2 questions, brainliest for both correct! please: only answer if you are sure of the answer!! :)

kykrilka [37]

Answer:

Both physical and chemical changes

Explanation:

From the definition of the law of conservation of mass, total mass of all the substances taking part in a chemical reaction is conserved and the number of atoms of each element in the reaction does not change when a new product is formed. Both physical and chemical changes follow the law since when the system is closed to all transfers of matter and energy, the mass of the system must remain constant over time, irrespective of the state.

3 0

2 years ago

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ignore this ignore this ignore this ignore this ignore this ignore this ignore this ignore this ignore this ignore this ignore t

nekit [7.7K]

Okay I will ignore this

4 0

3 years ago

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True or False: The height from which cliff divers jump affects the velocity at which they will fall. Velocity is speed in a cert

Juli2301 [7.4K]

Answer:

the diver's speed is independent of the launch height.

Explanation:

For this exercise we must use Newton's second law

fr - W = m a

the friction force has the general form

fr = b v

Let's analyze this equation to find out what happens with the speed of the distant club.

When jumping, the initial speed is zero, so the friction force is zero and has an acceleration equal to the acceleration of gravity, as the speed increases the friction force increases decreasing the acceleration until it becomes zero, when it arrives at this value the velocity it has is called terminal velocity and this velocity remains fixed in relation to the trajectory.

fr = W

v = cte

The distance or time in which this equilibrium is reached is relatively fast, so the diver's speed is independent of the launch height.

8 0

2 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.33 m, and is initially unc

Vesnalui [34]

A) $E=\frac{4.50\cdot 10^{10}}{r^2} V/m$

r < a

We can find the magnitude of the electric field by using Gauss theorem. Taking a Gaussian spherical surface of radius r centered in the centre of the sphere, the electric flux through the surface of the sphere is equal to the ratio between the charge contained in the sphere and the vacuum permittivity:

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

For r < a, the charge contained in the gaussian sphere is the point charge:

$q=5.00 C$

So the electric field in this region is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m$

B) E = 0

a < r < b

The region a < r < b is the region between the inner and the outer surface of the shell. We have to keep in mind that the presence of the single point charge +q = 5.00 C at the center of the sphere induces an opposite charge -q on the inner surface (r=a), and a charge of +q at the outer surface (r=b).

Using again Gauss theorem

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

this time we have that the gaussian sphere contains both the single point charge +q and the negative charge -q induced at r=a, so the net charge contained in the sphere is

q' = +q - q = 0

And so, the electric field in this region is zero.

C) $E=\frac{4.50\cdot 10^{10}}{r^2} V/m$

r > b

Here we are outside of the sphere. Using Gauss theorem again

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

this time we have that the gaussian sphere contains the single point charge +q, the negative charge -q induced at r=a, and the positive charge +q induced at r=b, so the net charge contained in the sphere is

q' = +q - q +q = q

And so the electric field is identical to the one inside the sphere:

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m$

D) -12.29 C/m^2

We said that the charge induced at the inner surface r=a is

-q = -5.00 C

The induced charge density is

$\sigma = \frac{-q}{A}$