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3 years ago

Explain how energy can escape from the water by convection,

1 answer:

6 0

As the molecules at the surface of a liquid absorb heat, they begin to move around more quickly. This gives them the energy to break the bonds that connect them to other water molecules. When the molecules are moving fast enough, they are able to "escape." They leave the surface of the liquid as gas molecules.

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A fire truck emits an 880Hz siren. As the truck approaches an observer on the sidewalk, he perceives the pitch to be 950Hz. Appr

Ludmilka [50]

Answer:

The pitch that he hears after the truck passes and is moving away is 819.6 Hz.

Explanation:

The pitch that he hears after the truck passes and is moving away can be calculated using the following equation:

$f = f_{0}*\frac{v_{s} \pm v_{o}}{v_{s} \pm v_{f}}$

Where:

$f$ : is the perceived frequency

$f_{0}$ : is the emitted frequency

$v_{s}$ : is the speed of sound = 340 m/s

$v_{o}$ : is the speed of the observer = 0 (he is not moving)

$v_{f}$ : is the speed of the fire truck

First, we need to find the speed of the fire truck. When it approaches the observer we have:

$f = f_{0}*\frac{v_{s} + v_{o}}{v_{s} - v_{f}}$

$950 Hz = 880 Hz*\frac{340 m/s}{340 m/s - v_{f}}$

$v_{f} = 340 m/s - \frac{880 Hz*340 m/s}{950 Hz}$

$v_{f} = 25.05 m/s$

Hence, the speed of the fire truck is 25.05 m/s.

Now, we can calculate the pitch that the observer hears after the truck passes:

$f = f_{0}*\frac{v_{s} - v_{o}}{v_{s} + v_{f}}$

$f = 880 Hz*\frac{340 m/s}{340 m/s + 25.05 m/s}$

$f = 819.6 Hz$

Therefore, the pitch that he hears after the truck passes and is moving away is 819.6 Hz.

I hope it helps you!

7 0

3 years ago

The speed for a car that went a distance of 125 miles in 2 hours time is ______. Question 2 options: 250 meters/sec 62.5 miles/h

Goryan [66]

Answer:

62.5 miles per hour

Explanation:

Speed = Distance travelled / Time taken

Speed = 125/2 = 62.5

You derive the units of the speed...by using the speed formula....,

Speed = Distance/Time

Speed = miles/hour

Hence the units for the speed = miles/hour

5 0

3 years ago

A motorist drives north for 36.0 minutes at 96.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km

Over [174]

Answer:

a) $d=187.6km$

b) $v=61.5km/h$

Explanation:

First we convert our minutes to hours so we work always in the same units.

$36min=36min(\frac{1h}{60min})=0.6h$

$15min=15min(\frac{1h}{60min})=0.25h$

Where we used the fact that 1 hour are 60 min, thus the multiplying factor is equal to 1 (not altering the time, just changing the units).

a) On the first part the motorist travels a distance $d_1=v_1t_1=(96km/h)(0.6h)=57.6km$ , and on the second part he travels $d_2=130km$ .

The total displacement is $d=d_1+d_2=57.6km+130km=187.6km$

b) The average velocity is the relation between the total displacement and the time taken to cover it. Our total time is t=0.6h+0.25h+2.2h=3.05h, thus we have:

$v=\frac{187.6km}{3.05h}=61.5km/h$

8 0

3 years ago

A car is traveling at a constant speed along the road ABCDE shown in the drawing. Section Ab and DE are straight. Rank the accel

vladimir2022 [97]

Answer:

AB = DE <CD <BC

Explanation:

This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.

Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.

Let's apply these reasoning to our problem.

AB Path

this path is straight and as they indicate that the constant speed the acceleration is zero

DE path

This path is straight and since the velocity is constant the zero steps

BC path

This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression

$a_{c}$ = v² / r

where r is the radius of the curve and the direction of acceleration is towards the center of the curve

CD path

This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less

$a_{BC}$ > $a_{CD}$