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Vinil7 [7]
3 years ago
15

A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 i + 1.00 j) m/s at a point in the ocean where the posit

ion relative to a certain rock is r with arrowi = (12.0 i − 2.00 j) m. after the fish swims with constant acceleration for 17.0 s, its velocity is v with arrow = (23.0 i − 1.00 j) m/s. (a) what are the components of the acceleration of the fish? ax = m/s2
Physics
1 answer:
jasenka [17]3 years ago
8 0
To determine the acceleration of the fish, we use one of the kinematic equation which relates velocity and acceleration. Using the given velocities, we substitute it to the equation. As we see, the velocities include their directions where i represents the x direction and j represents the y direction. We do as follows:

a = (v-vi) / t
a = [ (20.0 i - 5.00 j) - (4.00 i + 1.00 j)] / 23
Combining like terms,
a = [16i - 6j] / 23
TO determine the components of the acceleration, we do as follows:
 
<span>ax = 16.0/23.0 m/s2 = 0.70 m/s^2 ( to the right)</span>
<span>ay = -6.0/23.0 m/s2  = -0.26 m/s^2 ( accelerating down)</span>
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Answer:

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6 0
2 years ago
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

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3 years ago
A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.
Savatey [412]
We are given the equation:

<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,    
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At t=2.95 sec,   
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Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s )  = 56.65 m/s

</span>For (b)

At t=2.20 sec,    
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At t=2.40 sec,   
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Explanation:

I hope I helped , Have a great day or night . Xoxoxo.

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Answer:

A

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