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disa [49]
3 years ago
9

Copper exists in nature as two isotopes. The atomic masses and relative abundance of these isotopes is given in the table. What

is the average atomic mass of copper?
Physics
1 answer:
Alona [7]3 years ago
7 0
<span>The correct answer should be B) 63.55. That's because the most precise number is 63.546, but you would write 55 because 46 is rounded that way in the equation. The others are a bit higher, while E is a completely different element, Iodine. This isn't the most precise piece of data because in reality there would be a slight differentiation of +- 0,003u</span>
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Mester Exam 1 11 of 35
frez [133]
Please show picture of diagrams
8 0
3 years ago
The birth defect caused by the Zika Virus to unborn infants is called__________?
lions [1.4K]
It is B, Microcephaly. Hope I helped! :))
3 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?​
Ilya [14]

The frequency of the wave is 1\cdot 10^9 Hz

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

c=f \lambda

where

c is the speed of the wave

f is the frequency

\lambda is the wavelength

For the radio wave in this problem,

\lambda = 0.3 m

c=300,000,000 m/s = 3\cdot 10^8 m/s

Therefore, the frequency is:

f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0
3 years ago
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = \left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0
3 years ago
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