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2 years ago

Who ever gets it right I will give Brainly and 20 points.

Physics

2 answers:

svetlana [45]2 years ago

3 0

Answer:

Explanation:

Lera25 [3.4K]2 years ago

3 0

amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. ... Waves are generated by vibrating sources, their amplitude being proportional to the amplitude of the source.

Period refers to the time that it takes to do something. ... The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years.

.Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. ... An object with no movement at all has a zero speed.

wavelength, distance between corresponding points of two consecutive waves.

frequency refers to the number of waves that pass a fixed point in unit time. It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.

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Which of the following most directly shows how physics affects society?

Nataliya [291]

the answer is definitely A.

5 0

3 years 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:

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

6 0

2 years ago

What are all the<br> invertebrates with a<br> large foot

lyudmila [28]

Answer:

Explanation:

Bobbitt worm ( Eunice aphroditois). This segmented polychaete marine worm can attain lengths of 10 feet. It bristles...

Goliath beetle ( Goliathus species). African goliath beetle ( Goliathus giganteus ). Five species of goliath beetle...

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5 0

3 years ago

A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber

Kipish [7]

Answer:

The correct answers are

(a) It decreases to 1/3 L

(ii) is (c) It is constant

Explanation:

to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem

The given variables are as follows

Initial volume V1 = 1L

V2 = Unknown

Initial Temperature T1 = 300K

let us assume that the balloon is perfectly elastic

At 300K the balloon is filled and it stretches to maintain 1 atmosphere

at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

or V2 = (V1/T1)×T2 = $\frac{1L}{300K}$ × $100K$ = $\frac{1}{3}$ × L = L/3 hence the correct answer to (i) is 1/3L

8 0

3 years ago

Which two changes would decrease the electric force between two charged

ch4aika [34]

Answer:

it is b and e

Explanation:

<h2>if u look at the words twice you will notice that b and e are both saying the same meanings just in diff rent words way u need to look close on things like that and u will get passing grades </h2>

6 0

2 years ago