Negatively charged particles that orbit the nucleus of an atom are ________

Assume it takes 7.00 minutes to fill a 30.0 gal gas tank. (a) Calculate the rate at which the tank is filled in gallons per seco

Maslowich

A)
R1 = 30/(7*60)
We are multiplying 7 with 60 because there are 60 seconds in 1 minute.
R1 = 30/420 = 0.0714 gallons/second

b) For this we need to express gallons to cubic meters.
1 gallon = 0.003785 m^3
R2 = 0.0714*0.003785 = 0.00027 m^3/s

c) V = R2*t where V is volume of some tank.
which means that
t=V/R2
t = 3698.8s or
t = 1.0274 hours

3 0

4 years ago

Which of the following types of energy is potential energy A. gravitational energy

katovenus [111]

There are many forms of energy, but they can all be put into two categories: kinetic and potential. Kinetic energy is motion––of waves, electrons, atoms, molecules, substances, and objects. Potential energy is stored energy and the energy of position––gravitational energy

4 0

3 years ago

How do you use physics?

Firlakuza [10]

Answer:

Physics extends well into your everyday life, describing the motion, forces and energy of ordinary experience. In actions such as walking, driving a car or using a phone, physics is at work. For everyday living, all the technologies you might take for granted exploit the rules of physics.

Explanation:

3 0

3 years ago

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What is the thinnest film of a coating with n = 1.39 on glass (n=1.52) for which destructive interference of the red component (

Illusion [34]

In optics, you have to create a layer of coating that is approximately 1/4 of the light's wavelength. The working equation for this problem is:

d = λ₀/4n,
where
λ₀ is the wavelength of the incident light
n is the refractive index of the coating

Substituting the values,

d = (650 nm)/4(1.39)
d = 116.9 nm

8 0

4 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

Negatively charged particles that orbit the nucleus of an atom are _________. A. electrons B. protons C. neutrons