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MrRissso [65]
3 years ago
10

3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal

displacement of 3.00 meters, what is its launch velocity
Physics
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:

35.6 m

Explanation:

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Tanya [424]

D. 51 N. The minimum applied force that will cause the television slide is 51 N.

In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.

With μs = 0.35, and n = 15kg*9.8m/s² = 147 N

Fs = (0.35)(147 N)

Fs = 51.45 N

Fs ≅ 51 N

3 0
3 years ago
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If y^2= 3.249 x 10^-11, y = ?
Leni [432]

Answer:

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

5 0
3 years ago
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Indicate the result of each of the following unit vector cross products (unit vector hat-symbols not shown): . kxi = . jxi= -jxk
notka56 [123]

Answer:

Explanation:

The cross product of two vectors is given by

\overrightarrow{A}\times \overrightarrow{B}=\left | \overrightarrow{A} \right |\left | \overrightarrow{B} \right |Sin\theta \widehat{n}

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.

Here, K x i = - j

As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along  axis.

The direction of cross product of two vectors is given by the right hand palm rule.

So, k x i = j

j x i = - k

- j x k = - i

i x i = 0

4 0
3 years ago
Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference
strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

n_{air}=1

\theta_{liquid} = 19.38\°

\theta_{air}35.09\°

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

n_1sin\theta_1 = n_2sin\theta_2

n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}

n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}

Replacing the values we have:

n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}

n_liquid = 1.7323

Therefore the refractive index for the liquid is 1.7323

6 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
Read 2 more answers
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