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3 years ago

A 500-watt vacuum cleaner is plugged into a 120-volt outlet and used for 30 minutes. How much current runs through the vacuum?

Physics

1 answer:

drek231 [11]3 years ago

3 0

P = IV,

P = Power in Watts, I = Current in Ampere, V = Voltage in Volts.

Current, I = P/ V = 500 w/ 120v ≈ 4.17 A

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Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time

Ber [7]

Answer:

The value is $\Delta t = 4.0 *10^{-7} \ s$

Explanation:

From the question we are told that

The current is $\Delta I = 0.100 \ A$

The inductor is $L = 2.0mH = 2.0*10^{-3} \ H$

The voltage induced is $\epsilon = 500 V$

Generally the emf induced is mathematically represented as

$\epsilon = L * \frac{\Delta I }{\Delta t }$

Here $\Delta t$ is the time taken

=> $\Delta t = \frac{L * \Delta I }{\epsilon }$

=> $\Delta t = \frac{2*10^{-3} * 0.100 }{500 }$

=> $\Delta t = 4.0 *10^{-7} \ s$

8 0

3 years ago

Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2

posledela

A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.

8 0

3 years ago

Read 2 more answers

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

3 years ago

A 0.0600-kilogram ball traveling at 60.0 meters

Vikki [24]

The momentum of ball is given by:

$\Delta Q=mv \\ \Delta Q=6*10^{-2}*60 \\ \Delta Q=3.6kg*m/s$

Since both have the same momentum, we have:

$\Delta Q=MV \\ 3.6=10^{-2}V \\ \boxed {V=360m/s}$

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

7 0

3 years ago

A piston releases 18 J of heat into its surroundings while expanding from 0.0002 m^3 to 0.0006 m^3 at a constant pressure of 1.0

Elodia [21]

Answer:

value of heat is 18 J

2. step by step

formular w=p(volume1-volume2)

w= 1.0×10^5(0.0006-0.0004)

w= 40 J

8 0

3 years ago