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djyliett [7]
3 years ago
9

In Figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 N and the distance it acts over is 0.

25 m, what is the final speed of the backpack?
Physics
1 answer:
pshichka [43]3 years ago
4 0
I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
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Answer:

The value of the distance is \bf{14.52~cm}.

Explanation:

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v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

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Answer:

Write the following Quantitiesin scientific notation.

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Answer

5.0/5

2

kobenhavn

Expert

5.5K answers

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Answer: a.

b.  

c.  

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as

a. 10130 Pa to 2 decimal place is written as

b.  to 1 decimal place is written as

c.   to 3 decimal places is written as

Explanation:

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Answer:

Explanation:

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