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2 years ago

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Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c

ompletely filled with milk (no air space), and it accelerates at 4 m/s . If the minimum pressure in the tanker is 100 kPa, determine the maximum pressure difference and the location of the maximum pressure. Answer: 66.7 kPa

1 answer:

jeka57 [31]2 years ago

4 0

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

$=-17850 - 30018.6$

$=-47868.6 \ Pa$

$=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

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Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

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Gravitational acceleration is approx 9.8 m/s
Time is 7s

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A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m

Mariana [72]

Answer:

(a) 13.795 m/s.

(b) -3140.28 N.

Explanation:

(a) Using newton's equation of motion,

v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

Hence the velocity of the driver when he hits the water = 13.795 m/s.

(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

Substitute into equation 3

0² = 13.795²+2(2a)

0 = 190.30203 + 4a

-4a = 190.30203

a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

Substitute into equation 3

F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

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