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3 years ago

12

Is it plagiarism/copying to do a science fair project that is similar to one that has been done already, if you modify certain a

spects and you do not know the results?

2 answers:

Yanka [14]3 years ago

5 0

It is not as plagiarism as it is different project. You are even allowed to repeat past experiments, although you should have a category in your presentation of "background research". Science fair projects are about obtaining research on your own through you personal experiments. It sounds acceptable, but it would be advisable to double check it with your teacher or supervisor.

Ierofanga [76]3 years ago

3 0

If you cite your sources correctly and make sure to credit everything I'm p sure it's cool. You might lose points on originality but it depends on the teacher

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You can answer this to get points its not an actual question

larisa86 [58]

Answer:

the electromagnetic pulse

Explanation:

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3 years ago

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A horse walks along a straight path. it travels 3 in one second, 4 m in the next second, and 5 m in the third second. describe t

Temka [501]

Answer: The horse is moving with a uniform acceleration

Explanation:

According to the described situation, we are dealing with a <u>constant acceleration</u> (also called <u>uniform acceleration</u>), since the horse's velocity is changingn by a constant rate.

Let's prove it:

Firstly we are told the horse follows a straight path. In addition we are given its velocity:

3 m/s, then 4 m/s and 5 m/s

Since acceleration $a$ is defined as the change of velocity in time we can calculate it:

$a\frac{4 m/s-3 m/s}{1 s}=1m/s^{2}$ This is the horse's constant acceleration

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3 years ago

An object experiences an acceleration of -6.8m/s^2. As a result, it accelerates from 54m/s to a complete stop. how much distance

Kipish [7]

Answer:

210 m

Explanation:

Given:

a = -6.8 m/s²

v₀ = 54 m/s

v = 0 m/s

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (54 m/s)² + 2 (-6.8 m/s²) Δx

Δx ≈ 210 m

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3 years ago

at 25°c three moles of gas occupy a volume of 4.0L at 1.9 kpa. if the volume increases to 8.0L what will the new pressure be

ki77a [65]

Answer:0.95kPa

Explanation:

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3 years ago

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s$

D) 5.6 rad

The angular displacement of the wheel is given by the equation

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where we have

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel

$\omega_f = 4.00 rad/s$ is the final angular velocity

$\alpha=+1.67 rad/s^2$ is the angular acceleration

Solving for $\theta$ ,

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad$

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3 years ago