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2 years ago

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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?

1 answer:

kotykmax [81]2 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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The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

6 0

3 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

3 years ago

Easy question. Answer should include one line reasoning

KATRIN_1 [288]

C.figure 3 is the answer had the same and got is right

7 0

2 years ago

Read 2 more answers

Which of the following provides evidence that Earth's Moon is rotating and revolving at the same rate?

kkurt [141]

B. We can see only one side of the Moon from Earth.

( we only see one side of the moon because the moon rotates around the Earth)

6 0

2 years ago

tickets at a museum cost 17 dollars each for a field trip, the museum offers a 4 doller discount on each ticket. How much will t

Marizza181 [45]

Answer:

$416 sounds like the best answer.

Explanation:

The tickets started off at $17, but a $4 discount for EACH ticket bought.

so 17-4 is 13

now tickets cost $13 each.

Multiply it by 32.

13x32=416

5 0

2 years ago