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zysi [14]
3 years ago
7

Calculate the density of a rod of metal in g/cm3, with a mass of 9.58g, a diameter of 8 mm and a height of 3.5cm

Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

5.448 g/cm³

Explanation:

Density: This is defined as the ratio of the mass of a body to its volume.

The unit of density is kg/m³ other sub units are g/cm³, mg/mm³.

From the question,

D = m/πr²h......................... Equation 1

Where D = density, m = mass, r = radius, h = height.

Given: m = 9.58 g, r = 8/2 mm = 4 mm = 0.4 cm, h = 3.5 cm

Substitute this values into equation 1

D = 9.58/(3.14×0.4²×3.5)

D = 5.448 g/cm³

Hence the density of the metal rod is 5.448 g/cm³

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
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<u>Given:</u>

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  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
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  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
3 years ago
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