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Norma-Jean [14]
3 years ago
14

a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t

emperature remains constant (T1=T2), what is its volume just before it reaches the surface where the pressure is 1.0 atmosphere?
Physics
1 answer:
lys-0071 [83]3 years ago
8 0

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

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Write down one fact about each of the following crust, mantle, outer, core, & inner core
Oksanka [162]

Answer:

Hope this help you!!

Explanation:

Crust : The crust is the thinnest layer of the Earth. It has an average thickness of about 18 miles below land, and around 6 miles below the oceans. The crust is the layer that makes up the Earth's surface and it lies on top of a harder layer, called the mantle.

Mantle : The mantle is the mostly-solid bulk of Earth's interior. The mantle lies between Earth's dense, super-heated core and its thin outer layer, the crust. The mantle is about 1,802 miles thick, and makes up a whopping 84% of Earth's total volume

Outer Core : The outer core is the third layer of the Earth. It is the only liquid layer, and is mainly made up of the metals iron and nickel, as well as small amounts of other substances. The outer core is responsible for Earth's magnetic field. As Earth spins on its axis, the iron inside the liquid outer core moves around.

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5 0
3 years ago
the electron are accelrated to a speed of 2.40*10^7 in 1.8*10^-9, the force experinced by an electron ?
dalvyx [7]
Force = mass × acceleration

To find acceleration, we can divide the speed by the time it took:

acceleration = 2.40×10^7 / 1.8×10^-9

acceleration = 1.33×10^16

the mass is equal to the mass of an electron

force = (9.11×10^-31)(1.33×10^16)

force = 1.21×10^-14 N
3 0
3 years ago
Whiteout friction you could not, write, drive or fly and airplane. Why not?
aleksklad [387]

friction is the resistance that one surface or object encounters when moving over another. Due to gravity pulling everything down things need to friction in order to move

i hope this helps :/


7 0
3 years ago
A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
3 years ago
A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on M
Pie

Answer:

Gravitational field strength =weight/mass

Explanation:

14.8N/4.0kg

3.7N/kg

3 0
2 years ago
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