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3 years ago

The linear expansion of a material depends on which of the following?

Physics

1 answer:

nordsb [41]3 years ago

7 0

Answer:

1. The change in temperature of the material.

2. The length of the material.

3. The type of material.

Explanation:

The term linear expansion is defined as the increase in length of any conductor when there is a change in temperature. The new length of any wire or rod is given by :

$L=L_o(1+\alpha \Delta T)$

$L_o$ is the initial length of the rod

$\alpha$ is the coefficient of linear expansion

$\Delta T$ is the change in temperature

It is clear that the linear expansion of a material depends on :

1. The change in temperature of the material.

2. The length of the material.

3. The type of material.

So, all options are correct.

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A block of wood 3.0 cm on each side has a mass of 27g. what is the density of this block?

schepotkina [342]

The block of wood is 3cm on each side so it is a cube. The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3. density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3

8 0

3 years ago

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

$W=1.04.10^{-20}$

To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

8 0

3 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

3 years ago

How will the amount of power change if less work is done in more time?

yKpoI14uk [10]

The amount of power change if less work is done in more time"then the amount of power will decrease".

<u>Option: B</u>

<u>Explanation:</u>

The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt

$Power = \frac{Work}{Time}$

Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.

This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but when the work is done in a shorter period of time more power is required.

7 0

3 years ago

Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.

ludmilkaskok [199]

The correct answer is: Angular velocity = $1.208 * 10^{-9}$ rad/s

Explanation:
The angular velocity is given as:
ω = $\frac{2\pi}{T}$ --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = $\frac{2\pi}{5203440000} = 1.208 * 10^{-9}$ rad/s

7 0

3 years ago