Which of the following is a valuable part of the scientific method?

Can someone please help me out with this quiz will give brainiest and thanks to people

Virty [35]

Answer:

Energy transferred = 28.8 Joules.

1. Energy transferred = 144 Joules.

2. The unit of potential difference, volts can also be described as Joules per Coulombs.

3. Current, I = 6.945 Amperes.

Explanation:

Part A.

Given the following data;

Current, I = 1.2A

Time, t = 2 minutes

Potential difference, V = 12 volts.

To find the energy transfered;

Energy transferred = charge moved * potential difference

E = Q * V

Substituting into the equation, we have;

Energy transferred = (1.2 * 2) * 12

Energy transferred = 2.4 * 12

Energy transferred = 28.8 Joules.

Part B.

1. Given the following data;

Charge, Q = 24C

Potential difference = 6V

To find the energy transferred;

E = Q * V

Substituting into the equation, we have;

E = 24 * 6

E = 144 Joules.

2. Since we know that, Energy transferred = charge moved * potential difference

$Potential \; difference = \frac {Energy \; transferred}{Charged \; moved}$

The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.

Therefore, the unit of potential difference becomes Joules per Coulomb.

3. Given the following data;

Potential difference = 18V

Energy transferred = 500J

Time, t = 4 minutes.

To find the current;

E = Q * V

Substituting into the equation, we have;

500 = Q*18

Q = 500/18

Q = 27.78C

But, Charge moved (Q) = current (I) * time (t)

Current, I = Q/t

Substituting into the equation, we have;

Current, I = 27.78/4

Current, I = 6.945 Amperes..

3 0

3 years ago

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

mihalych1998 [28]

We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2

8 0

3 years ago

A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?

Kruka [31]

4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4

4 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

3 years ago

A garrafa térmica (também conhecida como "vaso de Dewar") é um dispositivo extremamente útil para conservar, no seu interior, co

igor_vitrenko [27]

Answer:

A opção A está correta.

O sistema formado pela garrafa térmica e a água perde 400 cal de calor para o meio ambiente.

Option A is correct.

The system formed by the thermos and the water loses 400 cal of heat to the environment.

Explanation:

Quando a temperatura de um sistema reduz, fica claro que o sistema perdeu calor ou energia térmica. Como a temperatura é um dos indicadores mais claros disso, esta conclusão é hermética e correta.

Mas, para saber a quantidade de calor perdida para o meio ambiente, agora fazemos alguns cálculos de energia térmica.

Transferência de calor de ou para o sistema de água e garrafa térmica = c × ΔT

c = capacidade térmica do sistema de água e garrafa térmica = 80 cal /°C

ΔT = Alteração da temperatura do sistema de água e garrafa térmica = (temperatura final) - (temperatura inicial) = 55 - 60 = -5°C

Calor transferido = 80 × -5 = -400 cal.

O sinal de menos mostra que o calor é transferido para fora do sistema, ou seja, o calor é perdido no sistema.

Espero que isto ajude!!!

English Translation

The thermos (also known as "Dewar vase") is an extremely useful device to conserve bodies (essentially liquid) at high temperatures, minimizing energy exchanges with the environment, which is generally colder. A thermos contains water at 60 o C. The thermos + water set has a thermal capacity of C = 80 cal / o C. The system is placed on a table and, after a considerable period of time, its temperature decreases to 55 o C. In this case, it is concluded that the system formed by the thermos and the water inside:

a) lost 400 cal. B) gained 404cal. C) lost 4 850 cal. D) gained 4 850 cal. E) did not exchange heat with the external environment.

Solution

When a system's temperature reduces, it is clear to conclude that the system has lost heat or thermal energy. Since temperature is one of clearest indicators of this, this conclusion is airtight and correct.

But, to know the amount of heat lost to the environment, we now do some thermal energy calculations.

Heat transferrred from or to the water and thermos system = c × ΔT

c = heat capacity of the water and thermos system = 80 cal/°C

ΔT = Change in temperature of the water and thermos system = (final temperature) - (initial temperature)

= 55 - 60 = -5°C

Heat transferred = 80 × -5 = -400 cal.

The minus sign shows that the heat is transferred out of the system, that is, the heat is lost from the system.

Hope this Helps!!!

7 0

3 years ago