All categories

1 year ago

Emotional elevation only occurs in animal kin groups. please select the best answer from the choices provided t f

2 answers:

8 0

Emotional elevation can occur in individuals or in groups, therefore, it is false that emotional elevation only occurs in animal kin groups.

<h3>What is emotional elevation?</h3>

Emotional elevation is the process whereby the moods of individuals are elevated due to their interactions with others or due to other factors.

Emotional elevation can occurs in individual animals or those on animal kin groups.

Therefore, it is false that emotional elevation only occurs in animal kin groups.

Learn more about emotional elevation at: brainly.com/question/8099018

#SPJ4

Pie1 year ago

3 0

Answer:

Explanation:

You might be interested in

What is the equivalent resistance of a circuit that contains four 75.0 resistors connected in parallel with a 100.0 V battery?

zhannawk [14.2K]

To get the total resistance in a parallel circuit, you need to remember that unlike in a series, you do not just merely add the resistances. You need to get the reciprocal first of each resistance and add them together.

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}... +\frac{1}{R_{n}}$

After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.

$\frac{1}{R_{T}} = \frac{1}{75}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}$

Add up the numerator and copy the denominator:

$\frac{1}{R_{T}} = \frac{4}{75}$

Then get the reciprocal to get the total resistance:

$R _{T} = \frac{75}{4} = 18.75$

The answer to your question then is A. 18.8.

5 0

3 years ago

Read 2 more answers

an electron, a proton and a deuteron move in a magnetic field with same momentum perpendicularly. the ratio of the radii of thei

wel

If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:

1: √2 : 1

<h3>How is the ratio of the perpendicular parts obtained?</h3>

To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.

The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:

r = √m/e : √2m/e : √4m/2e

When resolved, the resulting ratios will be:

1: √2 : 1

Learn more about the radii of their circular paths here:

brainly.com/question/16816166

#SPJ4

6 0

1 year ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .