I need the correct answer for this question please

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Soloha48 [4]

Kinetic energy-flashlight, and guitar
potential-coal,and composed spring

7 0

3 years ago

if a tank filled with water contains a block and the height of the water above point A within the block is 0.6meter, what's the

Fittoniya [83]

The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:

Pressure = density x g x height

In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:

P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa

6 0

3 years ago

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta

Alex

Answer:

The depth will be equal to 6141.96 m

Explanation:

pressure on the submarine $P_{sea}$ = 62 MPa = 62 x 10^6 Pa

we also know that $P_{sea}$ = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

$P_{sea}$ = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine $P_{g}$ = 101 kPa = 101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure $P_{sea}$ , we have

62 x 10^6 = 10094.49h

depth h = 6141.96 m

7 0

3 years ago

Apply Pascal's law to a ketchup packet. Explain what will happen if you tear a small

vivado [14]

Answer:

its will explode because it's like a soda with baking soda added together and you shake it the small hole can make a huge explosion

6 0

3 years ago

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I need the correct answer for this question please​

I need the correct answer for this question please