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Alinara [238K]
4 years ago
6

Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

Physics
1 answer:
skelet666 [1.2K]4 years ago
5 0

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:  

a) The vectorial product, a×b  is:

a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k

b) The escalar product a·b is:

a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17

d) The component of a along the direction of b is:

a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66

I hope it helps you!                        

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A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c
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865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

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Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

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Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

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Take the square root of both side.

t = √(62/4.9)

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Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

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The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

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We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

s=u_y t + \frac{1}{2}at^2

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u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

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u = 23.5 m/s is the initial velocity

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Substituting,

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We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

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Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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