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musickatia [10]

2 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.

1 answer:

Zigmanuir [339]2 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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forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s

zhenek [66]

Answer:

345453-5676

its the right answer

5 0

2 years ago

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0

2 years ago

The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who

densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

$L = \frac{y \times d}{1.22 \times \lambda} = \frac{1.2 \times 0.004}{1.22 \times 400 \times 10^{-9}} = 9836.066 \ km$

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)

$= 1.22 \times \frac{400 \times 10^{-9}}{0.004}$

sinθ = 1.22×10⁻⁴ rad

6 0

2 years ago

Read 2 more answers

You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

oksano4ka [1.4K]

As we know that reaction time will be

$t = 0.50 s$

so the distance moved by car in reaction time

$d = vt$

$d = 20 \times 0.50$

$d = 10 m$

now the distance remain after that from intersection point is given by

$d = 110 - 10 = 100 m$

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

$v_f = 0$

$v_i = 20 m/s$

now from kinematics equation we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2(a)100$

$a = \frac{-400}{200} = -2 m/s^2$

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

$v_f - v_i = at$

$0 - 20 = -2 (t)$

$t = 10 s$

total time to stop the car is given as

$T = 10 s + 0.5 s = 10.5 s$

3 0

2 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0

2 years ago