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2 years ago

Calculate the mass of -1.5C of electrons

1 answer:

algol132 years ago

8 0

The electron is a type of low-mass, very negatively charged with a particle. As such, it can easily be deflected by passing close to other electrons or the positive nucleus of an atom. m = mass of an electron in kg = 9.10938356 × 10-31 kilograms. e = magnitude of the charge of an electron in coulombs = 1.602 x 10-19 coulombs. Hope this helps!

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A small but bright light is at the bottom of a pool 2.2 m deep. How wide is the circle of light that exits the surface of the

mixas84 [53]

Answer: 1.65m

Explanation:

Refractive index in terms of the depth of liquid is the ratio of the real depth to the apparent depth of the liquid i.e Refractive index =Real depth/apparent depth

Refractive index of water given = 1.33

Real depth is the measure of how deep is the liquid while apparent depth is the depth at the surface of the liquid.

Real depth = 2.2m

Apparent depth =?

Applying the formula above

Apparent depth =Real depth/refractive index

= 2.2/1.33

= 1.65m

Therefore, the circle of light that exits the surface of the water when that light shines in the middle of the night is 1.65m wide

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2 years ago

If you need a wet paper towel out on the counter is a vintage I was change of state cuts is this

coldgirl [10]

As long as green Barghouti on painted charge. (Answer matches logic of the question.)

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2 years ago

A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struc

Ganezh [65]

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x) = 332.3

x = 2.7 Hz

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

3 0

2 years ago

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$