Why does a spherometer have three legs

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

solniwko [45]

Answer:

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

$a_t = R\alpha$

2) Centripetal acceleration

$a_c = \omega^2 R$

here we know that

$\omega = \alpha t$

$a_c = (\alpha t)^2 R$

now we know that net linear acceleration is given as

$a = \sqrt{a_c^2 + a_t^2}$

so we have

$a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}$

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

4 0

2 years ago

Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver

zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0

2 years ago

What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7

Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration $n= 2.7\times10^{18}\ cm^{-3}$

Electron mobility $\mu=1000 cm^2/Vs$

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

$\sigma=n\times q\times \mu$

$\rho=\dfrac{1}{\sigma}$

Put the value into the formula

$\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}$

$\rho=2\ \mu \Omega m$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

$R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}$

$R=0.4\ \Omega$

We need to calculate the voltage

Using formula of voltage

$V= IR$

Put the value into the formula

$V=0.17\times0.4$

$V=0.068\ V$

Hence, The voltage across a semiconductor bar is 0.068 V.

6 0

2 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$