When water flows in to the cracks of a rock freezes and breaks into smaller pieces is it a chemical or physical change?

Do geochemists need to have a knowledge of physical science? Explain your answer. please help:))

saul85 [17]

No they don’t lol I hate school

6 0

3 years ago

Which sentence provides evidence that heat travels from Earth’s mantle up to its crust?

olasank [31]

B. Earth’s outer surface is cooler than its interior layers.

Explanation:

The option given above is showing us that the temperature in the interior of the earth is higher than the temperature in the outer layer.
There is travel of heat from the inner core of the earth to the earth's crust. Due to the loss of heat when it reaches the outer layer, there arises a temperature difference.
The heat loss is due to the absorption of heat during its transfer. Hence, option B is the answer.

7 0

3 years ago

A major difference between radio waves, visible light, and gamma rays is the ____ of the photons, which results in the different

olga2289 [7]

The sentence can be completed as follows:
"A major difference between radio waves, visible light, and gamma rays is the energy of the photons, which results in the different photon frequencies and wavelengths."

In fact, gamma rays have greater energy than visible light and visible light has greater energy than radio waves. The energy E of a photon is related to its frequency, f, by
 $E=hf$
where h is the Planck constant. We see from this formula that the higher the frequency, the greater the energy. Instead, the wavelength is inversely proportional to the frequency:
 $\lambda= \frac{c}{f}$
where c is the speed of light. Since the frequency is directly proportional to the energy, this means that the wavelength is inversely proportional to the energy.

8 0

3 years ago

Read 2 more answers

Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt

Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

2d. $\rho gh$ again, what a surprise! $800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa$

3. Yet again, $\rho gh$ . $1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa$

4 0

2 years ago

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned

Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, $\omega _i$ = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, $\omega_f$ = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad$

Therefore, the angle through which the wheel turned is 947.7 rad.

5 0

3 years ago