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Ivanshal [37]
3 years ago
15

If you have an effusive personality, then it __________ of you. motor oil will __________ across the gravel if it spills out of

the can.
Physics
1 answer:
erica [24]3 years ago
6 0
The first answer in the blank is pours out of you, the second answer in the space provided is diffuse. It is because a person who has an effusive personality will likely pour out his or her emotions while the motor oil will likely diffuse if this has been spilled out of the can.
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Write with all the steps and formulas and drawing if needed.​
lianna [129]

<u>The answer is not contained detail explanation, just a solution and the required values. </u>

All the details are in the pictures, the answers are marked with orange colour.

Note,

in the task no 20.:

m_A - the \ mass \ of \ A; \ m_B-the \ mass \ of \ B \ balls.\\V_A \ and \ V_B-the \ velocities \ of \ the \ A&B \ balls \ before \ collision.\\V'_A \ and \ V'_B-the \ velocities \ of \ the \ A&B \ balls \ after \ collision.

V - the velocity of the pair of the balls after collision.

in the task no 21:

m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.

7 0
3 years ago
What kind of energy change occurs when a battery is operating a remote control toy?
solniwko [45]

Answer:

Explanation:

The correct answer is option C.

When the battery is operating a remote control toy the energy is converted from potential energy to the kinetic energy.

A battery stores electrical potential from the chemical reaction.

When the circuit is connected to the potential energy of the battery helps in the movement of the toy.

The energy produced by the movement of the control toy is kinetic energy.

Hence, we can say that Potential energy is changed to kinetic energy

7 0
3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
What are the units of radiation?
Ksenya-84 [330]

The radiation dose absorbed by a person is measured using the conventional unit rad or the SI unit gray. The biological risk of exposure to radiation is measured using the conventional unit rem or the SI unit sievert.

I hope this helps!

3 0
2 years ago
Predict the deformation or elongation of a spring that has a constant of elasticity of 400 N/m when a force of 75 N is applied i
morpeh [17]

Answer:

Explanation:

Give that,

Spring constant (k)=40N/m

Force applied =75N

Since the force is applied to the right, we don't know if it is compressing or stretching the spring

So let assume it compress

Using hooke's law

F=-ke

e=-F/k

Then, e=-75/40

e=-1.875m

The deformation is 1.875m.

Let assume it stretch

Using hooke's law

-F=-ke

e=F/k

Then, e=75/40

e=1.875m

The elongation is 1.875m

3 0
3 years ago
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