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LekaFEV [45]
2 years ago
8

Given the velocity v= ds dt and the initial position of a body moving along a coordinate line, find the body's position at time

t
Physics
2 answers:
kolezko [41]2 years ago
7 0

For  the​ body's position at time t  is mathematically given as

s(t)=4.9t^2+5t+16

<h3>What is the​ body's position at time t</h3>

Generally, the equation for the velocity is mathematically given as

v=ds/dt

v=9.8t+5 as velocity

s=4.9t^2+5t+c

s(0)=16

16=c

s(t)=4.9t^2+5t+16

In conclusion the​ body's position at time t

s(t)=4.9t^2+5t+16

Read more about Speed

brainly.com/question/4931057

nordsb [41]2 years ago
5 0

The body position at the given time t is s = 4.9t² + 5t + 16 if the velocity v= ds/dt and the initial position of a body moving along a coordinate line.

<h3>What is the distance?</h3>

A numerical representation of the distance between two items or locations is called distance. A physical length or an approximation based on other considerations in physics or common usage can be referred to as distance.

We have:

\rm v = \frac{ds}{dt}

\rm v  = 9.8t+5  is the velocity

\rm s = 4.9t^2+5t+c is the speed

At t = 0 s(0) = 16

16 = c

\rm s = 4.9t^2+5t+16

Thus, the body position at the given time t is s = 4.9t² + 5t + 16 if the velocity v= ds/dt and the initial position of a body moving along a coordinate line.

Learn more about the distance here:

brainly.com/question/26711747

#SPJ4

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A fixed coil of wire with 10 turns and an area of 0.055 m2 is placed in a perpendicular magnetic field. This field oscillates in
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Answer:

Part a)

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Part b)

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E_{avg} = 0

Part c)

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Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

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For average value of EMF from positive maximum to negative maximum which is equal to half cycle

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E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt

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Part b)

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
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Answer:

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\theta_{o}, \theta - Initial and final angular position, measured in radians.

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\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

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\bar \alpha = \frac{\omega-\omega_{o}}{t}

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