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3 years ago

An air bubble is at a depth of 3 m below

Physics

1 answer:

lisabon 2012 [21]3 years ago

4 0

Answer:

incorrect its 987 for exact

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Through the process of blank

maksim [4K]

Answer:

Subduction, Trench, Mantle

Explanation:

8 0

3 years ago

Read 2 more answers

Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g=9.8m/s^{2}$ the acceleration due gravity and $h=500m$ the height of the object.

Knowing this, let's begin with the calculaations:

For m=3kg

$U_{p}=(3kg)(9.8m/s^{2})(500m)$

$U_{p}=14700J=14.7kJ$

For m=6kg

$U_{p}=(6kg)(9.8m/s^{2})(500m)$

$U_{p}=29400J=29.4kJ$

For m=9kg

$U_{p}=(9kg)(9.8m/s^{2})(500m)$

$U_{p}=44100J=44.1kJ$

6 0

3 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago

System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas

Anna [14]

Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

$F=G \frac{m_1 m_2}{r^2}$

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

$F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}$

system B has masses m and 2m separated by a distance 2r:

$F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}$

system C has masses 2m and 3m separated by a distance 2r:

$F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}$

system D has masses 4m and 5m separated by a distance 3r:

$F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}$

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

5 0

3 years ago

AM radio signals have frequencies between 550 kHz and 1600 kHz (kilohertz) and travel with a speed of 3.0Ã108m/s. What are the w

Westkost [7]

Answer:

The wavelength of these signals is as follow:

Wavelength of 550 kHz is 545.45 m
Wavelength of 1600 kHz is 187.5 m

Explanation:

Given that:

Frequency = 550 kHz & 1600 kHz

Velocity = 3.0 x 10⁸ m/s

As we know that frequency is expressed by the following equation:

Frequency = Velocity / Wavelength ---- (1)

For 550 kHz:

The equation can be rearranged as

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (550 x 1000 Hz)

Wavelength = 545.45 m

For 1600 kHz:

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (1600 x 1000 Hz)

Wavelength = 187.5 m

5 0

3 years ago