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Ivenika [448]
3 years ago
11

How to calculate displacement?​

Physics
1 answer:
Blizzard [7]3 years ago
4 0

-- Take a straight ruler.

-- Lay it down with the 'zero' mark at the start point.

-- Rotate it around the start point until the end point is also touching the edge of the ruler.

-- From the marks on the ruler, read the straight-line distance from the start point to the end point.

-- Without moving the ruler, observe and write down the DIRECTION from the start point to the end point.

-- The Displacement is the straight-line distance and direction from the start point to the end point.

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block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

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Consider a golf club hitting a golf ball. To a good approximation, we can model this as a collision between the rapidly moving h
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Answer:

1. 8.0kg * m/s

2. The same as before the collision

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4. 14 m/s

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Answer:

KE = 225000000 J

Explanation:

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D.) Travels in a pattern of particle motion....
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A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
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Answer:

Part a)

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Part B)

v = 12.1 m/s

Explanation:

Part A)

At the top of the hump the force on the rider is

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so here we know that

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

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Part B)

At the top of the loop we will have

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so we will have

mg = \frac{mv^2}{R}

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v = \sqrt{15\times 9.81}

v = 12.1 m/s

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