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2 years ago

While buying a hot plate you notice the resistance of the hot

Physics

1 answer:

marissa [1.9K]2 years ago

7 0

Answer:

Current = 5.45amps

Power = 654 watts

Explanation:

E =120V

I =?

R = 22.02 Ohms

I= E/R

I= 120/22.02

I = 5.45AmPs

P = ? P= E x R

E = 120V P= 120x5.45

I = 5.45 AmPs P= 654W

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A passenger on a stopped bus notices that rain is falling vertically just outside the window. When the bus moves with constant v

nekit [7.7K]

Answer:

1)0.325

2) $6.17\ \rm m/s$

Explanation:

<u>Given:</u>

The angle that falling raindrops make with the vertical= $18^\circ$

Let $V_R$ be the velocity of the raindrops and $V_B$ be the velocity of the bus.

$\dfrac{V_R}{V_B}=\tan 18^\circ\\\dfrac{V_R}{V_B}=0.315\\$

2)Speed of the raindrops $=0.315\times 19$

$=6.17\ \rm m/s$

5 0

3 years ago

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

$T = \frac{2 \pi}{w}$

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

7 0

3 years ago

If the voltage output of a digital manifold absolute pressure (MAP)sensor is 2 volts, the approximate engine vacuum is _______ i

LenaWriter [7]

The correct answer to this question is 15

3 0

2 years ago

A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the

zysi [14]

As per the question, the mass of meteorite [ m]= 50 kg

The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

$Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2$

$=\frac{1}{2}50kg*[1000\ m/s]^2$

$=\frac{1}{2}50* 10^{6}\ J$

$=25*10^6\ J$

Here, J stands for Joule which is the S.I unit of energy.

$Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J$

4 0

3 years ago

Read 2 more answers

Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 494 Hz when air temperature i

elena-14-01-66 [18.8K]

Answer:

0.173 m.

Explanation:

The fundamental frequency of a closed pipe is given as

fc = v/4l .................. Equation 1

Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.

Making l the subject of the equation,

l = v/4fc ................ Equation 2

also

v = 331.5×0.6T ................. Equation 3

Where T = temperature in °C, T = 18.0 °c

Substitute into equation 3

v = 331.5+0.6(18)

v = 331.5+10.8

v = 342.3 m/s.

Also given: fc = 494 Hz,

Substitute into equation 2

l = 342.3/(4×494)

l = 342.3/1976

l =0.173 m.

Hence the length of the organ pipe = 0.173 m.

7 0

3 years ago