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3 years ago

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A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p

ressure along the height of the block is negligible.
The cylindrical block is fully immersed in water with density rw=1000 kg/m3 remaining suspended by the rope and in equilibrium. The top of the cylinder is at the surface level. What is the difference of pressure in the bottom to the pressure from the top?

2 answers:

agasfer [191]3 years ago

8 0

Answer:

$\Delta P = 1961.4\,Pa$

Explanation:

The difference of pressure is given by gauge pressure:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)$

$\Delta P = 1961.4\,Pa$

Over [174]3 years ago

3 0

Answer:

the difference of pressure in the bottom to the pressure from the top is 1960 Pa

Explanation:

Given data:

m = mass of the block = 50 kg

h = height = 0.2 m

ρ = density = 1000 kg/m³

Question: What is the difference of pressure in the bottom to the pressure from the top, ΔP = ?

Let P₁ the pressure on the top and P₂ the pressure in the bottom

$delta-P=P_{1} -P_{2}=h\rho g$

Here, g = gravity = 9.8 m/s²

$delta-P=0.2*1000*9.8=1960N/m^{2} =1960Pa$

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For first-order kinetics, the rate k can be found using the half-life:

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In other words, the half-life is inversely proportional with the rate.

At the lower temperature, the rate is reduced to a third, so the half-life increases by a factor of 3. Meaning that the new half-life is 170 × 3 = 510 days.

The "shelf life" is the time it takes to reduce the initial amount to 10%. We can solve for this using the half-life equation.

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Full solution calculation can be found in the attachment below.

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How do I go about this?

Anna71 [15]

Hi there!

(a)

Recall that:
$W = F \cdot d = Fdcos\theta$

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

$W =248(56)cos(30) = 12027.36 J$

To the nearest multiple of ten:
$W_A = \boxed{12030 J}$

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
$\boxed{W_g = 0 J}$

(c)
Similarly, the normal force is perpendicular to the displacement, so:
$\boxed{W_N = 0 J}$

(d)

Recall that the force of kinetic friction is given by:
$F_{f} =\mu_k mg$

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
$W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J$

In multiples of ten:
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(e)
Simply add up the above values of work to find the net work.

$W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}}$

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
$F_{net} = F_{Ax} - F_f$

$W = F_{net} \cdot d = (F_{Ax} - F_f)$

$W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}$

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