Use the term chemical change in a complete sentence

How many states of matter do you think are in the photo? Describe at least two.

laila [671]

Answer:

Three, the liquid in the man's cup, the stove is solid, and the air around them is a gas

Explanation:

3 0

4 years ago

Definition of definite

KIM [24]

Google definition: a statement of the exact meaning of a word, especially in a dictionary.

the degree of distinctness in outline of an object, image, or sound, especially of an image in a photograph or on a screen.

My definition: The meaning or deintion of any word.

5 0

4 years ago

A train is traveling at a speed of 80km/h when the conductor applies the brakes. The train slows with a constant acceleration of

melisa1 [442]

Answer:

d = 493.72 m

Explanation:

Given that,

Initial velocity of the train, u = 80 km/h = 22.22 m/s

Acceleration of the train, a = -0.5 m/s² (negative as it slows down)

Finally brakes are applied, v = 0

We need to find the distance the train travels. Let the distance be d. Using third equation of kinematics to find it.

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(22.22)^2}{2\times (-0.5)}\\\\d=493.72\ m$

So, the required distance is equal to 493.72 m.

3 0

3 years ago

What can you calculate using the equation P equals W/t

Natalka [10]

WE CALCULATE POWER AND RATE OF DOING WORK IS CALLED POWER

8 0

3 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q