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garri49 [273]
2 years ago
14

Many of the rocks of the canyon wall appear as horizontal bands. What do the bands indicate about the way the rocks formed?

Physics
1 answer:
Yuki888 [10]2 years ago
4 0

Given what we know, we can confirm that the bands indicate that the canyon was carved due to erosion caused by moving water.

<h3>What do these bands tell us about the canyon?</h3>

The bands are the result of sediments being deposited as the canyon was being formed. These sediments are deposited by the running water that carves the canyon. This results in horizontal bands that can tell us much about the history of the canyon as well as the river that formed it.

Therefore, we can confirm that the bands indicate that the canyon was carved due to erosion caused by moving water.

To learn more about erosion visit:

brainly.com/question/2635044?referrer=searchResults

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An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.
motikmotik

Answer:

5 i think

Explanation:

4 0
3 years ago
A bag containing 0ºC ice is much more effective in absorbing energy than one containing the same amount of 0ºC water.
rosijanka [135]

Answer:

No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a phase change). For example, consider water dripping from icicles melting on a roof warmed by the Sun. Conversely, water freezes in an ice tray cooled by lower-temperature surroundings.

Explanation:

Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart such that, in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Similarly, energy is needed to vaporize a liquid, because molecules in a liquid interact with each other via attractive forces. There is no temperature change until a phase change is complete. The temperature of a cup of soda initially at 0ºC stays at 0ºC until all the ice has melted. Conversely, energy is released during freezing and condensation, usually in the form of thermal energy. Work is done by cohesive forces when molecules are brought together. The corresponding energy must be given off (dissipated) to allow them to stay together Figure 2.

The energy involved in a phase change depends on two major factors: the number and strength of bonds or force pairs. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The strength of forces depends on the type of molecules. The heat Q required to change the phase of a sample of mass m is given by

Q = mLf (melting/freezing,

Q = mLv (vaporization/condensation),

where the latent heat of fusion, Lf, and latent heat of vaporization, Lv, are material constants that are determined experimentally.

6 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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What is one benefit of lifelong physical activity?
kvasek [131]
The answer would be C
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How would you describe the behavior of particles in a solid?
Minchanka [31]
How would you describe the behavior of particles in a solid?
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