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3 years ago

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The theory that attitudes will be formed and changed according to one’s actual behaviors is the __________ theory. A. learning B

. dissonance C. self-perception D. balance

2 answers:

Ivenika [448]3 years ago

7 0

Answer:

A is the answer

Explanation:

natulia [17]3 years ago

6 0

Learning. A. Actions and behaviors change when people learn

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You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

a = 9.8m/s2

Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

7 0

3 years ago

A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg

ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

$Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3$

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

$p=m/V\\\\ p=m/(L*A)\\p*A=m/L$

Now the equation for speed of traverse wave is calculated through:

$\sqrt \frac{F*L}{m}\\$

= $\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}$

Substituting the values

$\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3} \\$

=22.49 m/s

4 0

3 years ago

A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At

NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

$B=\frac{1}{2} \mu \epsilon_0 r$

Where,

$\mu =$ Permeability constant

$\epsilon_0 =$ Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

$B \propto r$

Then the distance relationship would be given by

$\frac{r}{R} = \frac{B}{B_{max}}$

$r =\frac{B}{B_{max}} R$

$r = \frac{0.5B_{max}}{B_{max}}R$

$r = 0.5R$

On the outside, however, it is defined by

$B = \frac{\mu_0 i_d}{2\pi r}$

Here the magnetic field is inversely proportional to the distance, that is

$B \not\propto r$

Then,

$\frac{r}{R} = \frac{B_{max}{B}}$

$r = \frac{B_{max}{B}}R$

$r = \frac{B_{max}{0.5B_{max}}}R$

$r = 2R$

7 0

3 years ago

elena55 [62]

Answer:

I took the test, and the correct answer is not Hydrosphere and Geosphere, but instead, its actually Hydrosphere and Atmosphere

Explanation:

4 0

3 years ago

A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.

ruslelena [56]

Answer:

C. $W = 115.12\,Btu$

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process ( $\eta$ ), no unit, is:

$\eta = \left(1-\frac{T_{L}}{T_{H}} \right)$ (1)

Where:

$T_{L}$ - Temperature of the cold reservoir, measured in Rankine.

$T_{H}$ - Temperature of the hot reservoir, measured in Rankine.

If we know that $T_{H} = 1809.67\,R$ and $T_{L} = 584.67\,R$ , then the energy efficiency of the ideal thermal process is:

$\eta = 0.678$

By First Law of Thermodynamics, we calculate the work output:

$W = Q_{H}-Q_{L}$

$W = \frac{W}{\eta} -Q_{L}$ (By definition of efficiency)

$Q_{L} = \frac{W}{\eta}-W$

$Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W$ (2)

Where:

$Q_{H}$ - Heat received by the engine, measured in Btu.

$Q_{L}$ - Heat rejected by the engine, measured in Btu.

$W$ - Work output, measured in Btu.

If we know that $\eta = 0.678$ and $Q_{L} = 55\,Btu$ , then the work output of the Carnot engine is:

$W = \frac{Q_{L}}{\frac{1}{\eta}-1 }$

$W = 115.807\,Btu$

The work output of the Carnot engine is 115.807 Btu. (Answer: C)

5 0

3 years ago