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3 years ago

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A track coach measures the 100-meter time of a track athlete. The runner completes the distance in 11.5 seconds. If the stopwatc

h has an uncertainty of 1.7%, what is the amount of uncertainty in the time?(1 point)
11.5 s ± 0.19 s
11.5 s plus or minus 0.19 s

11.5 s ± 0.20 s
11.5 s plus or minus 0.20 s

11.5 s ± 0.1955 s
11.5 s plus or minus 0.1955 s

11.5 s ± 0.196 s
11.5 s plus or minus 0.196 s

2 answers:

Andrei [34K]3 years ago

5 0

Answer:11.5 s plus or minus 0.196 s

Explanation:

Inessa05 [86]3 years ago

5 0

The uncertainty in the time of the runner to complete the distance is 11.5 s ± 0.1955 s

The given parameters;

length of the track, d = 100 m

time to complete the track, t = 11.5 s

the uncertainty, = 1.7%

The uncertainty in the time of the runner to complete the distance is calculated as;

the uncertainty = ± (0.017 x 11.5)

the uncertainty = ± 0.1955 s

Thus, the uncertainty in the time of the runner to complete the distance is 11.5 s ± 0.1955 s

Learn more here: brainly.com/question/22722990

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1 year ago

A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

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Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

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drag force $f_d=85\ pounds$

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Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

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3 years ago

Why are the orbits of planets only nearly circular and not perfectly circular?

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Explanation:

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2 years ago

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

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Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

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Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

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3 years ago

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

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Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

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Water flowing,

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Displacement,

d = 300 m

Now,

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Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

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Throughout the opposite direction, when the boat seems to be travelling then,

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On putting the values, we get

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hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

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3 years ago