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3 years ago

During any process, the net electric charge of an isolated system does not change.

1 answer:

7 0

<span>The law of conservation of electric charge states that the total electric charge of the universe is constant, so during any process the net electric charge of an isolated system remains constant.</span>

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an underground cannon launches a cannonball from ground level at a 35-degree angle. the cannonball is shot with an initial veloc

Allushta [10]

Answer:

$\displaystyle x_{max}=21.57\ m$

$\displaystyle y_{max}=3.78\ m$

Explanation:

<u>Motion in Two-Dimensions </u>

When an object is launched with a certain angle $\theta$ above ground level with an initial velocity $\vec v_o$ , it describes a curve called a parabola, defined by the force of gravity which will eventually make the object return to the ground. There are two overlapping motions, the horizontal motion, at a constant speed, and the vertical motion, at changing speed because the acceleration of gravity modifies it. The velocity $\vec v_o$ is split into two components x,y

$\displaystyle v_{ox}=|vo|\ Cos\theta$

$\displaystyle v_{oy}=|vo|\ Sin\theta$

The position of the object is also split into its components, assuming the object was launched from ground level

$\displaystyle x=v_{ox}.t$

$\displaystyle y=v_{oy}.t-\frac{g.t^2}{2}$

The maximum horizontal distance the object reaches (called range) is

$\displaystyle x_{max}=\frac{2v_{ox}\ v_{oy}}{g}$

The maximum height is given by

$\displaystyle y_{max}=\frac{v_{oy}^2}{2g}$

The question doesn't ask for anything in particular, but to guide you in the solution of your own problem, we'll compute $X_{max}$ and $Y_{max}$ for you. The data is

$\displaystyle |vo|=15\ m/s\ ,\ \theta =35^o$

Let's compute the range

$\displaystyle x_{max}=\frac{(2)(15)\ cos35^o\ (15)\ sen35^o}{9.8}$

$\displaystyle x_{max}=\frac{211.43}{9.8}=21.57\ m$

Now for the maximum height

$\displaystyle y_{max}=\frac{(15\ . \ Sin35^o)^2}{2(9.8)}$

$\displaystyle y_{max}=\frac{74.0227}{19.6}$

$\displaystyle y_{max}=3.78\ m$

6 0

3 years ago

What personal experiences have you had with watching a substance change from one phase to another?

Dmitry_Shevchenko [17]

Watching ice melt on a hot surface.

6 0

3 years ago

During which season at what latitude does the sun not set

Allisa [31]

The sun not set in the summer and greater than 66 N

I hope that's help !

8 0

3 years ago

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You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital spee

Alchen [17]

Answer:

The minimum total speed is 11.2km/s

Explanation:

We are been asked to find the escape velocity.

Escape velocity is defined as the minimum initial velocity that will take a body(projectile)away above the surface of a planet(earth) when it's projected vertically upwards.

The formula to calculate the escape velocity is Ve = √2gR

For the earth g = 9.8m/s2 , R = 6.4*10^6

Substituting into the equation Ve = √2*9.8*6.4*10^6 = 11.2*10^3m/s

=11.2km/s

5 0

3 years ago

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The diagram shows the electric field around two charged objects.

anyanavicka [17]

Here we have been given two charged body W and X.

We are asked to determine the nature of charge.

Before coming into conclusion, first we have to understand the electric field lines.

The electric field lines are pictorial representation of imaginary lines which are drawn to denote the electric field in a graphical way.

The electric field line starts from a positive charge and ends at a negative charge.It means for positive charges the electric field lines are outward and for negative charges the filed lines will be inward.

In the given diagram,the filed lines for W is towards W itself.The same is also in case of X.

Hence both the charges must be negative in nature.

Hence the correct answer to the question will be B i.e W negative and X: NEGATIVE.

8 0

3 years ago

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