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3 years ago

Which term denotes that the mere presence of other people in your environment affects your behavior

Physics

2 answers:

Thepotemich [5.8K]3 years ago

8 0

The term is Social facilitation.

Social facilitation can be defined as an improvement in performance caused by the mere presence of others, it can be further distinguished into two types:

co-action effects which is a phenomenon where increased task performance is caused by the mere presence of others doing the same task.

audience effect where performance is increased in the presence of a passive spectator/audience

Nadya [2.5K]3 years ago

7 0

The term is social facilitation.

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What is the only possible value of ml for an electron in an s orbital?

Archy [21]

Answer:

zero

Explanation:

$m_l$ is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

$l$ : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

This quantum number is related to the type (or shape) of the orbital:

For s orbitals $l=0$

For p orbitals $l=1$

For d orbitals $l=2$

For f orbitals $l=3$

In this case, it is an s orbital, so we have $l=0$ .

$m_l$ , the third quantum number can have integer values ${from-l}$ to ${+l}$

Since, for the s orbitals $l=0$ , the only possible value for ${m_l}$ is zero.

4 0

4 years ago

As part of his work for NASA, Dr. Murdock was asked to find out what percentage of people in the continental united States saw H

mario62 [17]

Answer:

Please, in the Explanation section you will find the explanation of the answer.

Explanation:

The exercise shows the continental United States and 3 cities used in the study carried out by Murdock. It can be said that the sample taken is part of the objective. There are several inconsistencies in Murdock's argument: the first has to do with the fact that the sample that was taken cannot represent the entire American population. A much larger, scientifically calculated sample would be required. The second is that their study did not take into account small cities or people living in the interior of the United States.

5 0

3 years ago

FREE POINTS!

san4es73 [151]

Answer:

Both conduction and convection require matter to transfer heat. ... Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas. Cooler liquid or gas then takes the place of the warmer areas which have risen higher. This results in a continuous circulation pattern.

Explanation:

HOPE THIS HELPS!!!

4 0

3 years ago

Read 2 more answers

A 60 kg man jumps down from a 0.8 m table. What is the speed when he

Rom4ik [11]

Answer: Speed = 4 m/s

Explanation:

The parameters given are

Mass M = 60 kg

Height h = 0.8 m

Acceleration due to gravity g= 10 m/s2

Before the man jumps, he will be experiencing potential energy at the top of the table.

P.E = mgh

Substitute all the parameters into the formula

P.E = 60 × 9.8 × 0.8

P.E = 470.4 J

As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.

When hitting the ground,

K.E = P.E

Where K.E = 1/2mv^2

Substitute m and 470.4 into the formula

470.4 = 1/2 × 60 × V^2

V^2 = 470.4/30

V^2 = 15.68

V = square root (15.68)

V = 3.959 m/s

Therefore, the speed of the man when hitting the ground is approximately 4 m/s

4 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

3 years ago