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Orlov [11]
2 years ago
14

You cheated on a test and you know it was the wrong thing to do according to the psychology theory what helped you know this

Physics
1 answer:
Alexandra [31]2 years ago
6 0

Answer:

You cheated on a test, and you know it was the wrong thing to do. According to the psychoanalytic theory, what helped you know this?

The id controlled your biological response and made you sweaty because you were scared of getting caught.

The superego acted as your moral conscience; you just knew that it wasn't right to cheat.

The ego made you feel guilty as a defense mechanism. both B and C

Explanation:

the  answer is both B,C

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The motion of a free falling body is an example of __________ motion​
swat32

Answer:

accelerated

Explanation:

The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.

4 0
2 years ago
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Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha
Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

v_{rms}=\sqrt{\frac{3RT}{M}}

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}

So, we have:

v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}

4 0
3 years ago
Explain why a person who has runny nose is unable to detect smells clearly.
Orlov [11]
If you have a runny nose then the mucus will block the path that smell goes through. So if you are trying to smell a flower then you will have a hard time because the mucus is in the way, or blocking the entrance path. Hope this helps:)

May I please have brainliest?
3 0
2 years ago
Read 2 more answers
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
2 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
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