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3 years ago

A 2 kg book is pushed from rest to a final velocity of 3 m/s. The book travels 2 m. How much force was the book pushed with

Physics

2 answers:

Studentka2010 [4]3 years ago

6 0

Explanation:

first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s

now using the 3rd law of motion

v^2=u^2+2as

0=9+2a2

a= -9/4m/s^2

now force=mass×accelration

=2kg×(-9/4)m/s^2

=4.5 N

4.5 newton force applied on the book!

✌️:)

Natali5045456 [20]3 years ago

6 0

Answer:

4.5N

Explanation:

Step 1:

Data obtained from the question:

Mass (m) = 2Kg

Initial velocity (U) = 0

Final Velocity (V) = 3m/s

Distance travelled (s) = 2m

Force (F) =?

Step 2:

Determination of the acceleration of the book. This is illustrated below:

Using V^2 = U^2 + 2as, the acceleration (a) can be obtained as follow:

V^2 = U^2 + 2as

3^2 = 0 + 2 x a x 2

9 = 4a

Divide both side by the coefficient of a i.e 4

a = 9/4

a = 2.25m/s2

Step 3:

Determination of the force applied. This is illustrated below:

Force = Mass x Acceleration

F = m x a

F = 2 x 2.25

F = 4.5N

Therefore, a force of 4.5N was used to push the book.

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Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

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To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

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$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

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$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

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$f_f = - \frac{20}{1} * \frac{1}{100}$

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$P = \frac{1}{f_f}$

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$- 5 \ D \le P \le - 2.667\ D$

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