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Neporo4naja [7]
3 years ago
7

The level of water in an olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2.00 meters deep) needs to be

lowered 6.50 cm. if water is pumped out at a rate of 4.20 liters per second, how long will it take to lower the water level 6.50 cm
Physics
1 answer:
Goshia [24]3 years ago
4 0
<span>We need to start by finding the surface area of the pool.
50 meters multiplied by 25 meters gives us 1250 square meters.
1250 square meters multiplied by .065 (6.5 cm in meters) gives us a volume of 81.25 cubic meters of water that needs to be pumped out of the pool.

There are 1000 liters in a cubic meter so this is 81250 liters. Divide by 4.2 to find the number of seconds required to pump out this much water and we get 19345.2 seconds. This equals approximately 5.37 hours.</span>
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A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot
sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0
3 years ago
A particle moves along the x axis from the origin. The magnitude of the position vector at time t is
Maurinko [17]

1) The average velocity is -2.1\cdot 10^5 m/s

2) The instantaneous velocity is 64t-260t^3

Explanation:

1)

The average velocity of an object is given by

v=\frac{d}{t}

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

x(t) = 32t^2 - 65t^4

where t is the time.

The position of the particle at time t = 6 sec is

x(6) = 32(6)^2 - 65(6)^4=-83,088 m

While the position at time t = 12 sec is

x(12)=32(12)^2-65(12)^4=-1,343,232 m

So, the displacement is

d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m

And therefore the average velocity is

v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

x(t) = 32t^2 - 65t^4

By differentiating with respect to t, we find the velocity vector:

v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

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6 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.
ziro4ka [17]

a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k = 4.3 is the dielectric constant

\epsilon_0 =8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=5.1\cdot 10^{-9} m^2 is the surface area

d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

+e=1.6\cdot 10^{-19} C

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

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6 0
3 years ago
The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a thi
Drupady [299]

The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a third quarter moon are the following

Explanation:

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waning gibbous phase

  • The waning gibbous phase occurs between the full moon and third quarter phases. The last quarter moon (or a half moon) is when half of the lit portion of the Moon is visible after the waning gibbous phase.

Time takes by the moon to go through all the phases

about 29.5 days

  • It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).
  • At 3rd quarter, the moon rises at midnight and sets at noon. Then we see only a crescent. At new, the moon rises at sunrise and sets at sunset, and we don't see any of the illuminated side!
6 0
3 years ago
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Select the three correct parts that make up the cell theory.
Lana71 [14]

Answer:

Options A, D and E....make up cell theory

8 0
3 years ago
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