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pickupchik [31]
3 years ago
12

When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

.09x10 ^-19 C 1.53x10 ^-19 C 6.39x10 ^-19 C 4.66x10 ^-19 C I used these measurements to find the charge on the electron
Physics
1 answer:
torisob [31]3 years ago
8 0

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

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During an experiment, Ellie records a measurement of 0.0034 m. How would
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Answer:

(A)   She needs to move the decimal point by 3 places

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3 years ago
The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,
Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

3 0
3 years ago
There isnt enough sanitidzer(alcohol)what solution to be replaced
Alex Ar [27]

For me: WASH OUR HANDS REGULARLY

3 0
2 years ago
A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i
andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) =  337.93

note : w = \sqrt{k/m}   = \sqrt{337.93} = 18.38 rad/sec

Period of oscillation =  2\pi  / w

                                  = 0.34 sec

8 0
3 years ago
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin
faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance L=0.450H=0.45\times 10^{-3}H

Current in the inductor i=1.90mA=1.90\times 10^{-3}A

Voltage v = 13 volt

Inductive reactance of the circuit X_l=\frac{v}{i}

X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm

We know that

X_l=\omega L=2\pi fL

2\times 3.14\times  f\times 0.45\times 10^{-3}=6842.10

f = 2421.127 kHz

6 0
4 years ago
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