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2 years ago

9

Flo is driving her sports car at 30 m/s when a ball rolls out into the street in front of her. Flo slams on the brakes and comes

to a stop in 3.0 seconds. What was the acceleration of flos car

1 answer:

butalik [34]2 years ago

6 0

Answer:

the acceleration of Flo's car is 10 m/s

Explanation:

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A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

7 0

2 years ago

Does every light source emit only one type of light?

blsea [12.9K]

Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.

3 0

2 years ago

What does friction mean?

WARRIOR [948]

Rubbing on something

5 0

2 years ago

If you were to drive an average velocity of 65 mi/hr to Traverse City, which is

Amanda [17]

Find the number of hours by dividing the distance by mph. The number of hours will be to the left of the decimal point:

250 miles / 65 mph
= 3.846153846
= 3 hours

2) Find the number of minutes by multiplying what is remaining from step 1 by 60 minutes. The minutes will be to the left of the decimal point:

0.846153846 x 60
= 50.76923076
= 50 minutes

3) Find the number of seconds by multiplying what is remaining from step 2 by 60 seconds. The seconds will be to the left of the decimal point:

0.76923076 x 60
= 46.1538456
= 46 seconds

So 3 hours 50 mins and 46 seconds

3 0

2 years ago

The membrane that surrounds a certain type of living cell has a surface area of 7.1 x 10-9 m2 and a thickness of 1.5 x 10-8 m. A

hoa [83]

Answer:

Q = +1.4 pC

Explanation:

The capacitance of any capacitor, by definition, is as follows:
$C = \frac{Q}{V} (1)$
Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
$C = \frac{\epsilon_{r}*\epsilon_{0} * A}{d} (2)$
Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
Replacing by the givens in (2) we can find the value of the capacitance C, as follows:

$C = \frac{\ 4.4*8.85e-12C2/N*m2*7.1e-9m2}{1.5e-8m} = 18.4e-12 F = 18.4 pF$

Replacing the values of C and V in (1), we can solve for Q, as follows:

$Q = C*V = 18.4e-12 F* 75.9e-3 V = 1.4e-12 C = +1.4 pC$

As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.

7 0

3 years ago