Which type of box will best protect a fragile object by minimizing the force on the object

Temperature is a measure of work. True False

Veseljchak [2.6K]

The answer is false.

5 0

3 years ago

Look at picture and please help if i get one more wrong i can’t get to 80%

NemiM [27]

A, B and C are all correct. They can be made from igneous, sedimentary or other metamorphic rock. I hope I’m not too late to answering!

3 0

2 years ago

An object is dropped from a height of 25 meters.at what velocity will it hit the ground?

Dmitry [639]

Answer:

C. 22 m/s

Explanation:

Ignoring air friction....all of the Potential energy will be converted to Kinetic Energy

mgh = 1/2 m ^v^2

gh = 1/2 v^2

sqrt (2 gh) = v = ~ 22.1 m/s

4 0

2 years ago

What is a risky behaviour

Sindrei [870]

Explanation:

Doing homework is risky behaviour broo

7 0

3 years ago

Read 2 more answers

A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell

LUCKY_DIMON [66]

Answer:

(a) We should apply Gauss' Law to each region to find the corresponding electric field. In order to use Gauss' Law, we should draw an imaginary spherical surface inside each region. The electric field passes through this imaginary surface is equal to the charge inside in this surface.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The integral in the left-hand side is equal to the area of the imaginary surface.

(i) r < a:

E = 0, since there is no charge inside the imaginary surface.

(ii) a < r < b:

E = 0, since the electric field is always zero within conductors. Another reason for this is that the inner shell has no charge, and the enclosed charge in the Gauss' Law is zero.

(iii) b < r < c:

The imaginary surface is drawn in this region and it contains +2q charge, which is the net charge of the smaller shell. Therefore, Gauss' Law gives

$E4\pi r^2 = \frac{2q}{\epsilon_0}\\E = \frac{q}{2\pi \epsilon_0 r^2}$

(iv) c < r < d:

The enclosed charge inside the imaginary surface in this region is the sum of the total charge of the smaller shell and the inner surface of the larger shell, which are +2q - 2q = 0. Equivalently, the electric field within a conductor is zero always.

E = 0.

(v) r > d:

The imaginary surface is now drawn to the outermost region and contains the total charge of both spheres, which is +6q. Therefore, Gauss' Law reads

$E4\pi r^2 = \frac{6q}{\epsilon_0}\\E = \frac{3q}{2\pi\epsilon_0 r^2}$

(b)

Since both shells are conductors, the charges are distributed around the inner and outer surfaces, so no charge within the conductor.

(i) Since the innermost region (r > a) has no charge, the inner surface of the smaller shell is also neutral.

(ii) Since the total charge of the smaller shell is +2q, and the inner surface is neutral, the outer surface has a charge of +2q.

(iii) The +2q charge on the outer surface of the smaller shell attracts an equal amount of opposite charges on the inner surface of the larger shell. So, the inner surface of the larger shell has a charge of -2q.

(iv) Since the total charge of the larger shell is +4q, and the inner surface contains -2q, then the outer surface has a charge of +6q.

8 0

4 years ago