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2 years ago

9

A large ebony wood log, totally submerged, is rapidly floating down a flooded river. If the mass of the log is 165 kg, what is t

he buoyant force acting on the submerged log ?

1 answer:

Sedbober [7]2 years ago

3 0

Answer:

F = 1618.65[N]

Explanation:

To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.

We know that the density of wood is equal to 750 [kg/m^3]

density = m / V

where:

m = mass = 165[kg]

V = volume [m^3]

V = m / density

V = 165 / 750

V = 0.22 [m^3]

The floating force is equal to:

F = density * g * V

F = 750*9.81*0.22

F = 1618.65[N]

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By law of refraction we know that image position and object positions are related to each other by following relation

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here we know that

$\mu_1 = 1.473$

$h_o = 5 cm$

$\mu_2 = 1$

now by above formula

$\frac{1.473}{5} = \frac{1}{h_i}$

$h_i = 3.39 cm$

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2 years ago

A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

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Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

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2 years ago

You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr

Nitella [24]

Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

First 60 miles with speed 55 mi/h

Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = $=\frac{100}{1.953}=51.2 mi/h$

Thus, the average speed of the journey is 51.2 mi/h.

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2 years ago

The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?

Fynjy0 [20]

The formula for potential energy is PE=mgh
It can have that high of a potential energy if it's relative height what super high.

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2 years ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Svetlanka [38]

<u>Answer:</u>

Mass of base ball $m_b=3.992*10^{14}kg$

<u>Explanation:</u>

Circumference of baseball = 2πr = 23 cm

So radius of baseball = 3.66 cm = $3.66*10^{-2}$ m

Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

Mass of proton = $10^{-27}$ kg

Diameter of proton = $10^{-15}$ m

Radius of proton = $5*10^{-16}$ m

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Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

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$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

So mass of base ball $m_b=3.992*10^{14}kg$

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2 years ago

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