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3 years ago

How much current will a 500 W vacuum cleaner draw if it has a resistance of 30Ω?

Physics

1 answer:

Strike441 [17]3 years ago

3 0

Answer:

Electric current, I = 4.08 A

Explanation:

We have,

Power of vacuum cleaner is 500 W

Resistance of vacuum cleaner is 30 ohms

It is required to find the current flowing through the vacuum cleaner. It can be calculated using the definition of power delivered. Its formula is given by :

$P=I^2R$

I is electric current

$I=\sqrt{\dfrac{P}{R}} \\\\I=\sqrt{\dfrac{500}{30}} \\\\I=4.08\ A$

So, the electric current flowing through the vacuum cleaner is 4.08 A.

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A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

4 0

3 years ago

The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func

Kitty [74]

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = $570 \times \dfrac{2\pi}{60}$

= 59.69 rad/s

1600 rpm = = $570 \times \dfrac{2\pi}{60}$

= 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt

167.6 = 59.7 + α x 7980

α x 7980 = 107.9

α = 0.0135 rad/s²

8 0

3 years ago

A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, $\Delta f = 4\ beats/s$

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

$\Delta f = f - f_{i}$

$f_{i} = f \pm \Delta f$

when

$f_{i} = f + \Delta f = 523 + 4 = 527 Hz$

when

$f_{i} = f - \Delta f = 523 - 4 = 519 Hz$

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0

3 years ago

Read 2 more answers

Interference is an example of which aspect of electromagnetic radiation

Phantasy [73]

Answer:

We experience interference while listening to the radio. A radio station works by sending and receiving radio waves. Since the radio waves are being interfered with other waves which must have a wave nature.

The interference is the net result of two individual waves. It can be constructive or destructive interference and is the property of waves and not particles.

This interference is an example of electromagnetic radiation. Thus we experience wave behavior of electromagnetic radiation in our daily communications.

7 0

3 years ago

Read 2 more answers

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago