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e-lub [12.9K]
1 year ago
14

2 a A pile of 60 sheets of paper is 6 mm high. Calculate the average thickness of a sheet of the paper.

Physics
1 answer:
fiasKO [112]1 year ago
6 0

The average thickness of a sheet of the paper is 0.1 mm.

The number of ice blocks that can be stored in the freezer is 80 blocks of ice.

<h3>Average thickness of a sheet of the paper</h3>

The average thickness of a sheet of the paper is calculated as follows;

average thickness = 6 mm/60 sheets = 0.1 mm /sheet

Thus, the average thickness of a sheet of the paper is 0.1 mm.

<h3>Volume of each block of ice</h3>

Volume = 10 cm x 10 cm x 4 cm

Volume = 400 cm³

<h3>Volume of the freezer</h3>

Volume = 40 cm x 40 cm x 20 cm = 32,000 cm³

<h3>Number of ice blocks that can be stored</h3>

n = 32,000 cm³/400 cm³

n = 80 blocks of ice

Thus, the number of ice blocks that can be stored in the freezer is 80 blocks of ice.

Learn more about average thickness here: brainly.com/question/24268651

#SPJ1

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The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

T = 81.52 Deg C

4 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f
stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2

When s = Db, t = 2t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2

Dividing the two equations

\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db

Hence, Da=(1/4)Db

3 0
3 years ago
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