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3 years ago

11

When an object slides across the floor, it is slowed down by friction and the floor surface gets warmer. what energy conversion

process takes place in this situation?

1 answer:

OverLord2011 [107]3 years ago

6 0

The movement of the object is considered to be kinetic energy while the object getting warmer indicates that there is thermal (heat) energy formed.

Based on this, as the object slides across the floor, friction slows down this motion and the object becomes warmer as kinetic energy is converted into thermal energy.

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A resistor with r = 340 ω and an inductor are connected in series across an ac source that has voltage amplitude 490 v. The rate

Arada [10]

The value of impedance Z of the circuit, when the rate at which electrical energy is dissipated in the resistor is 316 w, is 508 ohms.

<h3>What is impedance Z of the circuit?</h3>

The impedance Z of the circuit is the ratio of voltage amplitude to the maximum current.

$Z=\dfrac{V}{I}$

Here, <em>V </em>is voltage amplitude and<em> I</em> maximum current.

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 490V. The rate at which electrical energy is dissipated in the resistor is 316 W.

The rate at which electrical energy is dissipated in the resistor is the product of the resistance and the square of current. Thus,

$316=340\times I^2\\I=\sqrt{\dfrac{316}{340}}\\I=0.964\rm\; A$

The impedance Z of the circuit is,

$Z=\dfrac{V}{I}\\Z=\dfrac{490}{0.964}\\Z=508\rm\; ohm$

Thus, the value of impedance Z of the circuit, when the rate at which electrical energy is dissipated in the resistor is 316 w, is 508 ohms.

Learn more about the impedance Z of the circuit here:

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5 0

1 year ago

Can I improve the design of my simple machine? How?

blsea [12.9K]

Answer:

12345

Explanation:

yan na po answer ko hehehe

5 0

3 years ago

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

2 years ago

Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin

erastova [34]

Answer

(C).

When there is an angle between the two directions, the cosine of the angle must be considered.

Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is $W=F\times d$ .

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the $sin(90)=0$ meaning that there is no work done in the direction perpendicular to the motion.

The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is $F\times d\times \cos(\theta).$

4 0

3 years ago

Read 2 more answers

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago