

   The  speed of the fire engine is  v =  5\ m/s

    The frequency of the tone is  f =  500 \ Hz

    The speed of sound in air is v_s  =  340 \ m/s

The  beat frequency is mathematically represented as

     f_b  =  f_a  -  f

Where  f_a is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as

   f_a  =  [\frac{v_s + v }{v_s  -v} ]* f

substituting values

  f_a  =  [\frac{340 + 5 }{340  -5} ]* 500

  f_a  = 514.9 \  Hz

Thus  

      f_b  =514.9 -  500

      f_b  =14.9 \  beats/s

8 0
3 years ago
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an
Dominik [7]

Answer:

position as a function of time is y = 0.05 × cos(9.9)t

Explanation:

given data

mass = 5 kg

length = 10 cm = 0.1 m

displaced = 5 cm

to find out

position as a function of time

solution

we will apply here equilibrium that is

mass × g = k × length

put here value and find k

k = \frac{5*9.8}{.01}

k = 490 N/m

and ω is

ω = \sqrt{\frac{k}{m} }

ω = \sqrt{\frac{490}{5} }

ω = 9.9

so here position w.r.t  time is

y = 0.05 × cosωt

y = 0.05 × cos(9.9)t

so position as a function of time is y = 0.05 × cos(9.9)t

8 0
3 years ago
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