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mr_godi [17]
2 years ago
9

Distance travelled by a free falling object in the first second is: a) 4.9m b) 9.8m c) 19.6m d) 10m​

Physics
1 answer:
zvonat [6]2 years ago
6 0
  • Time=1s=t
  • Acceleration due to gravity=g=9.8m/s^2
  • Distance=s

In free fall

\boxed{\sf s=-\dfrac{1}{2}gt^2}

\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2

\\ \sf\longmapsto s=-4.9(1)

\\ \sf\longmapsto s=-4.9m

  • Take it positive

\\ \sf\longmapsto s=4.9m

  • Option a is correct
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Explanation:

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If we assume that the only force acting on the arm is the weight of the arm, and that this is concentrated in a point in the center of it (taking the arm as a solid bar with the center of mass at the mid-point), clearly the torque will be the greatest when the force be exactly perpendicular, which is the case of the arm placed straight out parallel to the ground (Case 2).

b)  As the torque and the angular acceleration are directly proportional each other (being the rotational inertia the proportionality constant) the angular acceleration will be maximum when torque be maximum also, which is the case that the arm begins to swim, due to the perpendicular distance to the shoulder is the maximum possible (Case 1).

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3 years ago
Below is an "oracle" function. An oracle function is a function presented interactively. When you type in a t value, and press t
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Answer:

The rate of change of height with respect to time is -10.64 feet/sec

Explanation:

Given that,

There are three lines, so you can calculate three different values of the function at one time.

The function f(t) represents the height in feet of a ball thrown into the air, t seconds after it has been thrown.

Given table is,

Time t = 0, 1, 1,02

Function is,F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015

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Using equation of motion

h=h_{0}+ut-\dfrac{1}{2}gt^2

Where, h₀ = initial height

u = initial velocity

t = time

g = acceleration due to gravity

At t = 0,

Put the value into the formula

-3.053113177191196\times10^{-18}=h_{0}+0-0

h_{0}=-3.053113177191196\times10^{-18}

We need to calculate the height of ball at t = 1

Using equation of motion

h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2

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6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32

6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}

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Velocity is the rate of change of height with respect to time

So, velocity at 1.02 sec is given

We need to calculate the height

Using equation of motion

h=ut-\dfrac{1}{2}gt^2

On differentiating w.r.to t

h'(t)=u-\dfrac{1}{2}g(2t)

Put the value into the formula

h'(t)=22-\times32\times(1.02)

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4 0
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In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.

Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.

For reference: brainly.com/question/20380620

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