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3 years ago

What does "Time (s)” represent in the graph? the y-axis the dependent variable an inverse relationship the independent variable

Physics

2 answers:

MArishka [77]3 years ago

7 0

Answer: Time represents independent variable.

Explanation:

Independent variable: These are the variables which not dependent on other variables. Independent variables are used to see the variation in dependent variables. For example age is independent variable.

Time is the example of Independent variable as we cannot fix time or we can say that time cannot stopped. It is variable which doesn't get affected by the other variables.

Dependent variable : These are the variables which are depends upon the independent variables. For example height is a dependent variable.

san4es73 [151]3 years ago

6 0

Independent, and inverse means if one variable increases then the other decreases, direct is when they both go up or down together.

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It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

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3 years ago

Read 2 more answers

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

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the correct answer is option E

4 0

3 years ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago