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3 months ago

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A snapshot of three racing cars is shown in the diagram. All three cars start the race at the same time, at the same place, and

move along a straight track. As they approach the finish line, which car has the lowest average speed?

1 answer:

Maru [3.3K]3 months ago

8 0

Answer:

The car that is the furthest from the finish line is: Car III (Choice C).

Explanation:

Here, we seek the car with the lowest overall average speed throughout the race. Thus, the one in last place inherently possesses the slowest average speed.

Since Car III is significantly behind Cars I and II, Choice A and B cannot be correct. Choice D is also not valid, as the positions of the cars are not the same. Lastly, Choice E is incorrect due to sufficient evidence demonstrating that Choice C has the lowest average speed.

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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.

8 0

3 months ago

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:

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

4 0

3 months ago

Using evidence from the article, defend the concept that Earth’s magnetic poles have swapped places over time.

Maru [3345]

Answer: Research has revealed signs of the Earth's magnetic field flipping in ocean floor rocks, especially at tectonic plate boundaries. These rocks exhibit alternating polarity because of the magnetization that happened as they solidified. Through radio metric dating, scientists suggest these reversals take place roughly every few hundred thousand years.

Explanation:

6 0

2 months ago