Ask question

Ask question

All categories

2 years ago

11

while looking at a graph, you notice a period of time where the line is perfectly horizontal what is most likely taking place du

ring this time period brainly

1 answer:

aev [14]2 years ago

7 0

Answer:

where the y axis is

Explanation:

In more simple terms, a horizontal line on any chart is where the y-axis values are equal. If it has been drawn to show a series of highs in the data, a data point moving above the horizontal line would indicate a rise in the y-axis value over recent values in the data sample.

You might be interested in

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

Svet_ta [14]

Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

E $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3

so if R become 2R

E becomes 1/8 of the initial electric field.

8 0

2 years ago

Read 2 more answers

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?

MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

<h3><u>Data:</u></h3>

v = 4.6 m/s
d = ¿?
t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:

$\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }$

$\bf{d=46 \ m}$

Therefore, the speed at 10 seconds is 46 meters.

$\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}$

6 0

2 years ago

What will happen if you light up a candle that is in a beaker?

salantis [7]

It will melt so it is c

8 0

3 years ago

If f(x)=4/x+2 and g is the inverse of f,then g'(10)=

ch4aika [34]

Answer:

g'(10) = $\frac{-1}{16}$

Explanation:

Since g is the inverse of f ,

We can write

g(f(x)) = x <em> </em><em>(Identity)</em>

Differentiating both sides of the equation we get,

g'(f(x)).f'(x) = 1

g'(10) = $\frac{1}{f'(x)}$ --equation[1] Where f(x) = 10

Now, we have to find x when f(x) = 10

Thus 10 = $\frac{4}{x}$ + 2

$\frac{4}{x}$ = 8

x = $\frac{1}{2}$

Since f(x) = $\frac{4}{x}$ + 2

f'(x) = - $\frac{4}{x^{2} }$

f'( $\frac{1}{2}$ ) = -4 × 4 = -16

Putting it in equation 1, we get:

We get g'(10) = - $\frac{1}{16}$

5 0

3 years ago