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olga nikolaevna [1]
3 years ago
12

A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the b

ottom moving along the ground when the bottom of the ladder is 5 ft from the wall
Physics
1 answer:
netineya [11]3 years ago
7 0

Answer: The bottom of the ladder is moving at 3.464ft/sec

Explanation:

The question defines a right angle triangle. Therefore using pythagorean

h^2 + l^2 = 10^2 = 100 ...eq1

dh/dt = -2ft/sec

dl/ dt = ?

Taking derivatives of time in eq 1 on both sides

2hdh/dt + 2ldl/dt = 0 ....eq2

Putting l = 5ft in eq2

h^ + 5^2 = 100

h^2 = 25 = 100

h Sqrt(75)

h = 8.66 ft

Put h = 8.66ft in eq2

2 × 8.66 × (-2) + 2 ×5 dl/dt

dl/dt = 17.32 / 5

dl/dt = 3.464ft/sec

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Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
5 0
3 years ago
The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o
Leya [2.2K]
Answer:
frequency = 5.52 * 10² Hz

Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength

We are given that:
velocity = 331 m/sec
wavelength = 0.6 m

Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz

Hope this helps :)
4 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
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Alisiya [41]

If my math is right its A) 7

because 189 divided by 27 is 7

7 0
3 years ago
Read 2 more answers
What is the right answer
DIA [1.3K]

Answer:

b

Explanation:

6 0
3 years ago
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