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dimulka [17.4K]
3 years ago
10

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

air and the depth of the water in the tank is 2.00 m. There is a hole with diameter 0.520 cm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole. Part A When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?
Physics
1 answer:
zysi [14]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

\Delta h = 2- 1.99963

\Delta h = 3.7*10^{-4}m=0.37mm

Therefore te change in the height of the water in the tank is 0.37mm

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An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x
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Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

v(x)=y'(x)

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

a(x)=v'(x)=y''(x)

We are given the function for y

y(x)=3.0+4.8x +6.4x^2

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

v(x)=y'(x)=4.8+12.8x

And the acceleration is a(x)=v'(x)=12.8

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3 years ago
A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

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How can humans increase the rate of endangered of extinct species
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I would think A. Reduce Environmental Impacts.

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Answer:

U₂ = 20 J

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Explanation:  

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H= 12 m

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h= 4 m

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U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

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From energy conservation

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By putting the values

60 - 20 = KE₂

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lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

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