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Elis [28]
3 years ago
12

A book has a mass of 5 lb what is the mass of the book kg

Physics
1 answer:
OverLord2011 [107]3 years ago
6 0
The mass of the Book is 2.27 kg.
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The average height of an apple tree is 4.00 meters. How long would it take an apple falling from that height to reach the ground
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Answer:

The time is 0.90 sec.

Explanation:

Given that,

Height = 4.00 m

We need to calculate the time

Using equation of motion

s = ut-\dfrac{1}{2}(-g)t^2

For free fall, u = 0

t =\sqrt{\dfrac{2s}{g}}

Where, s = distance

g = acceleration due to gravity

t = time

Put the value into the formula

t =\sqrt{\dfrac{2\times4.00}{9.8}}

t= 0.90\ sec

Hence, The time is 0.90 sec.

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The force required to compress a non-standard spring as a function of displacement from equilibrium x is given by the equation f
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Answer:

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3 years ago
A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
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Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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