According to Newton's first law, a moving odject acted on by a net force of zero-

A block of mass m rests on a frictionless, horizontal surface. A cord attached to the block passes over a pulley whose radius is

ra1l [238]

Answer:

Explanation:

Let the tension in the cord be T₁ and T₂ .

for motion of block placed on horizontal table

T₁ = m a , a is acceleration of the whole system .

for motion of hanging bucket of mass m

mg - T₂ = ma

adding the two equation

mg + T₁- T₂ = 2ma

for rotational motion of the pulley

torque = moment of inertia x angular acceleration

(T₂ - T₁) R = I x α , I is moment of inertia of pulley , α is angular acceleration .

(mg - 2ma ) R = I x α

(mg - 2ma ) R = I x a / R

(mg - 2ma ) R² = I x a

mgR² = 2ma R² + I x a

a = mgR² / (2m R² + I )

Since body moves by distance d in time T

d = 1/2 a T²

a = 2d / T²

mgR² / (2m R² + I ) = 2d / T²

mgR²T² = 2d x (2m R² + I )

mgR²T² - 4dm R² = 2dI

m R² ( gT² - 4d ) = 2dI

I = m R² ( gT² - 4d ) ] / 2d .

3 0

4 years ago

An electron is projected with an initial speed vi = 4.60 × 105 m/s directly toward a very distant proton that is at rest. Becaus

frutty [35]

Answer:

$2.99\times 10^{-19}\ m$

Explanation:

<u>Given:</u>

$u$ = initial velocity of the electron = $4.60\times 10^5\ m/s$

$v$ = final velocity of the electron = $3u$

$x$ = initial position of the electron from the proton = very distant = $\infty$

<u>Assume:</u>

$m$ = mass of an electron = $9.1\times10^{-31}\ kg$

$e$ = magnitude of charge on an electron = $1.6\times10^{-19}\ C$

$p$ = magnitude of charge on an proton = $1.6\times10^{-19}\ C$

$k$ = Boltzmann constant = $9\times 10^9\ Nm^2/C^2$

$y$ = final position of the electron from the proton

$\Delta K$ = change in kinetic energy of the electron

$W$ = work done by the electrostatic force

$F$ = electrostatic force

$r$ = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

$W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}$

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

$\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq$

Hence, the electron is at a distance of $2.99\times 10^{-10}\ m$ when the electron instantaneously has speed of three times the initial speed.

8 0

4 years ago

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

Elena L [17]

Answer:

It would take 8.22037 hrs away. Wouldn't it?

Explanation:

Because

4.11016

15

= 8.22037

5 0

4 years ago

Help for physics

12345 [234]

Answer:

A. 5 m/s

Explanation:

From the graph, for the first 2 seconds, the graph is a straight line meaning that the slope is a constant.

Average speed of an object is the rate of change of position. Here, the position of the object changes from 0 m to 10 m for a time interval of 2 seconds.

The change in position ( $\Delta x$ ) and time interval ( $\Delta t$ ) are given as:

$\Delta x=10-0=10\ m\\\Delta t=2-0=2\ s$

Therefore, the average speed ( $s_{avg}$ ) is given as the ratio of the total change in position and the time interval for the change.

$s_{avg}=\frac{\Delta x}{\Delta t}=\frac{10-0}{2-0}=\frac{10}{2}=5\ m/s$

Hence, the average speed is 5 m/s.

8 0

3 years ago

Read 2 more answers

A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 31.7◦ and the angle of refraction

Andrei [34K]

Media 1 and 2 are air and liquid. By Snell's law;
n1/n2 = Sin ∅2/Sin ∅1

Then,
n2 = (n1* Sin Ф1)/Sin ∅2 = (1*Sin 31.7)/Sin 21.3 = 1.4466.

When the light travels in the opposite direction and at critical angle, media 1 and 2 are liquid and air respectively while ∅2 = 90°

Therefore,
n1/n2 = Sin 90 / Sin ∅c => ∅c = Sin ^-1[n2*Sin 90]/n1 = Sin ^-1[1*Sin 90]/1.4466 = 43.73°

The critical angle (∅c) is 43.73°.

8 0

3 years ago