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Ann [662]
3 years ago
15

A book having 50 sheets of paper has a thickness of 0.5cm. What will be the thickness of a single sheet in mm

Physics
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

we have to find the thickness of a single sheet in mm.

so,

50 = 0.5 cm

we know that

0.5 cm = 5 mm

NOW THICKNESS OF 50 SHEETS OF PAPER IS 5mm

•°• thickness of one sheet of paper will

5/50

= 0.1 cm

we have to find the answer in mm so

0.1 cm = 1 mm

Explanation:

hope it may help you

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Statement B is true.
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Whats a good string length for a parachute
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Answer: Hope This Helps!

Explanation:

The length of the string should be equal to the radius of the desired circle. Attaching the suspension lines: Creator of parachutes Use 4 suspension lines for each parachute. And Attatch the suspension lines onto the canopy.

3 0
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An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas
jek_recluse [69]

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

6 0
3 years ago
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
Nana76 [90]

Answer:

E

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice  the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

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
3 years ago
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