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3 years ago

A hair dryer draws a current of 10 A plugged into a 120 V outlet. What is the resistance of the hair dryer?

2 answers:

8 0

This problem can be solved by the formula used to find resistance. The formula is R=V/I which basically means divide the Voltage by the Current to find the Resistance in an object. Ohm's law.

Arada [10]3 years ago

8 0

Answer: The resistance of the hair dryer is $12\Omega$

Explanation:

Ohm's law is defined as the law which gives us the relationship between voltage, resistance and current.

Mathematically,

$V=I\times R$

where,

V = voltage of the hair dryer = 120 V

I = current of the hair dryer = 10 A

R = resistance of the hair dryer = ?

Putting values in above equation, we get:

$120=10\times R\\\\R=\frac{120}{10}=12\Omega$

Hence, the resistance of the hair dryer is $12\Omega$

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You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun us

ZanzabumX [31]

Answer:

The answer is

A. Pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

Explanation:

The question is incomplete, here is a complete question with full options

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun uses a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because:_________.

A. pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

B. viscous drag between the walls of the tip and the caulk causes the caulk to swirl around chaotically.

C. Newton’s third law requires most of the energy in the caulk to be used to push back on the plunger rather than moving it through the tip.

D. the high density of the caulk impedes its flow through the small opening.

Since the caulk is thick and the exit nozzle is small, the pressure needed to deliver the caulk will be very high as pressure is uniformly distributed at the plunger side at every part of the caulk, hence very high pressure is needed to deliver the caulk which is why the handle needed the very hard squeeze

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3 years ago

Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the

WINSTONCH [101]

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

$t_A=0.55s$

$t_B=0.45s$

(a) The distance from the kicker to each of the 2 spectators is given by:

$d_A=v \times t_A$

where,

v= speed of sound

$t_A$ =time taken for the sound waves to reach the ears

$d_A=343\times 0.55=188.65$ m

(b) $d_B=v \times t_B$

where,

v= speed of sound

$t_B$ =time taken for the sound waves to reach the ears

$d_B=343\times 0.45=154.35m$

(c)As the angle b/w slight lines from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

$D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m$

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3 years ago

As you approach a roundabout, you should __________________.

motikmotik

The answer should be B

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3 years ago

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A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

Diagonal Launch

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$