The Rorschach inkblots and the TAT (Thematic Appreciation Test) both rely on providing the subject with ambiguous visual stimuli and assessing the subject's state of mind using the subject's interpretation of the stimuli.
Both use cards, although not all of the cards are used in the TAT. Moreover, the TAT cards contain sketches, while the Rorschach inkblots contain patterns of ink.
The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°
cos θ = Adjacent side / Hypotenuse
θ
= 47°
Hypotenuse = Length of ladder = 8.5 m
cos 47° = Adjacent side / 8.5
Adjacent side = Initial distance of base of ladder from the building = 5.8 m
Adjacent side 2 = Final distance of base of ladder from the building
Adjacent side 2 = 5.8 - 0.82 = 4.98 m
cos θ
= Adjacent side 2 / Hypotenuse
cos θ
= 4.98 / 8.5 = 0.59
θ
=
( 0.59 )
θ
= 53.84°
The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.
Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°
To know more about trigonometric ratios
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The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 ×
×
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss =
×
×
× 
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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<span>160 Joules
For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance:
80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.
So the cart has a final kinetic energy of 160 Joules.</span>
<h2>F = kAρv²</h2>
Explained in the attachment !
<h3>Hope it helps you!!</h3>