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Hoochie [10]
3 years ago
6

how do i.... Find the angle between each pair of vectors below. (1 point each) a. a = 2x + 3y a = 4x + 2y

Physics
1 answer:
Shalnov [3]3 years ago
6 0

Recall that the dot product between two vectors \mathbf a and \mathbf b satisfies

\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta

where \|\mathbf x\| denotes the norm of a vector \mathbf x, and \theta is the angle between the two vectors.

So if \mathbf a=(2,3) and \mathbf b=(4,2), then

(2,3)\cdot(4,2)=\|(2,3)\|\|(4,2)\|\cos\theta

\implies2\cdot4+3\cdot2=\sqrt{2^2+3^2}\sqrt{4^2+2^2}\cos\theta

\implies\cos\theta=\dfrac{14}{\sqrt{13}\sqrt{20}}

\implies\theta\approx29.7^\circ

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A. -30 N

Explanation:

not sure but if the box isn't moving then the force opposite of Polly would be equal to the force she's exerting.

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Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to t
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3 years ago
The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

v_y^2 =v_0^2 +2ay

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

 v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s  v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0
4 years ago
An object has an initial velocity of 15 m/s. How long must it accelerate at a constant rate of 3.0 m/s^2 before its final veloci
Anuta_ua [19.1K]

Answer: v=u + at

V= Final velocity =30m/s

U= initial velocity = 15m/s

a= acceleration = 3m/s^2

t= time taken

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t=5s

Explanation:

3 0
3 years ago
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