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sleet_krkn [62]
3 years ago
14

When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels

the grains, and causes sugar to go flying out in all directions. If each of two sugar grains acquires a charge of 1.79 x 10-11 C at a separation of 3.45 x 10-5 m, with what force will they repel each other?
Physics
1 answer:
Afina-wow [57]3 years ago
3 0

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

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A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

7 0
3 years ago
Which wave characteristic has the unit hertz??
kotegsom [21]
Frequency has the unit hertz.
8 0
3 years ago
A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo  = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

-64 = -16.1*t'^2 + 48.5*t'

By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

3 0
3 years ago
Which landform represents the boundary between the land and an ocean or a lake?
Likurg_2 [28]
Coastline or seashore
3 0
3 years ago
Read 2 more answers
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
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