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3 years ago

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According to Jax's pedometer, he walked 2.783\,\text{km}2.783km2, point, 783, start text, k, m, end text while taking care of th

e livestock in the morning and 3.124\,\text{km}3.124km3, point, 124, start text, k, m, end text taking care of the livestock in the evening.

1 answer:

Ray Of Light [21]3 years ago

7 0

Answer:

5.907 kilometers.

Step-by-step explanation:

Please consider the complete question.

According to Jax's pedometer, he walked 2.783 km while taking care of the livestock in the morning and 3.124 km taking care of the livestock in the evening. How far did Jax walk while taking care of the livestock today?

To find the total distance covered by Jax, while taking care of the livestock, we will add distances covered by Jax in the morning and evening.

$\text{Total distance covered by Jax}=2.783+3.124$

$\text{Total distance covered by Jax}=5.907$

Therefore, Jax walked for 5.907 kilometers, while taking care of the livestock.

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Not sure if any of this is correct, but it’s what I got so far

Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

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Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote $\frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}$

I would write it as $\frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}$

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For problem 3, we first need to convert the runner's speed from mph to feet per second.

$17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}$

Since the runner needs to travel 90-12 = 78 ft, this means $\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}$

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

-------------

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

$\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}$

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0

3 years ago

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave

erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒ $\lambda_2=\frac{4L}{3}$ ⇒ $\lambda_3=\frac{4L}{5}$

⇒ $\lambda_2=\frac{4(120)}{3}$ ⇒ $\lambda_3=\frac{4(120)}{5}$

⇒ $\lambda_2=\frac{480}{3}$ ⇒ $\lambda_3=\frac{480}{5}$

⇒ $\lambda_2=160\ cm$ ⇒ $\lambda_3=96\ cm$

So the pipe is open at one end and closed at one end .

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3 years ago

What’s 2 + 2? <br> A. 0<br> B. 1<br> C. 2<br> D. 4

snow_tiger [21]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Help how do I solve it

Mila [183]

When two lines intersect, opposite angles are equal. This means that angle 1 equals angle 4. We can use that information to find their values.
Angle 1 = Angle 4
6n+1 = 4n+19
2n=18
n=9

6(9)+1=54+1=55

Angle 1 and 4 equal 55 degrees.

Two angles that form a straight line together have a total sum of 180 degrees. Angles 1 and 5 are like this, as well as Angles 4 and 5, and Angles 4, 3, and 2 added together.

Therefore, 180 = (angle 4) + (angle 3) + (angle 2)
180= 55+(angle 3) + (angle 2)
125= angle 3 + angle 2

I'm not sure what else can be extrapolated from this. There doesn't seem to be a way to find out what the measure of angle 2 is without angle 3 as well. I hope this helps and you can figure it out from the answer choices!

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3 years ago

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AleksandrR [38]

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3 years ago