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Marina CMI [18]

2 years ago

10

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 6.00 m/

s. The velocity of the ball relative to Mia is 5.00 m/s in a direction 30.0 east of south. What are the magnitude and direction of the velocity of the ball relative to the ground?

1 answer:

Airida [17]2 years ago

4 0

Answer:

v_b = 10.628 m /s (13.605 degrees east of south)

Explanation:

Given:

- Velocity of mia v_m = + 6 j m/s

- Velocity of ball wrt mia v_bm = 5.0 m/s 30 degree due east of south

Find:

What are the magnitude and direction of the velocity of the ball relative to the ground? v_b

Solution:

- The relation of velocity in two different frame is given:

v_b - v_m = v_bm

- Components along the direction of v_b,m:

v_b*cos(Q) - v_m*cos(30) = 5

v_b*cos(Q) = 5 + 6 sqrt(3) / 2

v_b*cos(Q) = 5 + 3sqrt(3)

- Components orthogonal the direction of v_b,m:

-v_m*sin(30) = v_b*sin(Q)

-6*0.5 = v_b*sin(Q)

-3 = v_b*sin(Q)

- Divide two equations:

tan(Q) = - 3 / 5 + 3sqrt(3)

Q = arctan(- 3 / 5 + 3sqrt(3)

Q = -16.395 degrees

v_b = -3 / sin(-16.395)

v_b = 10.63 m/s

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It is given that,

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The frequency in the spring is given by :

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On rearranging, we get the value of m as follows :

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Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?

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A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the

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An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

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a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

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s = 18 m is the displacement along the ramp

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a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

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$g=9.8 m/s^2$ is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

b)

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

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