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cluponka [151]
3 years ago
8

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

m three times that height, how long (in terms of T) would it take to reach the ground?
Physics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

h=ut+\frac{1}{2}at^2

where  

h=displacement

u=initial\ velocity

a=acceleration

t=time

here u=0 because it dropped from a certain height

h=\frac{1}{2}gT^2

T=\sqrt{\frac{2h}{g}}

When height is increases to three times of original height

i.e. h'=3 h

then time period becomes

T'=\sqrt{\frac{2\times 3h}{g}}

T'=\sqrt{3}T

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Boyle's law balloon was filled to a volume of 2.50 l when the temperature was 30.0∘
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7 0
3 years ago
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
2 years ago
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