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atroni [7]
2 years ago
5

PLEASE HELP John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the

5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. Assuming that all the numbers given are exact, what is John's position at a time of 4.79 s ? Enter your answer to at least three significant digits.

Physics
1 answer:
Reptile [31]2 years ago
8 0

Answer:

PLEASE HELP John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. Assuming that all the numbers given are exact, what is John's position at a time of 4.79 s ? Enter your answer to at least three significant digits.

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Determine the values of m and n when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,00
yuradex [85]

Explanation:

Mass of the Earth is equal to,

m=5,970,000,000,000,000,000,000,000\ kg

Any number can be written in the form of scientific notation as :

N=m\times 10^n

m is the real number

n is any integer

Mass of the earth can be written in the form of scientific notation as :

m=5.97\times 10^{24}\ m

Here,

m = 5.97

n = 24

Hence, this is the required solution.                                                      

7 0
2 years ago
How much power does a machine have that does 15204 J of work in 40 seconds?
pentagon [3]

Explanation:

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P=15204/40

P=380.1 Watt

5 0
1 year ago
Which statement correctly describes the relationship between frequency and wavelength?
Len [333]
The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency. 

If we divide the equation by f we get:

λ=v/f

From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases. 

So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.  
3 0
2 years ago
Read 2 more answers
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
2 years ago
A swimming pool contains x (less than 0.02) grams of chlorine per cubic meter. the pool measures 5 meters by 50 meters and is 2
zubka84 [21]
The solution for this problem would be:(10 - 500x) / (5 - x) 
so start by doing: 
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V = (10 - 500x) / (5 - x) 
(V stands for the volume, but leaves us with the expression for x)
3 0
2 years ago
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