You drop a penny from the top of the Salesforce building in San Francisco. If the

Which type of acceleration would be best if you were merging onto the interstate?

kondaur [170]

Answer:

If merging onto an interstate, it is best to merge at the speed of the moving traffic.

Explanation:

As a driver, you cannot always count on the other drivers to see you wanting to merge onto the interstate, and even if they see you, they might not be willing to stop just so you merge, therefore it is best advised to look for a gap in the traffic and merge at the traffic speed.

4 0

3 years ago

What happens when a substance undergoes a physical change

zysi [14]

Physical changes are those which do not affect the chemical properties of a substance. For example, when a paper is cut, the two pieces obtained are still paper. The chemical properties have not changed. Only, physically when one piece is cut to two. A chemical change occurs when the chemical properties change and a new substance with different chemical properties is obtained. For example, when the paper is burnt, it turns to ashes. Hence, this is a chemical change.

When a substance undergoes a physical change, the physical properties of the substance changes like the state (solid to liquid during melting), shape, size, volume etc. Many physical changes are reversible whereas, in general many chemical changes are irreversible.

5 0

3 years ago

Read 2 more answers

A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

chubhunter [2.5K]

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

a)

For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance