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musickatia [10]
3 years ago
9

Focus groups are one of the most widely used ________ methods to gain greater understanding of a current problem or to develop p

reliminary knowledge to guide in the design of descriptive or causal research.
Physics
1 answer:
Zigmanuir [339]3 years ago
8 0

Answer:

Exploratory

Explanation:

<u>Focus groups</u>

It is a small group of 8-12 respondents guided by a moderator through a thorough debate on a specific subject or idea.It's great for generation of ideas, brainstorming, insight into motives, attitudes, and perceptions. it can  show likes, dislikes,emotional requirements and prejudices.

Exploratory methods are used to gain initial insights that could pave the way for further investigation.

Some of the exploratory methods are focus groups, Key informant,case studies,secondary data and observational data.

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Which of the following contains the majority of the mass in the solar system?
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First think which has less mass in the solar system. The sun is the largest object in the solar system, so the answer is C. the sun. Hope I helped! :P
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3 years ago
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Light travels at a speed of 2.998*108 m/s. Light takes approximately 3.25 minutes to travel from the Sun to reach a planet. Calc
ANTONII [103]

Answer:

585×10⁸ m

Explanation:

Distance = rate × time

d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

d = 585×10⁸ m

5 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
4 years ago
1. The gravitational force acting on a falling body and its weight is constant. But the law of universal gravitation tells us th
vagabundo [1.1K]
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3 0
3 years ago
1) An ice skater with a moment of inertia of 2.2 kg m^2 rotates at a frequency of 0.8 rotations per second. The ice skater tucks
trapecia [35]

Answer:

Explanation:

2.3 kg·m/s²

4 0
3 years ago
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