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sammy [17]
3 years ago
9

A new school is being built in a field of wildflowers. A class wanted to develop a research project to predict the effects of th

e new school on wildflower species. When would be the best time for the class to conduct the study?
Physics
1 answer:
kotykmax [81]3 years ago
4 0
A new school is being built in a field of wildflowers. A class wanted to develop a research project to predict the effects of the new school on wildflower species. Given the background of the class, then probably, the best time for them to do the study is during right after the school is being built. Through that, they can get some data how the school affects the flowers. So, this helps the flowers in the future.


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Identify each statement as an example of melting or sublimation,
xeze [42]

1st is sublimation

2and is melting

3red is melting

4th is sublimation

sublimation is just "skipping" the liquid phase / state

5 0
3 years ago
The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th
DaniilM [7]

Answer:

\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}

Explanation:

The vertical component of the initial velocities are

v_v = v_0sin\theta

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

v_vt - gt^2/2 = s = 0

t(v_v - gt/2) = 0

v_v - gt/2 = 0

t = 2v_v/g = 2v_0sin\theta/g

So the ratio of the times of the flights is

t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}

8 0
3 years ago
A ship anchored at sea is rocked by waves that have crests 14 m apart the waves travel at 7.0 m/s how often do the wave crest re
xeze [42]

Answer:

I think its 2 seconds

Explanation:

14/7

3 0
3 years ago
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Tju [1.3M]
I feel like it could be A
8 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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