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LuckyWell [14K]
4 years ago
16

The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

object is 150 feet. After how many seconds does the object reach its maximum height? 2 seconds 3 seconds 6 seconds 9 seconds
Physics
2 answers:
Rainbow [258]4 years ago
6 0

Answer:

3 seconds

Explanation:

Since h(t) represents the height and t represents the time, we can set the equation equal to 150 to find t.

-16t^2+96t+6=150

Subtract 150 from both sides to set the equation equal to 0, to find the solutions.

-16t^2+96t-144=0

Factor out -16, because all of the terms are divisible by it.

-16(t^2+6t+9)=0

Now we can focus on the terms inside the parenthesis and factor it again.

t^2-6t+9=0

We need to find two value that can be multiplied to get 9 and added to get -6.

-3 and -3 works.

Thus, we get (x-3)(x-3).

Now solve for 0.

x-3=0

x=3

The object reaches its maximum height after 3 seconds.

svp [43]4 years ago
6 0

Answer:

3 seconds

Explanation:

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You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler
scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

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we have F = W

so\frac{KQq}{r^2} =mg

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3 years ago
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
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Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

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