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3 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]3 years ago

7 0

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

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If six moles of hydrogen chloride (HCl) react with plenty of aluminum, how many moles of aluminum chloride (AlCl3) will the reac

AlexFokin [52]

Answer:

Two moles of aluminum chloride $(AlCl_3)$ are produced when six miles of hydrogen Chloride $(HCl)$ react with plenty of aluminum

Explanation:

6 Moles of $HCl$ will only react with 2 moles of $Al$ irrespective of the number of moles of each compound present. The reaction wiil take place in this ratio only. The products produced will be 2 moles of $AlCl_3$ and 3 moles of $H_2$ this ratio will also be constant.

So, six moles of hydrogen chloride $(HCl)$ will react with plenty of aluminum to produce many 2 moles of aluminum chloride $(AlCl_3)$ .

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3 years ago

Two compounds are standing at the same temperature. Compound "A" is evaporating more slowly than compound "B." According to the

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3 years ago

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A 0.30 kg yo-yo consists of two solid disks of radius 5.10 cm joined together by a massless rod of radius 1.0 cm and a string wr

docker41 [41]

Answer:

0.70046 m/s²

2.732862 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of yo-yo = 0.3 kg

R = Radius of rolling disk = 5.1 cm

r = Radius of rod = 1 cm

For a rolling disks the acceleration is given by

$a=\frac{g}{1+\frac{R^2}{2r}}\\\Rightarrow a=\frac{9.81}{1+\frac{0.051^2}{2\times 0.01^2}}\\\Rightarrow a=0.70046\ m/s^2$

The acceleration of the yo-yo is 0.70046 m/s²

The tension in the string will be

$T=m(g-a)\\\Rightarrow T=0.3\times (9.81-0.70046)\\\Rightarrow T=2.732862\ N$

The tension in the string is 2.732862 N

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3 years ago

a motorcycle is trying to leap across the canyon by driving horizontally off a cliff 38 m/s. Ignoring air resistance, find the s

vaieri [72.5K]

Answer:

46.15m/s

Explanation:

We are given that

Initial speed,u=38 m/s

Height of cliff,h=70 m

We have to find the speed with which the cycle strikes the ground at a height 35 m on the other side.

h'=35 m

Using conservation law of energy

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2+mgh'$

$\frac{1}{2}u^2+gh=\frac{1}{2}v^2+gh'$

Where $g=9.8m/s^2$

Substitute the values

$\frac{1}{2}(38)^2+9.8\times 70=\frac{1}{2}v^2+9.8\times 35$

$\frac{1}{2}v^2=\frac{1}{2}(38)^2+9.8\times 70-9.8\times 35=1065$

$v=\sqrt{2\times 1065}$

$v=46.15m/s$

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3 years ago

A plane flying horizontally at an altitude of 1 mi and a speed of 490 mi/h passes directly over a radar station. Find the rate a

Len [333]

Answer: 438.3 mph

Explanation:

Let z be the distance from the plane to the station. You should draw a right triangle

for the diagram with z on the hypotenuse, 1 on the vertical side, and x on the horizontal

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Find the attached file for the solution.

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3 years ago

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