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3 years ago

What does frequency mean in science

2 answers:

6 0

There's nothing mysterious about it at all. "Frequency" simply means
"often-ness" ... how often or how frequently something happens.

-- The frequency of traditional meals is 3 per day.
-- The frequency of an equinox is 2 per year.
-- The frequency of my sleeping really late is 1 per week.
-- The frequency of my intense desire to sleep late is 30 per month.
etc.

-- The standard unit of frequency in the SI system is "per second".
The special name for that unit is "Hertz". (Hz)

OleMash [197]3 years ago

5 0

A rate at which a cycle or phenomenon is repeated.

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Which particle rarely interacts with matter?

Marysya12 [62]

D. Neutrino
Neutrinos are particles that rarely interact with matter.

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4 years ago

On which moment of inertia of an object does depend on?

Likurg_2 [28]

Additional Information:

I couldn't get your question very clearly. In order to solve the question, I will define moment of inertia, state the formula and factors that the moment of inertia of a body depends and does not depend on.

Answer:

Moment of inertia depends on;

1. Mass of the body

2. Axis of rotation and

3. Distribution of the body

Moment of inertia does not depend on;

1. Angular velocity of the body.

Explanation:

The moment of inertia is defined as a quantity that determines the torque needed for a desired angular acceleration or a property of a body due to which it resists angular acceleration about a rotational axis.

Moment of Inertia, I = ∑mr²

Where,

I is the moment of Inertia

m is the mass

r is the distance from the axis of the rotation

The moment of inertia of a body depends on distribution of body, axis of rotation and mass of the body. However, the moment of Inertia of a body is not dependent on angular velocity of the body.

4 0

4 years ago

Solids have a definite shape and volume

Black_prince [1.1K]

Yes that is true, toothpaste has been very debatable over the years. (I really need Brainliest.) Hope this helps!!!!

7 0

3 years ago

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

Explanation: (a) Potential difference is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

5 0

4 years ago

If you are in a spaceship that is sitting on the surface of a planet, you feel your weight. How does this compare to the weight

nordsb [41]

Answer:

You will feel more weight if it is accelerating out of the planet.

You will feel less weight if it is accelerating towards the planet.

Explanation:

The weight that you are observing or feeling is basically due to the change in acceleration of your fall or rising up in the spaceship. When the acceleration is stationary on the surface, you experience your normal weight due to the gravitational acceleration of that planet.

When the spaceship accelerates above or out of the planet you experience acceleration more than the acceleration of gravity hence more weight.

When the spaceship accelerates towards the planet you experience acceleration less than the acceleration of gravity hence less weight.

If the spaceship is free falling at the gravitational acceleration you experience a zero weight

8 0

4 years ago