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nevsk [136]
3 years ago
6

Light with a frequency of 7.30 x 1014 hz lies in the violet region of the visible spectrum. What is the wavelength of this frequ

ency of light?
Physics
1 answer:
Leona [35]3 years ago
6 0

From the theory we know that:

c = λ / T

f = 1 / T

Where:

c = 3.10^{8} / m   (the speed of light)

λ is the wavelengh (in meters)

T is the period (in seconds)

f is the frequency (in Hz)

We were told that:

f = 7.30 . 10^{14}

And we want to find out the value of λ.

c = λ / T  

c = λ . 1/T

Swaping 1/T = f

c = λ . f

λ = c / f

λ = 3 . 10^{8} /  7.30 . 10^{14}

λ = 4.12 10^{-7} m

Response: 4.12 10^{-7} m = 412 nm

:-)

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A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
 We have then:
 1 = jug
 2 = skateboard + boy
 By conservation of the linear amount of movement:
 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
 -24.96 = (-0.65) M2
 M2 = (-24.96) / (- 0.65) = 38.4 kg
 Then, the child's mass is:
 M2 = Mskateboard + Mb
 Clearing:
 Mb = M2-Mskateboard
 Mb = 38.4 - 1.9
 Mb = 36.5 Kg
 answer:
 the boy's mass is 36.5 Kg
4 0
3 years ago
A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits
Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300)                ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0
3 years ago
Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o
Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.  

\lambda_{proton} = 6.05x10^{-14}m < \lambda_{electron} = 1.10x10^{-10}m

Part B:

The proton has a smaller wavelength than the electron.

\lambda_{proton} = 1.29x10^{-13}m < \lambda_{electron} = 5.525x10^{-12}m

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

\lambda = \frac{h}{p} (1)

Where h is the Planck's constant and p is the momentum.

\lambda = \frac{h}{mv} (2)

Part A

Case for the electron:

\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

But J = Kg.m^{2}/s^{2}

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

\lambda = 1.10x10^{-10}m

Case for the proton:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}

\lambda = 6.05x10^{-14}m

Hence, the proton has a smaller wavelength than the electron.  

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

KE = \frac{1}{2}mv^{2}

2KE = mv^{2}

v^{2} = \frac{2KE}{m}

v = \sqrt{\frac{2KE}{m}}  (3)

Case for the electron:

v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}

but 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}

v = 1.316x10^{8}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}    

\lambda = 5.525x10^{-12}m

Case for the proton :

v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}

But 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}

v = 3.07x10^{6}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}

\lambda = 1.29x10^{-13}m    

Hence, the proton has a smaller wavelength than the electron.

7 0
3 years ago
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
Which is a correct step in a scientific experiment involving acids and bases?
Gre4nikov [31]
The answer, using an indicator to measure the hydrogen ion concentration of a solution, is correct
4 0
3 years ago
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