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3 years ago

The elementary charge equals 1.6 x 10^-9 C. a. True b. False

Physics

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

False

Explanation:

The magnitude of electric charge carried by an electron as well as proton is called elementary charge. Both the electron and proton have the same charge but the difference is that the electrons are negatively charged species and the protons are positively charged.

The elementary charge is equal to $1.6\times 10^{-19}\ C$ . Hence, the given statement "The elementary charge equals $1.6\times 10^{-9}\ C$ " is false.

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1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of

Masja [62]

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^{2} +2^{2} }$ = $\sqrt{13}$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^{2} + 3^{2} }$ = $\sqrt{25}$ = 5cm = 0.05m

electric potential V = $\frac{kq}{r}$

change in potential ΔV = $V_{1} - V_{2}$

ΔV = $\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }$ , where $q_{1} = q_{2}=$ 2.00μC

ΔV = $2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $(\frac{1}{0.036} - \frac{1}{0.05} )$

ΔV= 2.789×10⁵

$\frac{1}{2}mv^{2}$ = ΔV × q₃

$\frac{1}{2}$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

3 0

3 years ago

What is the definition of dynamics in physics

grin007 [14]

A part of mechanics dealing with the movement of bodies under the activity of power.

5 0

3 years ago

A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinde

Romashka [77]

Answer:

(a)10.5 rad/s2

(b) 20.9 rev

Explanation:

As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law

T = mg = 53*9.81 = 519.93 N

Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is

To = TR = 519.93 * 0.36 = 187.17 Nm

This solid cylinder would have a moment of inertia around it's rotating axis of:

$I = \frac{mR^2}{2} = \frac{275 * 0.36^2}{2} = 17.82kgm^2$

(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder

$\alpha = \frac{To}{I} = \frac{187.17}{17.82} = 10.5 rad/s^2$

(b) With such constant angular acceleration, the angle it would make after 5s is

$\theta = \frac{\alphat^2}{2} = \frac{10.5*5^2}{2} = 131.3 rad$

Since each revolution equals to $2\pi rad$ of angle, we can calculate the number of revolution it makes

$\frac{\theta}{2\pi} = \frac{131.3}{6.28} \approx 20.9 rev$

(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad

$\theta R = 131.3*0.36 = 47.27 m$