A Review | Constants Periodic Table

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

What do you mean by kinetic energy and potential energy ?

Verizon [17]

Kinetic Energy of an object is the measure of the work an object can do by virtue of its motions..

Formula - KE = 1/2 mv^2

Where KE is the kinetic energy, m is the body’s mass, and v is the body’s velocity.

Potential energy is the stored energy in any object or system by virtue of its position or arrangement of parts..

Formula - W = mgh

Where,

. m is the mass in kilograms

. g is the acceleration due to gravity

. h is the height in meters

Hope it helpz~ uh..

4 0

2 years ago

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

AnnZ [28]

Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

5 0

3 years ago

Read 2 more answers

Please help!

andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0

2 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago