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3 years ago

5

Oque ocorre com a velocidade se diminuímos o tempo, mantendo a distância constante? E se aumentarmos o tempo? Que tipo de grande

za sao tempo e velocidade?

1 answer:

spin [16.1K]3 years ago

4 0

No se
Pero if I did I would help you

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After working in a fast-food restaurant for three years to pay for your college tuition, you vowed never to work in a restaurant

devlian [24]

Answer:

The correct answer is Cognitive.

Explanation:

The psychologist Leon Festinger proposed the theory of cognitive dissonance, which explains how people try to maintain their internal consistency. He suggested that individuals have a strong inner need that pushes them to ensure that their beliefs, attitudes and behavior are consistent with each other. When there is inconsistency between them, the conflict leads to a lack of harmony, something that people strive to avoid.

This theory has been widely studied in the field of psychology and can be defined as the discomfort, tension or anxiety that individuals experience when their beliefs or attitudes conflict with what they do. This displeasure can lead to an attempt to change behavior or defend their beliefs or attitudes (even reaching self-deception) to reduce the discomfort they produce.

4 0

3 years ago

At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this

Anastasy [175]

Answer:

Part A:

$\rm 2.8\times 10^3\ Volts.$

Part B:

$\rm 5.6\times 10^{-5}\ J.$

Explanation:

<u> Part A:</u>

Potential energy of charge at the given point, $\rm U=42\ \mu J=42\times 10^{-6}\ J.$

Charge, $\rm q=15\ nC = 15\times 10^{-9}\ C.$

The potential energy at a point due to a charge is defined as

$\rm U=qV$ .

<em>where</em>,

V = electric potential at that point.

Therefore,

$\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.$

<u>Part B:</u>

Now, if the charge at that point is replaced with $\rm q_1 = 20\ nC = 20\times 10^{-9}\ C.$ , then the electric potential energy at that point is given by

$\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.$

5 0

3 years ago

What is the electric force on a proton 2.5 fm from the surface of the nucleus? hint: treat the spherical nucleus as a point char

Whitepunk [10]

In this case, you need the formula below where:

F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges

$F = k((q1.q2)/r^{2})$

pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).

Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...

6 0

3 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago

Which of the following Resistors A, B, C or D, would use the least power? Make

Westkost [7]

i think the answer is D 10.0

4 0

3 years ago