In the car experiment you completed what part of the scientific method would the following statement be classified as?

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

3 years ago

At an air show a jet flies directly toward the stands at a speed of 1200 km/h, emitting a frequency of 3500 Hz, on a day when th

kenny6666 [7]

Answer:

(a): The frequency received by the observers is f'= 138,062.28 Hz.

(b): When the plane flies directly away from them, they receive a frequency of f'= 1772.46 Hz.

Explanation:

Vf= 333.33 m/s

Vo= 0 m/s

V= 342 m/s

f= 3500 Hz

(a) f' = f * ( V / (V - Vf) )

f'= 138062.28 Hz

(b) f'= f* ( V / (V - (- Vf) )

f'= 1772.46 Hz

4 0

3 years ago

A string is wound around a uniform disc of radius 0.68 m and mass 3.7 kg. The disc is released from rest with the string vertica

Zolol [24]

Answer:

3.962

Explanation:

mass = 1.2

gravity g = 9.81

velocity = ?

mgh = 1/2mv² + 1/2(1/2)mv²

= 9.81 x 1.2 = 3/4v²

11.772 x 4 = 3/4v²

47.088 = 3v²

divide through both sides by 3

15.696 = v²

take the square root of both sides

√15.696 = √v²

= 3.6962 = v

the speed is therefore 3.692

8 0

3 years ago

What is the electric potential 12 cm away from a charge of –5.2 × 10^–6C?

MAXImum [283]

The equation for electric potential of a point charge is:

V=(k*q)/r where k=9*10^11 N m²/C², q is the charge and r is the distance from the charge to the point in which we are calculating the potential.

q=-5.2*10^-6 C
r=12 cm = 0.12 m

Now we plug in the numbers and get:

V=-3.9*10^5 Volts, so the correct answer is the second one.

8 0

3 years ago

If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th

Diano4ka-milaya [45]

Answer:

$\frac{K.E_r}{K.E}=2.875$