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Sloan [31]
3 years ago
6

I need help i dont have a caculator that need to use for it sadly

Physics
2 answers:
Bond [772]3 years ago
7 0

Answer:

A

Explanation:

because the initial value is 10 and 1.3 is the slope

Amiraneli [1.4K]3 years ago
3 0

Answer:

A) y = 1.3x + 10

Explanation:

<u>2 Points on the graph by using x and y values:</u>

(0, 10) and (10, 23)

<u>Slope:</u>

m= (y2-y1)/(x2-x1)

m= (23 - 10)/(10 - 0)

m= 13/10

m = 1.3

<u>Slope-intercept:</u>

y - y1 = m(x - x1)

y - 10 = 1.3(x - 0)

y - 10 = 1.3x

y = 1.3x + 10

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8 0
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WILL GIVE BRAINLIEST IF CORRECT!
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Answer:

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Explanation:

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Which of the following about Dalton's theory of matter is true?
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6 0
3 years ago
Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1
zheka24 [161]

Answer:

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Explanation:

Given data

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As we know that:

dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\    (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s

5 0
3 years ago
A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done in the process
ivann1987 [24]
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4 0
2 years ago
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