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2 years ago

Write the formula of Lever, Pulleys, wheel and axle and inclined plane.

Physics

1 answer:

maxonik [38]2 years ago

6 0

Answer:

Lever => $d_{e} = d_{r}$

Pulley => G = M x n (gravitational acceleration)

Wheel and axle => M.A = Radius of the wheel/radius of the axle = R/r

Inclined plane => It can be divided into two components: Fi = Fg * sinθ - parallel to inclined plane. Fn = Fg * cosθ - perpendicular one.

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The electric field strength in the space between two closely spaced parallel disks is 1.0 10^5 N/C. This field is the result of

alex41 [277]

To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.

The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:

$C = \frac{\epsilon_0 A}{d}$

Where,

$\epsilon_0$ = Permeability constant

A = Cross-sectional Area

d = Diameter

Potential difference between the two disks,

V = Ed

Where,

E = Electric field

d = diameter

Q = Charge on the disk equal to $\rightarrow Q=ne=(3.9*10^9)(1.6*10^{-19})= 6.24*10^{-10}C$

Through the value found and the expression given for capacitance and potential, we can define the electric charge as

$Q = CV$

$Q = \frac{\epsilon A}{d}(Ed)$

$Q = \epsilon_0 AE$

$Q = \epsilon_0 \pi(\frac{d}{2})^2E$

$Q = \frac{\epsilon \pi d^2E}{4}$

Re-arranging the equation to find the diameter of the disks, the equation will be:

$d = \sqrt{\frac{4D}{\epsilon_0 \pi E}}$

Replacing,

$d = \sqrt{\frac{4(6.24*10^{-10})}{(8.85*10^{-12})\pi(1*10^{5})}}$

$d = 0.0299m$

Therefore the diameter of the disks is 0.03m

8 0

3 years ago

A bird flies overhead from where you stand at an altitude of 270.0ĵ m and at a velocity horizontal to the ground of 14.0î m/is.

Varvara68 [4.7K]

Answer:

L = 8694 Kg.m²/s

Explanation:

r = 270 ĵ m

v = 14 î m/s

m = 2.3 kg

θ = 90º

L = ?

We can apply the equation

L = m*v*r*Sin θ

L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s

8 0

3 years ago

Insulators have very high .

Vlad1618 [11]

Answer:

Resistance to electrical currents

Explanation:

Conductors have low resistance to electrical currents, and are used to "conduct" the flow of electricity.

Insulators have very high resistance and are used to protect us from the flow of electricity.

5 0

2 years ago

Generally the length of metre is equal all over the world why?

erastova [34]

Answer:

It is an SI unit

Explanation:

The metre is defined as the length of the path travelled by light in a vacuum in 1299 792 458 of a second. The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole

8 0

3 years ago

Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t

Tamiku [17]

Answer:

Approximately $9.62$ .

Explanation:

$y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0]$ .

$y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)]$ .

Notice that sine waves $y_1$ and $y_2$ share the same frequency and wavelength. The only distinction between these two waves is the $(-0.250)$ in $y_2\!$ .

Therefore, the sum $(y_1 + y_2)$ would still be a sine wave. The amplitude of $(y_1 + y_2)\!$ could be found without using calculus.

Consider the sum-of-angle identity for sine:

$\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)$ .

Compare the expression $\sin(a + b)$ to $y_2$ . Let $a = (4.35\, x - 1270)$ and $b = (-0.250)$ . Apply the sum-of-angle identity of sine to rewrite $y_2\!$ .

$\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Therefore, the sum $(y_1 + y_2)$ would become:

$\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Consider: would it be possible to find $m$ and $c$ that satisfy the following hypothetical equation?

$\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Simplify this hypothetical equation:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}$ .

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}$ .

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real $x$ and $t$ , the following should be satisfied:

$\displaystyle 1 + \cos(-0.250) = m\, \cos(c)$ , and

$\displaystyle \sin(-0.250) = m\, \sin(c)$ .

Consider the Pythagorean identity. For any real number $a$ :

${\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2$ .

Make use of the Pythagorean identity to solve this system of equations for $m$ . Square both sides of both equations:

$\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2$ .

$\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2$ .

Take the sum of these two equations.

Left-hand side:

$\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}$ .

Right-hand side:

$\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}$ .

Therefore:

$m^2 = 2 + 2\, \cos(-0.250)$ .

$m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98$ .

Substitute $m = \sqrt{2 + 2\, \cos(-0.250)}$ back to the system to find $c$ . However, notice that the exact value of $c\!$ isn't required for finding the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ .

(Side note: one possible value of $c$ is $\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125$ radians.)

As long as $\! c$ is a real number, the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ would be equal to the absolute value of $(4.85\, m)$ .

Therefore, the amplitude of $(y_1 + y_2)$ would be:

$\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}$ .

8 0

3 years ago

Write the formula of Lever, Pulleys, wheel and axle and inclined plane.​

Write the formula of Lever, Pulleys, wheel and axle and inclined plane.