A meter stick of uniform material has some dough stuck on one end. The stick has a mass of 0.1 kg. It will balance when supported at a point 0.33m from the dough. How much mass does the dough have?

Answers

Answer 1

At the equilibrium or balance point the meter stick stationary and the sum of the moments and forces acting on it is zero

The mass of the dough is .[tex]\underline {51.\overline{51}}[/tex] grams

Reason:

Known;

Length of the stick = 1 meter

Mass of the stick, m = 0.1 kg

Point the stick will balance when the dough is at one end = 0.33 m

Required:

How much mass does the dough have

Solution:

The meter stick is balanced when the clockwise moment and the anticlockwise moment about the balance point are equal

Let M, represent the mass of the dough, we have;

The center of mass of the meter stick = The 0.5 m point

Taking moment, we have;

[tex]\mathbf{\sum M_{clockwise} = \sum M_{anticlockwise}}[/tex]

Distance from the center of mas of the ruler to the fulcrum, d₁ = (0.5 - 0.33) m

Distance from the dough to the fulcrum, d₂ = 0.33 m

Clockwise moment, [tex]\mathbf{\sum M_{clockwise} }[/tex] = m × d₁

Anticlockwise moment, [tex]\mathbf{\sum M_{anticlockwise} }[/tex] = M × d₂

Therefore;

(0.5 - 0.33) × 0.1 = M × 0.33

Therefore, we get;

[tex]M = \dfrac{(0.5 - 0.33) \times 0.1}{0.33} =\dfrac{17}{330} =0. 0\overline{51}[/tex]

The mass the dough has, M = 0.0[tex]\overline{51}[/tex] kg = 51.[tex]\underline{\overline{51}}[/tex] grams

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Answer:

Explanation:

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Now

[tex]\\ \sf\longmapsto s=ut+\dfrac{1}{2}gt^2[/tex]

[tex]\\ \sf\longmapsto s=0t+\dfrac{1}{2}(10)t^2[/tex]

[tex]\\ \sf\longmapsto s=5t^2\dots(1)[/tex]

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[tex]\\ \sf\longmapsto 30-s=(24)t+\dfrac{1}{2}(10)t^2[/tex]

Using eq(1)

[tex]\\ \sf\longmapsto 30-5t^2=24t-5t^2[/tex]

[tex]\\ \sf\longmapsto 24t=30[/tex]

[tex]\\ \sf\longmapsto t=\dfrac{30}{24}[/tex]

[tex]\\ \sf\longmapsto t=1.2s[/tex]

After 1.2s they will meet.

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putting t in eq(1)

[tex]\\ \sf\longmapsto s=5t^2=5(1.2)^2=5(1.44)=7.2m[/tex]

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Answer:

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Explanation:

ytkjsgxhxkvgakavsvdjddhdbiwbe vs

The pressure due to water at the bottom of the tank  is 200000 Pascal.

What is pressure?

It is the ratio between the force being applied and the surface area being applied to. The force delivered perpendicularly to an object's surface per unit area across which that force is dispersed is known as pressure.

Given parameters:

Depth of the tank = 25 meter.

Depth of water in the tank: d = 25×4/5 meter = 20 meter.

Density of water: ρ = 1000 kg/m³.

Acceleration due to gravity: g = 10 m/s².

Hence, the pressure due to water at the bottom of the tank

= depth of water (d) × density of water (ρ) × acceleration of the gravity (g)

=  20 meter × 1000 kg/m³ × 10 m/s²

= 200000 Newton/meter²

=200000 Pascal.

Hence, the pressure due to water at the bottom of the tank  is 200000 Pascal.

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The distance traveled by the second ball is calculated as follows;

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