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Svetlanka [38]
3 years ago
11

A concave mirror has a focal length of 11 cm . What is its radius of curvature?

Physics
1 answer:
OverLord2011 [107]3 years ago
6 0

Answer:

22cm

Explanation:

focal length = 11cm

radius of curvature,r = 2f

r= 2 x 11

r=22cm

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An 1700 kg car is moving to the right at a constant speed of 1.50 m/s. (a) what is the net force on the car
cluponka [151]
I am sorry if I am wrong but, the net force would be zero. 0
8 0
3 years ago
What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength
ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

\lambda  = \frac{h}{\sqrt{2 m K}}

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}

λ = 1.2826 x 10^-13 m

6 0
3 years ago
Perform the calculation and report your answer using sig figs. 657.70 - 26.543
Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70  - 26.543  = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0
2 years ago
I need help on this question!
Mashutka [201]

Explanation:

click the link you can get answer of the question

3 0
3 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

#SPJ1

7 0
2 years ago
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