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3 years ago

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With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

tmosphere. If an electron at that point has a charge of -1.6 × 10-19 coulombs, what is the magnitude of the force on the electron as a result of the electric field?
A.
2.0 × 10-6 newtons
B.
3.2 × 10-17 newtons
C.
5.8 × 105 newtons
D.
6.4 × 10-3 newtons
E.
1.7 × 104 newtons

1 answer:

Aleks04 [339]3 years ago

5 0

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, $E=2\times 10^2\ N/C$

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

$F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N$

So, the required force on the electron is equal to $3.2\times 10^{-17}\ N$ .

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Answer:

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Explanation:

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6 0

3 years ago

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it

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Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, $m=0.02 kg$

Kinetic energy imparted, $K=1200 J$

Length of rifle barrel, $d=1 m$

(a)

Let the speed of bullet when it leaves the barrel is $v$ .

Kinetic energy, $K=\frac{1}{2} mv^{2}$

$v=\sqrt{\frac{2K}{m} }$

$=\sqrt{\frac{2\times1200}{0.02} }$

$=346.4m/s$

Initial speed of bullet, $u=0$

The average speed in the barrel, $v_a_v_g=\frac{u+v}{2}$

$=\frac{0+346.4}{2} \\=173.2 m/s$

Time taken by bullet to cross the barrel, $t=\frac{d}{v}$

$=\frac{1}{173.2}\\ =0.00577 second$

Power, $P_a_v_g=\frac{W}{t}$

$=\frac{1200}{0.00577} \\=207.97kW$

(b)

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Maximum height, $H_m=\frac{v^2\sin^2\theta}{2g} \\$

Range, $R=\frac{v^2\sin2\theta}{g}$

given that, $H_m=R$

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5 0

3 years ago

Calculate the moment of inertia for each scenario: (a) An 80.0 kg skater is approximated as a cylinder with a 0.140 m radius. (b

Zina [86]

Answer:

a) the moment of inertia is 0.784 Kg*m²

b) the moment of inertia is with arms extended is 1.187 Kg*m²

c) the angular velocity in scenario (b) is 4.45 rad/s

Explanation:

The moment of inertia is calculated as

I = ∫ r² dm

since

I = Ix + Iy

and since the cylinder rotates around the y-axis then Iy=0 and

I = Ix = ∫ x² dm

if we assume the cylinder has constant density then

m = ρ * V = ρ * π R²*L = ρ * π x²*L

therefore

dm = 2ρπL* x dx

and

I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 = mR² /2

therefore

I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²

b) since the arms can be seen as a thin rod

m = ρ * V = ρ * π R²*L = ρ * π R²*x

dm =ρ * π R² dx

I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)

= ρ * π R²*2*L³/24 = mL²/12

therefore

I skater 2 = I1 + I skater = mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²

c) from angular momentum conservation

I s2 * ω s2 = I s1 * ω s1

thus

ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s

4 0

3 years ago