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3 years ago

A book weighing 12 n is placed on a table. how much support force does a table exert on the book?

Physics

1 answer:

Alenkasestr [34]3 years ago

7 0

It should be 12 N. the force of the book on the table should be the same as the force of the table on the book.

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3 years ago

Equation of motion description is v=20+2t.How big is the initial speed,acceleration?

trasher [3.6K]

Answer:

the initial velocity is 20 m/s and the acceleration is 2 m/s²

Explanation:

Given equation of motion, v = 20 + 2t

If V represents the final velocity of the object, then the initial velocity and acceleration of the object is calculated as follows;

From first kinematic equation;

v = u + at

where;

v is the final velocity

u is the initial velocity

a is the acceleration

t is time of motion

If we compare (v = u + at) to (v = 20 + 2t)

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2 years ago

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3 years ago

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2 years ago

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A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its ra

mr_godi [17]

<h2>Answer:</h2>

The rate of deceleration is -0.14 $m/s^{2}$

<h2>Explanation:</h2>

Using one of the equations of motion;

v = u + at

where;

v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)

u = initial velocity of the boat = 25m/s

a = acceleration of the boat

t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes

<em>Convert the time t = 3 minutes to seconds;</em>

=> 3 minutes = 3 x 60 seconds = 180seconds.

<em>Substitute the values of v, u, t into the equation above. We have;</em>

v = u + at

=> 0 = 25 + a(180)

=> 0 = 25 + 180a

<em>Make a the subject of the formula;</em>

=> 180a = 0 - 25

=> 180a = -25

=> a = -25/180

=> a = -0.14 $m/s^{2}$

The negative value of a shows that the boat is decelerating.

Therefore, the rate of deceleration of the speed boat is 0.14 $m/s^{2}$

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3 years ago