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Wewaii [24]
2 years ago
8

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

on the book, perpendicular to the wall, to prevent the book from slipping down the wal
Physics
1 answer:
Vika [28.1K]2 years ago
8 0

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

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<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) f=1617.9\ Hz

(b) T=0.618\ ms

(c) A=20 \ Volts

(d) \varphi=60^o

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

V(t)&=A\cdot \sin(2\pi ft+\varphi )

Where

A=Amplitude

f=frequency

\varphi=Phase\  angle

The voltage of the question is the sinusoid expression  

V(t)=20cos(5\pi\times 103t+60^o)

(a) By comparing with the general formula we have

f=5\pi\times 103=1617.9\ Hz

\boxed{f=1617.9\ Hz}

(b) The period is the reciprocal of the frequency:

\displaystyle T=\frac{1}{f}

\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec

Converting to milliseconds

\boxed{T=0.618\ ms}

(c) The amplitude is

\boxed{A=20 \ Volts}

(d) Phase angle:

\boxed{\varphi=60^o}

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2 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
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Answer:

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b) v₂fin = 1.0588 m/s     (→)

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⇒  v₂fin = 1.0588 m/s     (→)

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