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sukhopar [10]
3 years ago
7

The length of a soild of a metallic cube at 20°C is 5•0cm. Given that the linear expansivity of the metal is 4×^-5k^-1. Find the

volume of the cube at 120°C. Find the volume of the cube at 120°C
Physics
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer:

Explanation:

Length of one side of cube = 5 x 10⁻² m .

volume = ( 5 x 10⁻² )³

= 125 x 10⁻⁶ m³

Expression for thermal volume expansion is as follows :

V₂ = V₁ ( 1 + γ Δt )

V₂ and  V₁ are volume before and after rise in temperature , γ is coefficient of volume expansion Δt is increase in temperature .

linear expansivity = 4 x 10⁻⁵ K⁻¹

coefficient of volume expansion γ  = 3 x 4 x 10⁻⁵

= 12 x 10⁻⁵ K⁻¹

Putting the value

V₂ = 125 x 10⁻⁶ ( 1 + 12 x 10⁻⁵ x 100 )

=  125 x 10⁻⁶  +  125 x 10⁻⁶  x 12 x 10⁻³

= 125 x 10⁻⁶  +  1.5  x 10⁻⁶  

= 126.5 x 10⁻⁶

= 126.5 cm³

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Explanation:

→ Volume of cone = πr² × h/3

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→ Volume of cone = 1521 × 22/7 cm³

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Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>

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1 year ago
The earth has a mass ME = 5.98 · 1024 kg and the moon has a mass MM = 7.36 · 1022 kg. The distance from the center of the earth
ivann1987 [24]

Answer:

2 x 10^20 N

Explanation:

Me = 5.98 x 10^24 kg

Mm = 7.36 x 10^22 kg

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The gravitational force between earth and moon is

F = G Me x Mm / r^2

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3 years ago
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
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
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<u>Assume:</u>

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

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