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Savatey [412]
3 years ago
8

A brass bar, density 9.87g/cm3, has a volume of 20.25cm3. What is the mass of this brass bar?

Physics
1 answer:
gregori [183]3 years ago
4 0

Answer:

The mass of the bar = 199.87 grams

Explanation:

Given:

Density of a brass bar = 9.87 g/cm^3

Volume of bar = 20.25 cm^3

To find the mass of the brass bar.

Solution:

Density of a body is defined as mass of the body per unit volume of the body.

Thus, we have:

\rho=\frac{m}{V}

where:

<em />\rho\rightarrow<em> density of the body</em>

<em />m\rightarrow<em> mass of the body</em>

<em />V\rightarrow<em> Volume of the body</em>

Plugging in the given values.

9.87\ g/cm^3=\frac{m}{20.25\ cm^3}

Solving for m

Multiplying both sides by 20.25 cm^3.

20.25\ cm^3\times 9.87\ g/cm^3=\frac{m}{20.25\ cm^3}\times 20.25\ cm^3

199.87\ g =m

∴ m=199.87\ g

Thus, the mass of the bar = 199.87 grams

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Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2

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3 years ago
A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plan
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(a) L = 0·73 m

(b) 4·39 × 10^{-3} J

Explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

<h3>For small oscillations, α = ω² × β</h3>

After substituting the value of α and solving we get

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Time period = (2 × π) ÷ ω =  (2 × π) ÷ √((3 × g) ÷ (2 × L))

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Maximum potential energy = 4·39 × 10^{-3} J

∴ Maximum kinetic energy = 4·39 × 10^{-3} J

4 0
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Answer:

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