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3 years ago

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A 60.0 kg soccer player kicks a 0.4000 kg stationary soccer ball with 6.25 N of force. How fast does the soccer ball accelerate,

assuming that no friction acts on the ball?
a)2.50 m/s2
b)3.84 m/s2
c)15.6 m/s2
d)97.7 m/s2

1 answer:

frozen [14]3 years ago

8 0

F = ma
6.25 N = 0.4 kg · a
a = (6.25/0.4) m/s² since N=kg·m/s²
a = 15.625 m/s²

The answer is c) 15.6 m/s²
(Note that the mass of the soccer player is irrelevant.)

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Answer:

(a) , . and .

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Explanation:

Given that,

,

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.

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$|\vec B|=\sqrt {55^2+ (-12)^2}$

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.

And, the magnitude of the is

$|\vec C|=\sqrt {2^2+ 43^2}$

$\Rightarrow |\vec C|=\sqrt {1853}$

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(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

$\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)$

$\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j$

$\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j$

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$\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)$

$\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j$

$\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)$

The magnitude of is

$|\vec A - \vec B|=\sqrt {(-31)^2+55^2}$

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Now, if a vector $\vec V= -\alpha \hat i +\beta \hat j$ in 3rd quadrant having direction $\theta$ with respect to $\hat i$ direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector $\vec A - \vec C$ , $\alpha=31$ and $\beta=45$ .

$\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)$

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(d) Vector diagrams for $\vec A +\vec B$ and $\vec A - \vec B$ has been shown

in the figure(b) and figure(c) recpectively.

Vector $\vec A - \vec B$ is in 3rd quadrant as calculated in part (c).

While Vector $\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)$

$\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j$ , which is in 1st quadrant as both the components are position has been shown in figure(b).

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