Ask question

Ask question

All categories

3 years ago

12

Stu wanted to calculate the resistance of a light bulb connected to a 4.0-V battery, with a resulting current of 0.5 A. He used

the formula R = VI and obtained an answer of 2 . Was Stu’s answer correct? How do you know?

2 answers:

Alexxandr [17]3 years ago

7 0

Answer:

<u><em>Sample Response</em></u>: Stu's answer was incorrect. According to Ohm's law, R = V/I. This is the equation that should be used to obtain the correct answer, which is 8 ohms.

Explanation:

Straight copied and pasted from edge. 2020

Margarita [4]3 years ago

6 0

Answer:

The answer is not correct.

Explanation:

Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.

$V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\$

In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.

$R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]$

That is the reason that the result found by Stu is not correct.

You might be interested in

A bird flew16 km west in 5 hours, then flew 20 km east in 6 hours. what was the birdâs velocity?

Mademuasel [1]

Solve for Speed: 16+20=36 km in 11 hrs
Answer: speed = 3.27273 kilometers per hour

To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.

<span>speed = distance/time</span>

4 0

4 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

Which of the following statements about energy in systems is true?

Elena-2011 [213]

Option number three is correct energy can be transformed and moved and released but it can't be destroyed and doesn't disappear.

8 0

3 years ago

A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

$v=at+v_o$ where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

$v=at+v_o$

$(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})$

$1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t$

$\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}$

$10\text{ [s]}=t$

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is $a=\dfrac{100 [\frac{m}{s}]}{1[s]}$ , the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0

2 years ago

Explain why the same side of the moon is always facing Earth.

Ad libitum [116K]

Answer:

The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.

7 0

3 years ago