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Ede4ka [16]
3 years ago
13

Answer the attached question.No Spam!⚔️⚔️⚔️​

Physics
1 answer:
RoseWind [281]3 years ago
6 0
<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

= (4-0)/(4 - 0)

=4/4

= 1/1

<h3> =1 m/s</h3>

b) Average velocity for 4 to 8 seconds

Average velocity = Total Displacement/ Time taken

= (4 - 4)/(8-4)

= 0/4

<h3> = 0</h3>

c) Average velocity for last 6 seconds

Last 6 seconds = from 8 to 14 seconds

Average velocity = Total Displacement/Time taken

= (2 - 4)/(14 - 8)

= -2/6

<h3> = -1/3 m/s</h3>

<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

=20×20

<h3> = 400 m</h3>

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Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

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v² = u² + 2as

Making a the subject, we have;

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F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

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