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ladessa [460]
3 years ago
11

A ____________ sloped line means the object is returning to the starting point. A Upward B Backwar C Missing D Downward

Physics
1 answer:
Damm [24]3 years ago
4 0

Answer:

A downward sloped line means the object is returning to the starting point.

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lions [1.4K]

i dont get it so much but

The weight of the bag pack is 8.2 N. g = 1.64 m/s2. Hence, the acceleration due to gravity on moon is 1.64 m/s2. sooo? is it right

7 0
2 years ago
Your friend asks you to describe pseudoscience. You reply pseudoscience is
oksian1 [2.3K]
The answer is most likely D. hope that helped 
5 0
3 years ago
Read 2 more answers
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
in positive numbers less than 1, the zeros between the decimal point and a non-zero number are _______ significant?
DedPeter [7]

Answer:

Explanation:

If a number of less than 1, then the number has a decimal point like

0.085, 0.008 e.t.c.

The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.

But if there a zero between the none zero e.g. 0.0087056

Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6

But if the zero is not in between the none zero digit, then the zero is insignificant

E.g 0.05800

The last two zero is insignificant, the significant number is 5 and 8

So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.

8 0
3 years ago
Explain why selling cereal by mass rather then by volume be more fair to customers
Orlov [11]

<em>Answer:</em>

Simple, there could be air in the package and volume would record that, whereas mass would count the mass of the cereal and discount the air.

7 0
2 years ago
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