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shtirl [24]
3 years ago
5

Which statement is true with respect to an atom?

Physics
2 answers:
maria [59]3 years ago
7 0
The statement that is true concerning an atom would be C. The atomic number of an atom is equal to the number of protons.
Allushta [10]3 years ago
6 0
<span>C.) The atomic number of an atom is equal to the number of protons.

The atomic number represents the equal number of protons and electrons in an element.</span>
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Saying english so we can help u
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3 years ago
What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?
ivann1987 [24]

Answer:

Explanation:The simple pendulum calculator finds the period and frequency of a ... Acceleration of gravity (g) ... Pendulum length (L) ... First of all, a simple pendulum is defined to be a point mass or bob (taking ... For example, it can be equal to 2 m. ... Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz

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g In 1920, Stern and Gerlach performed an experiment that first demonstrated Group of answer choices energy quantization. space
anyanavicka [17]

Answer:

B. space quantization.

Explanation:

In 1921, Otto Stern developed the idea behind this experiment, while Walther Gerlach performed the actual experiment in 1922. The Ster-Gerlach experiment provides prove to the fact that the spatial orientation of angular momentum is quantized. To demonstrate the experiment, silver atoms were made to travel through a magnetic field path.

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4 0
3 years ago
A deer with a mass of 146kg is running head on toward you with a speed of 17 m/s. You are going north. Find the momentum of the
Pavel [41]
Momentum = Mass * Velocity = 146 * 17 = 2482  kgm/s
6 0
3 years ago
Read 2 more answers
Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the
Assoli18 [71]

Answer:

Part a)

\rho = 3\epsilon_0 k r^2

Part b)

Q = 4\pi \epsilon_0kR^5

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

E = kr^3 \hat r

now electric flux through a spherical surface of radius r is given as

\phi = E. A

\phi = kr^3(4\pi r^2)

now by Guass law we know that

E.A = \frac{q}{\epsilon_0}

q = 4\pi \epsilon_0kr^5

now volume charge density is given as

\rho = \frac{q}{\frac{4}{3}\pi r^3}

\rho = 3\epsilon_0 k r^2

Part b)

Total charge inside the radius R is given as

Q = 4\pi \epsilon_0kR^5

7 0
3 years ago
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