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2 years ago

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An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce

nter of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed

1 answer:

Morgarella [4.7K]2 years ago

3 0

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = $\sqrt{} 4ghi/R^{2}$

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= $\sqrt4* 9.81* 0.5/0.5^{2}$ = 8.86 rad/s

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What would you do to increase resistance

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Answer:

If this is electrical currents , make the wire longer, smaller diameter wires, heat it up

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2 years ago

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Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t

fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, $M_O=32$

Molar mass of hydrogen, $M_H=2$

We know ideal gas law as:

$PV=nRT$

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵ $n=\frac{m}{M}$

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

$PV=\frac{m}{M}.RT$

$P=\frac{m}{V}.\frac{RT}{M}$

as: $\frac{m}{V}=\rho\ (\rm density)$

So,

$P=\rho.\frac{RT}{M}$

Now, according to given we have T,P,R same for both the gases.

$P_O=P_H$

$\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}$

$\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}$

$\rho_O=16\rho_H$

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

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3 years ago

Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

2 years ago

A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result

Over [174]

B) law of conservation of momentum

It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.

In this case total momentum before impact:

10kg*5m/s + 5kg * 0m/s = 50 kg m/s

After Impact:

10kg*0m/s + 5kg*10m/s = 50 kg m/s

You can see the momentum before and after impact is same as 50 kg m/s

Of course we assumed that the first cart stopped after the impact, and there are no energy losses.

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3 years ago

2. A student drives 7.8-km trip to school and averages a speed of

Alekssandra [29.7K]

Answer:

<em>The total time is: t=451.22 sec</em>

<em>The average speed is: V=34.57 m/s</em>

Explanation:

<u>Average speed</u>

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

$\displaystyle V=\frac{x}{t}$

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

$\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec$

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

$\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec$

The total time is:

$t=239.26\ sec+211.96\ sec=451.22\ sec$

$\boxed{t=451.22\ sec}$

The average speed is:

$\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s$

$\boxed{\displaystyle V=34.57\ m/s}$

6 0

3 years ago