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3 years ago

6

How many neutrons does C-13 have? A. 6 B. 7 C. 13 D. 19

1 answer:

Mrac [35]3 years ago

6 0

The answer is B. 7.

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What are the similarities and differences between hydraulic and pneumatic systems?

sergey [27]

Answer:

Pneumatics use easily compressible gas like air or pure gas. Meanwhile, hydraulics utilize relatively-incompressible liquid media like mineral oil, ethylene glycol, water, synthetic types, or high-temperature fire-resistant fluids to make power transmission possible.

Explanation:

hope this helps

6 0

3 years ago

Star A and star B appear equally bright in the sky. Star A is twice as far away from Earth as star B. How do the luminosities of

DerKrebs [107]

Answer:

The answer to the question is

The luminosity of stars A is four times that of star B

Explanation:

Flux (F) produced by a source of light is directly proportional to the brightness or Luminosity (L), and varies inversely to its distance d, that is $F \alpha \frac{L}{d^2}$

Therefore if the two stars present the same flux then we have

$\frac{L_1}{d_1^2} = \frac{L_2}{(2d_1)^2}$ then crossing out like terms gives $\frac{L_1}{1} = \frac{L_2}{2^2}$ or 4·L₁ = L₂

The luminosity of star A is 4 times the that of star B

7 0

3 years ago

What is the kinetic energy of the rock just before it hits the ground <br> What is the equation ?

gtnhenbr [62]

Using every bit of the information that's given in the question, we can see that there is no rock, and it's not moving.

6 0

3 years ago

What is the energy of a photon with a frequency of 2.2 x 1016 Hz? Planck's constant is 6.63 x 10-34 Jos.

aleksandr82 [10.1K]

Answer: 1.5 × 10^-17

Explanation:

Given the following :

Frequency(f) = 2.2 × 10^16 Hz

Planck's constant(h) = 6.63 × 10^-34

The energy of a photon 'E' is given as the product of frequency and the planck's constant

E = hf

E = (6.63 × 10^-34) × (2.2 × 10^16)

E = 6.63 × 2.2 × 10^(-34 +16)

E = 14.586 × 10^-18

E = 1.4586 × 10^-17

E = 1.5 × 10-17 (2 S. F)

9 0

3 years ago

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

Ghella [55]

Answer:

Total electric potential, $V=1.32\times 10^6\ volts$

Explanation:

It is given that,

First charge, $q_1=3\ \mu C=3\times 10^{-6}\ C$

Second charge, $q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C$

Distance of first charge from origin, $r_1=1.15\ cm=0.0115\ m$

Distance of second charge from origin, $r_2=2\ cm=0.02\ m$

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

$V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

$V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})$

$V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})$

V = 1321826.08 V

or

$V=1.32\times 10^6\ volts$

So, the total electric potential at the origin is $1.32\times 10^6\ volts$ . Hence, this is the required solution.

3 0

3 years ago