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3 years ago

7

A batter hits a 0.140-kg baseball that was approaching him at 40.0 m/s and, as a result, the ball leaves the bat at 30.0 m/s in

the direction of the pitcher. what is the magnitude of the impulse delivered to the baseball

1 answer:

mamaluj [8]3 years ago

8 0

p=mv, sothat change of v =-10m/s. 0.14kg* -10m/s= -1.4kgm/s

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Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

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Answer:

(a) , . and .

(b) $\vec A - \vec C=22 \hat i -10 \hat j$ .

(c) $|\vec A - \vec B|=63.13$ and the direction $\theta =$ 124.56°.

Explanation:

Given that,

,

and

$\vec {C}=2 \hat i +43 \hat j$

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

$|\vec A|=\sqrt {24^2+ 33^2}$

$\Rightarrow |\vec A|=\sqrt {1665}$

.

The magnitude of the is

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

$\Rightarrow |\vec B|=\sqrt {3169}$

.

And, the magnitude of the is

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

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

.

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

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

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

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

(c) The expression of is

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

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

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

The magnitude of is

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

$\Rightarrow |\vec A - \vec B|=\sqrt {3986}$

$\Rightarrow |\vec A - \vec B|=63.13$

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 $\beta=45$ .

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

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(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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