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3 years ago

Which best describes what occurs when an object takes in a wave as the wave hits it?

Physics

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

Explanation:

ABSORPTION

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Help needed

MakcuM [25]

Answer: Atmosphere and geosphere.

Geo-sphere is the solid part of the earth. Hydrosphere is the water part. The living things on the earth make up the biosphere. The gases, water vapors etc make up the atmosphere. When a volcanic eruption adds carbon dioxide to the air, the carbon-dioxide is being added from geosphere to the atmosphere. Hence, there is interaction between atmosphere and geosphere.

6 0

4 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

7 0

3 years ago

If the temp and pressure stay the same, 3 different pieces of the same material will have the same__________

valentina_108 [34]

The answer is Volume.

3 0

3 years ago

If the frequency of the 13C signal of TMS is 201.16 MHz, the two 13C signals of acetic acid at 179.0 and 20.0 ppm are separated

Lelu [443]

The difference in frequency of the two signals is $1.33 \times 10^{10} \ kHz$ .

The given parameters;

<em>frequency of the 13 C signal = 201.16 MHz</em>

The energy of the 13 C signal located at 20 ppm is calculated as follows;

$E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 = \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J$

The energy of the 13 C signal located at 179 ppm is calculated as follows;

$E_2 = \frac{hc}{\lambda} \\\\E_2 = \frac{(6.626\times 10^{-34})\times (3\times 10^{8})}{179 \times 10^{-6} } \\\\E_2 = 1.11 \times 10^{-21} \ J$

The difference in frequency of the two signals is calculated as follows;

$E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz$

Thus, the difference in frequency of the two signals is $1.33 \times 10^{10} \ kHz$ .

Learn more here:brainly.com/question/14016376

4 0

3 years ago

Which of these would have a volume equal to about 2 cm³?

postnew [5]

Answer could be2 grains of rice

5 0

3 years ago