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pashok25 [27]
3 years ago
10

A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

is 1.6 m/s^2 due north. (Part B) Find the velocity of the car after 5.6 s if its acceleration is 1.5 m/s2 due south.
Physics
1 answer:
galben [10]3 years ago
5 0

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

b)

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

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Do not forget that mass = <span>volume x density
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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
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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

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