All categories

4 years ago

A pure substance can be a

Physics

1 answer:

valina [46]4 years ago

4 0

Hey there! A pure substance can be a element.

Hope this helps! :)

You might be interested in

Two point charges of -7uC and 4uC are a distance of 20

aivan3 [116]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy $\mathrm{EPE}$ .

$\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}$ ,

where

The coulomb's constant $k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2}$ ,
$q_1$ and $q_2$ are the sizes of the two charges, and
$r$ is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

$q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C$ ;
$q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C$ ;

Initial separation: $\rm 20\; cm = 0.20\; cm$ ;
Final separation: $\rm 90\; cm = 0.90\; cm$ .

Apply Coulomb's law:

Initial potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}$ .

Final potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}$ .

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

$\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}$ .

8 0

3 years ago

Read 2 more answers

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

$E_{k1}+W=E_{k2}\\$

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

$\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]$

We know that work is defined as the product of force by distance.

$W=F*d$

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

$f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]$

Now solving the equation for the work.

$d=W/F\\d = 50/19.62\\d = 2.54[m]$

4 0

3 years ago

The time between a lightning flash and the following thunderclap may be used to estimate, in kilometers, how far away a storm is

Kruka [31]

Given Information:

Elapsed time = t = 6 seconds

Required Information:

Distance = d = ?

Answer:

Distance = d = 2.058 km

Explanation:

We know that the speed of sound in the air is given by

v = 343 m/s

The relation between distance, speed and time is given by

distance = speed*time

substituting the given values yields,

distance = 343*6

distance = 2058 m

There are 1000 meters in 1 km so

d = 2058/1000

d = 2.058 km

Therefore, the storm is about 2.058 km away when elapse time between the lightning and the thunderclap is 6 seconds.

3 0

3 years ago

How do you calculate the electric force by object a exerted on object a (itself)?

max2010maxim [7]

I don't think an object can exert a force on itself.

Try it: Get up on a skateboard, and see if you can do anything to yourself that makes you start moving ... without touching anything else.

It'll be easy to tell if you succeed. If you actually do exert an unbalanced force on yourself, then you'll begin to accelerate.

5 0

3 years ago

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star syste

Slav-nsk [51]

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

$1Ly =9.4605284*10^{15}m \rightarrow 'Ly'$ means Light Year