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alukav5142 [94]
2 years ago
9

A 5kg object is moving downward at a speed of 12m/s. If it is currently 2.6m above the ground, what is its potential energy? Use

g=10m/s^2
Physics
1 answer:
motikmotik2 years ago
4 0
Don’t trust the people that give you these file things to click on it’s a scam!
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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

We need to calculate the initial height of ball

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

Put the value in the equation

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

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

u_{0}=22\ feet/s

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)

h'(t)=-10.64\ feet/sec

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

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A car with a velocity of 6.4 m/s, forward, accelerates to a velocity of 10.6 m/s, forward, in 16 s. What is the card acceleratio
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Answer:

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