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3 years ago

10

Based on the graph what kind of relationship exists between the change in temperature of the water in the gravitational potentia

l energy of the cylinder
a. it is an inverse relationship
b. it is a direct relationship
c. it is an indirect relationship
d. it is a cyclic relationship

2 answers:

Kisachek [45]3 years ago

8 0

Answer:

The answer is C

Explanation:

just got it right on edge

if this helped leave a thanks and go follow me

Musya8 [376]3 years ago

6 0

Answer: It is a direct relationship

Explanation: I just did that question.

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shannon dropped his ICP textbook. how fast was it going after 3 seconds? how far did it fall? please guys help mee quickly pleas

Mekhanik [1.2K]

¡Hola!

For this problem, first, lets recabe information:

v = 0 m/s

t = 3 s

g = 9,82 m/s²

v' = ?

d = ?

First, for calculate the final velocity:

v' = v + g * t

v' = 0 m/s + 9,82 m/s² * 3 s

v' = 29,46 m/s $\leftarrow \boxed{\text{Velocity} }$

Now, for calculate how far did it fall:

d = v * t + g * t^2 / 2

Like v = 0 m/s, we can simplificate equation:

d = g * t^2 / 2

d = 9,82 m/s² * (3 s)^2 / 2

d = 9,82 m/s² * 9 s² / 2

d = 88,38 m / 2

d = 44.19 m $\leftarrow \boxed{\text{Distance} }$

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3 years ago

A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t

Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

Electric field $E=7.80\times10^{2}\ N/C$

We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

We need to calculate the acceleration

Using newton's law

$F = ma$

$a = \dfrac{F}{m}$

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m

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Leokris [45]

Answer:

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explanation:

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