Which Of The Following Research Methods Are Widely Used By Psychologists?

Define the term sorting

Nikolay [14]

Answer: arrange systematically in groups; separate according to type, class, etc.

Explanation:

i hooked this up btw

6 0

3 years ago

Providing subgoals for individuals helps them solve problems more quickly.<br> True or False

trasher [3.6K]

True because they can focus on small goals to eventually solve their big problems.

3 0

3 years ago

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A ball is dropped from the top of a building that is known to be 400 feet high. The formula for finding the height of the ball a

Ira Lisetskai [31]

Answer:

3 seconds

Explanation:

Height of the building = 400 feet

Height of the ball from the ground is given by

h=400−16t²

This formula has been derived from

$s=ut+\frac{1}{2}at^2$

a = Acceleration due to gravity = 32 ft/s²

u = Initial velocity = 0

t = Time taken

Substituting all the values we get

$s=0t+\frac{1}{2}32t^2\\\Rightarrow s=16t^2$

This is the height of the ball from the top of the building

The height of the ball from the ground will be

h = 400-s

⇒h = 400−16t²

When h = 256 ft

$256=400-16t^2\\\Rightarrow t=\sqrt{\frac{256-400}{-16}}\\\Rightarrow t=3\ s$

Time taken by the ball to reach a height of 256 feet above the ground is 3 seconds

4 0

3 years ago

Two narrow, parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the

CaHeK987 [17]

Answer:

a) 7.947 radians

b) $\mathbf{\frac{I}{I_{max}}=0.4535}$

Explanation:

y = Distance from central bright fringe = 2.5 mm

λ = Wavelength = 600 nm

L = Distance between screen and source = 2.8 m

d = Slit distance = 0.85 mm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{2.5}{2800}=0.000892\\\Rightarrow \theta=tan^{-1}0.000892=0.05115^{\circ}$

$\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589\times10^{-7}$

a) Phase difference

$\phi=\frac{2\pi}{\lambda}\Delta r\\\Rightarrow \phi=\frac{2\pi}{600\times 10^{-9}}7.589\times10^{-7}=7.947\ rad$

∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians

b) $\frac{I}{I_{max}}=cos^2\frac{\phi}{2}\\\Rightarrow \frac{I}{I_{max}}=cos^2\frac{7.947}{2}\\\Rightarrow \mathbf{\frac{I}{I_{max}}=0.4535}$

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe $\mathbf{\frac{I}{I_{max}}=0.4535}$

8 0

4 years ago

A garden hose has a radius of 0.0120 m, and water comes out at a speed of 4.88 m/s. How much time does it take to fill up a kids

Doss [256]

The radius of the garden hose is $r=0.0120 m$ , so its cross-sectional area is
$A=\pi r^2 = \pi (0.0120 m)^2 = 4.52 \cdot 10^{-4} m^2$

The amount of water (in $m^3$ ) that comes out from the hose in one second is given by the product between the speed of the water and the cross-sectional area of the hose:
$V_w = A v = (4.52 \cdot 10^{-4}m^2)(4.88 m/s)=2.2 \cdot 10^{-3} m^3/s$

The time needed to fill the pool is equal to the volume of the pool divided by the amount of water that comes out every second:
$t= \frac{V}{V_w}= \frac{3.88 m^3}{2.2 \cdot 10^{-3}m^3}=1758 s$

6 0

4 years ago

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