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Oxana [17]
4 years ago
7

Part E

Physics
2 answers:
Elodia [21]4 years ago
7 0

Answer:  

416,500 J – 67,200 J = 349,300 J

The potential energy at the top of the second hill is 349,300 joules.

miv72 [106K]4 years ago
4 0

Answer:

The potential energy at the top of the second hill is the potential energy at the top of the first hill minus the potential energy lost from friction:

416,500 J – 67,200 J = 349,300 J

The potential energy at the top of the second hill is 349,300 joules.

Explanation:

This answer came straight from the website of Edmentum.

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I have these questions I need to find the acceleration, Please explain how to do them, Don't have to explain all xD
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1. a=(v-v0)/t
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4 years ago
After fertilization in the fallopian tube, how many days will the zygote travel before arriving at uterus?.
natta225 [31]

After fertilization in the fallopian tube, It will take 6-12 days for the zygote to travel before arriving at the uterus.

<h3>What is a zygote?</h3>

A zygote is, generally speaking, a cell created by the fusion of two gametes; the growing person is created from such a cell.

It takes the zygote around 6–12 days following fertilization in the fallopian tube for the fertilized egg to travel to the uterus and attach to the uterus, a process known as implantation.

Hence It will take 6-12 days for the zygote to travel before arriving at the uterus.

To learn more about the zygote refer;

brainly.com/question/465851

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6 0
2 years ago
Two musicians are comparing their trombones. The first produces a tone that is known to be 438 Hz. When the two trombones play t
galben [10]

Answer:

It is producing either a 435-Hz sound or a 441-Hz sound.

Explanation:

When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:

f_B = |f_1 -f_2| (1)

In this problem, we know that:

- The frequency of the first trombone is f_1 = 438 Hz

- 6 beats are heard every 2 seconds, so the beat frequency is

f_B=\frac{6}{2 s}=3 Hz

If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:

f_2 = f_1 - f_B = 438 Hz - 3 Hz = 435 Hz\\f_2 = f_1+f_B = 438 Hz+3 Hz=441 Hz

7 0
3 years ago
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