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2 years ago

11

The concept of aggression cues is associated with A. frustration-aggression theory. B. the work of animal behaviorist Konrad Lor

enz. C. social learning theories. D. observational learning theories.

1 answer:

leva [86]2 years ago

7 0

<span>A. frustration-aggression theory.
Good Luck!!</span>

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5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-

Airida [17]

D. 980, this is the best answer because 35 x 7 is 980 :)

3 0

2 years ago

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4

dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0

3 years ago

A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of

Anna007 [38]

Answer:

The ratio is $\frac{F_{T1}}{F_{T2}} = 2$

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

Considering the second car

The force causing it s movement is mathematically represented as

$F_{T2} = ma$

Considering the first car

The force causing it s movement is mathematically represented as

$F = F_{T1} -F_{T2} = ma$

=> $F_{T1} -ma = ma$

=> $F_{T1} = 2 ma$

=> $\frac{F_{T1}}{ma} = 2$

=> $\frac{F_{T1}}{F_{T2}} = 2$

7 0

3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

$d_2$ = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

$d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m$

Time

$t=\dfrac{Distance}{Speed}$

Time difference would be

$\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s$

The time difference is 0.0000109261200583 s

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s$

The ratio is

$\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583$

The ratio is 0.0109261200583

7 0

3 years ago

If you are driving your car at 45.0 miles per hour, and you have been driving 2.50 hours, how far has your car gone?

Greeley [361]

Answer:

113 miles

Explanation:

45.00 x 2.50= 1.12.5 so 113 miles in 2.50 hours

5 0

3 years ago