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rusak2 [61]
1 year ago
11

lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu

lling to the right, a 36 N force that makesan angle of 30 deg with the horizontal-right.Calculate the net work done on the object. (1 point)A. O 462.747 JB. O1096.557 JC. O151.221 JD. O 675.066 J

Physics
1 answer:
Oksanka [162]1 year ago
6 0

The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

Therefore, the total work done is 578.123 J and the answer is option E

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k=\frac{0.693}{t_{1/2}}

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k=2.47\times 10^{-2}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.47\times 10^{-2}\text{ years}^{-1}

t = time passed by the sample  = ?

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Now put all the given values in above equation, we get

t=\frac{2.303}{2.47\times 10^{-2}}\log\frac{1.00}{0.600}

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3 years ago
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An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take
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Answer:

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When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

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