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3 years ago

When should a line graph be used?

Physics

1 answer:

OLEGan [10]3 years ago

4 0

Answer:

A. When the independent variable is continuous and shows a causal link to the dependent variable.

Explanation:

A graph can be defined as the graphical representation of data (informations) on horizontal and vertical lines i.e x-axis and y-axis respectively.

In an experiment , the variable being manipulated by an experimenter is known as an independent variable while the dependent variable is the event expected to change when the independent variable is manipulated

Generally, a line graph should be used when the independent variable is continuous and shows a causal link to the dependent variable.

This ultimately implies that, a line graph should be used when the data changes continuously over time and as such there exist a linear relationship between the data (variables).

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

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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.

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3 years ago

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A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two

stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

$T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s$

FOR THE WAVE TRAVELING THROUGH AIR:

$T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s$

The separation in time between two pulses can now be given as follows:

$\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\$

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3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

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4 years ago

As an emergency vehicle approaches Bob and moves away from Jill, how does the actual frequency of the siren change? A) As an eme

Tamiku [17]

The correct answer is:
<span>C) The actual frequency of the siren does not change despite appearances.

In fact, Bob will observe an increase in the apparent frequency as the emergency vehicle approaches him, while Jill will observe a decrease in the apparent frequency as the emergency vehicle moves away from him, because of the Doppler effect (the relative velocity between the observer and the source of the sound is changing), but this effect involves the apparent frequency, while the real frequency of the siren will remain the same.</span>

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vovangra [49]

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Explanation:

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