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Flauer [41]
3 years ago
10

Which of the following correctly represents an element?

Physics
2 answers:
eduard3 years ago
5 0

Answer:

A water H2O

Explanation:

H2O is composed of 2 hydrogens which is represented H2 and the hydrogen is combined with one oxygen which forms water which is H2O

garik1379 [7]3 years ago
3 0
A water - H20 because you can touch it
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For maximum benefit, you should exercise 5-7 times a week for 60 minutes and your heart rate should be between _____________ of
lana [24]

Answer:

60 to 80%

Explanation:

For maximum benefit, the heart rates during exercise regimes should be between<u> 60 to 80% </u>of an individual's maximum heart rate.

<em>The maximum heart rate varies from individuals to individuals and largely depends on age. It is calculated by subtracting one's age from 220. For example, a person that is 40 years old would have a maximum heart rate of;</em>

<em>   220 - 40 = 180 beats per minute.</em>

Such a person's heart rates during exercises should be between 108 and 144 beats per minute for maximum benefits.

5 0
3 years ago
In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of
Naddika [18.5K]

Answer:

Probably none of the above. it's says directly proportional so there would be a proportionality constant.

4 0
4 years ago
A parallel-plate capacitor is constructed from two 6.0 cm × 6.0 cm electrodes spaced 1.5 mmapart. The capacitor plates are charg
Brilliant_brown [7]

A) 2.4\cdot 10^{-6} J

The energy stored in a capacitor is given by:

E=\frac{1}{2}\frac{Q^2}{C}

where

Q is the charge stored

C is the capacitance

The capacitance of a parallel-plate capacitor is

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=6.0 cm \cdot 6.0 cm=36.0 cm^2=36\cdot 10^{-4} m^2 is the area of each plate

d=1.5 mm=0.0015 m is the distance between the plates

Substituting,

C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0015 m}=2.1\cdot 10^{-11} F

The charge stored on the capacitor is

Q=10 nC=10\cdot 10^{-9}C

So, the energy stored is

E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{2.1\cdot 10^{-11} F}=2.4\cdot 10^{-6}J

B) 2.6\cdot 10^{-6}J

This time, the separation between the plates is

d = 1.7 mm = 0.0017 m

So, the new capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0017 m}=1.9\cdot 10^{-11} F

And so, the new energy stored is

E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{1.9\cdot 10^{-11} F}=2.6\cdot 10^{-6}J

C)

Energy must be conserved, so the difference between the initial energy of the capacitor and its final energy is just equal to the work done to increase the separation between the two plates from 1.5 mm to 1.7 mm (in fact, the two plates of the capacitor attract each other since they have opposite charge, so work must be done in order to increase their separation)

7 0
3 years ago
How are reflecting telescopes similar to devices that produce laser light?
Novosadov [1.4K]

Answer:a

Explanation:

Because its light

7 0
3 years ago
A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λ
cluponka [151]

Answer:

Explanation:

Given

N=3680 cm^{-1}

therefore slit spacing d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm

since d\sin \theta =n\lambda

for n=1

d\sin \theta =\lambda

Now,at \theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm

at \theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm

at \theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm

5 0
3 years ago
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