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1 month ago

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Your friend states in a report that the average time required to circle a 1.5-mi track was 65.414 s. This was measured by timing

7 laps using a clock with a precision of 0.1 s. How much confidence do you have in the results of the report? Explain.

1 answer:

Yuliya22 [3.3K]1 month ago

5 0

The average time per lap was determined by timing seven laps and averaging the total time over seven. It’s noted that the precision is 0.1 seconds, indicating accuracy to one decimal place. Therefore, the true value might fluctuate between 457.800 seconds and 457.899 seconds, suggesting that lap times could range from 65.400 seconds to 65.414 seconds. Hence, the recorded time of 65.414 seconds has a maximum error of 0.021%, indicating that the measurement is fairly precise.

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The accurate choice is option C. <span>This serves as an illustration of Boyle’s law. As the volume expands, the pressure diminishes, resulting in the marshmallow increasing in size.
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1 month ago

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The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

kicyunya [3294]

To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
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2 months ago

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as $w_{h}$ .

We aim to determine the heart's weight for a human

Using the provided information

$w_{h}=0.5\times w$

Where, h = heart weight

w = human weight

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

The final weight of a human heart is 0.93 lbs.

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3 months ago

Potential energy matter has a result of its ____ or ____.

Yuliya22 [3333]

Position or composition

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2 months ago

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A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

Ostrovityanka [3204]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Utilizing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Now let’s solve for t.

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

The computed velocity magnitude is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The ball's angle is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

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3 months ago