PHYSICS COLLEGE

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer 1

Answer:

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

Related Questions

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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

Answers

Answer

given,

Tension of string is F

velocity is increased and the radius is not changed.

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone

= $(mv^2)/(r)$

now centrifugal force is balanced by tension

T = $(mv^2)/(r)$

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?

Answers

Answer:

a) x = 5.48 10⁻² m and b) 0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

A = √ (2Em / k)

A = √ (2 0.12 / 80)

A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

Em = K = ½ m v²

Final

Em = Ke = ½ k x²

½ m v² = ½ k x²

x = √(m/k) v

x = 2 √(0.50 /800.0)

x = 0.05 m

Answer:

a) The greatest extension of the spring is 0.055 m

b) The spring compress 0.05 m

Explanation:

Please look at the solution in the attached Word file

A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and  the reflected wave and the transmitted wave.

Answers

Answer:

Explanation:

Suppose a boat moves at 16.4 m/s relative to the water. If the boat is in a river with the current directed east at 2.70 m/s, what is the boat's speed relative to the ground when it is heading east, with the current, and west, against the current? (Enter your answers in m/s.)

Answers

Answer:

with the current: 19.1 m/s eastern

against the current: 13.7 m/s western

Explanation:

The boat speed relative to ground is the sum of the boat speed relative to water + the water speed relative to ground.

Suppose eastern is positive, water speed due east is 2.7m/s.

When the boat is heading east, its speed relative to water is 16.4m/s, its speed relative to ground is 16.4 + 2.7 = 19.1 m/s eastern

When the boat is heading west, its speed relative to water is -16.4m/s, its speed relative to ground is -16.4 + 2.7 = -13.7 m/s western

When a box is placed on an inclined surface with no friction, it will:

Answers

Answer: With no friction, the box will accelerate down the ramp

Explanation:

It will gain speed down the ramp

What is the magnitude of the force you must exert on the rope in order to accelerate upward at 1.4 m/s2 , assuming your inertia is 63 kg ? Express your answer with the appropriate units.

Answers

Answer:

The magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Explanation:

The magnitude of force, you must exert can be estimated as follows;

Since it is upward motion, we must consider acceleration due to gravity which opposes the upward motion.

F = m(a+g)

where;

F is the magnitude of the upward force

m is your mass, which is the measure of inertia = 63kg

a is the acceleration of the rope = 1.4 m/s²

F = 63(1.4 + 9.8)

F = 63(11.2)

F = 705.6 N

Therefore, the magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Answer:

705.6 N

Explanation:

Force: This can be defined as the product of mass a acceleration.

The S.I unit of force is Newton.

The expression for the force on the rope in order to accelerate upward is given as,

F-W = ma .......................... Equation 1

Where F = Force exerted on the rope, W = weight of the rope, m = mass of the rope, a = acceleration.

But,

W = mg........................ Equation 2

Where g = acceleration due to gravity

substitute equation 2 into equation 1

F-mg = ma

F = ma+mg

F = m(a+g).............. Equation 3

Given: m = 63 kg, a = 1.4 m/s²

Constant: g = 9.8 m/s²

Substitute into equation 3

F = 63(1.4+9.8)

F = 63(11.2)

F = 705.6 N

The magnitude of the force exerted on the rope = 705.6 N

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