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wariber [46]
2 years ago
7

. A milk truck carries milk with density 64.6 lbyft3 in a horizontal cylindrical tank with diameter 6 ft. (a) Find the force exe

rted by the milk on one end of the tank
Physics
1 answer:
Goryan [66]2 years ago
8 0

Answer:

Explanation:

one end of tank will be circular in shape . Area of circle A

= π r² , r is radius of the circle

= 3.14 x 3²

A = 28.26 ft³

To calculate force  on the circular area , we first find pressure at the center of the circle which is at depth equal to r

pressure at the center = h d g ' here h = depth = r   , d = density of milk

pressure = 3 x 64.6 x 32 poundal / ft²

= 6201.6 poundal / ft²

total force on circular face = pressure at the center x area of circle

= 6201.6 x 28.26

= 175257.21 poundal .

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A ball rolls up a ramp with a velocity of 8.0 m/s. How high up the ramp does it travel?
Flura [38]

The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>
  • Initial velocity (u) = 8 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 8² – (2 × 9.8 × h)

0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

Thus, the greatest height the ball can attain is 3.27 m

Learn more about motion under gravity:

brainly.com/question/13914606

5 0
2 years ago
Which astronomer gave his name to a telescope launched into space in 1990?
Mnenie [13.5K]

Answer:

Edwin Hubble

Explanation:

5 0
3 years ago
A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal
zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity u_{y}=0

The initial velocity is the horizontal component u_{x}

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = u_{y} t + \frac{1}{2} gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , u_{y}=0

Substitute these values in the rule

→ -60 = 0 + \frac{1}{2} (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component u_{x}

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = u_{x} t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = u_{x} (3.5)

Divide both sides by 3.5

→ u_{x} = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is v_{y}

→ v_{y} = u_{y} + gt

→ u_{y} = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ v_{y} = 0 + (-9.8)(3.5)

→ v_{y} = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of  v_{x} and v_{y}

→ Its magnetude is v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}

→ Its direction tan^{-1}\frac{v_{y}}{v_{x}}

→ v_{y} = 28.6 , v_{y} = -34.3

Substitute this values in the rules above

→ v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66

→ Its direction tan^{-1}\frac{-34.3}{28.6}=-50.18

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

7 0
2 years ago
Help me please !!!!!
bezimeni [28]

Answer:

confounding cause they had exposure to many programmes

8 0
3 years ago
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Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0
3 years ago
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