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3 years ago

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What are some ways that scientists would collect data and make observations to help them learn more about the severity of the li

on fish problem

1 answer:

nika2105 [10]3 years ago

6 0

Experiments, tests, and trials

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a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and

Klio2033 [76]

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately $\left(-6.0\; \rm m \cdot s^{-2}\right)$ .

It took the bus approximately $1.7\;\rm s$ to come to a stop.

Explanation:

Quantities:

Displacement of the bus: $x = 10\; \rm m$ .
Initial velocity of the bus: $\displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}$ .
Final velocity of the bus: $v = 0\; \rm m\cdot s^{-1}$ because the bus has come to a stop.
Acceleration, $a$ : unknown, but assumed to be a constant.
Time taken, $t$ : unknown.

Consider the following SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t$ .

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

$\displaystyle t = \frac{v - u}{a}$ .

Therefore, replace the quantity $t$ with the expression $\displaystyle \left(\frac{v - u}{a}\right)$ in that SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right)$ .

Simplify this equation:

$\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}$ .

Therefore, $\displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)$ .

In this question, the value of $x$ , $u$ , and $v$ are already known:

$x = 10\; \rm m$ .
$\displaystyle u =10\; \rm m \cdot s^{-1}$ .
$v = 0\; \rm m\cdot s^{-1}$ .

Substitute these quantities into this equation to find the value of $a$ :

$\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}$ .

(The value of acceleration $a$ is less than zero because the velocity of the bus was getting smaller.)

Substitute $a \approx -6.0\; \rm m \cdot s^{-2}$ (alongside $u = 10\; \rm m \cdot s^{-1}$ and $v = 0\; \rm m \cdot s^{-1}$ ) to estimate the time required for the bus to come to a stop:

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}$ .

8 0

3 years ago

The positions of four objects as a function of time are shown

KengaRu [80]

There is no movement in line C and the greatest velocity occurs at line D. The answers are:

1. 0.5 m/s

2. 0.25 m/s

3. 14m and -2m

4. -1 m/s

<h3>What is Position - time Graph ?</h3>

Position time graph is the graph of distance or displacement against time. The slope of the graph is velocity.

The given positions of four objects as a function of time are shown

on the graph to the right.

1.) The velocity of object A will be the slope m of the line A.

Slope m = Δx / Δt

m = (4 - 0) / (8 - 0)

m = 4 / 8

m = 0.5 m/s

Velocity at A = 0.5 m/s

2.) The average velocity of object B will be the slope m of the line B.

Slope m = Δx / Δt

m = (6 - 4) / (8 - 0)

m = 2 / 8

m = 0.25 m/s

The average velocity of object B is 0.25s

3.) The object moved a total distance during the first eight seconds will be 4m for A, 2m for B, and 8m for D

Total distance = 4 + 2 + 8 = 14m

It’s net displacement during the same time will be 2. That is,

Displacement = 8 - 6 = -2m

4.) The greatest speed occurred at line D. The velocity of the object moving at the greatest speed will be the slope of the line D

V = -Δx / Δt

V = -8/8

V = -1 m/s

Therefore, there is no movement in line C and the greatest velocity occurs at line D.

Learn more about velocity time graph here :brainly.com/question/769606

#SPJ1

6 0

1 year ago

What do you think are some differences between sound waves traveling through air versus traveling through water?

mel-nik [20]

Answer:

in water its more slower because of the liquid matter but in air its a gas formaiton so its more quicker

Explanation:

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3 years ago

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The model of the atom proposed by Greek philosophers appears similar to the model proposed centuries later by Dalton. What was t

Sladkaya [172]

The main difference between the model of the atom proposed by Greek philosophers and the model proposed centuries later by Dalton is that the Greek one was mainly speculative and philosophical - it wasn't based on real evidence, but on their suggestions and thoughts about the matter. On the other hand, Dalton had the means to prove his theory using viable evidence, not just speculations.

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3 years ago

What causes the granular structure on the sun's photosphere?

Charra [1.4K]

Something to do with how the suns magnetic field interacts with the surface plasmas I think.

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3 years ago