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mixer [17]
3 years ago
5

Lana wanted to do an investigation to see what kind of birdseed the birds in her neighborhood like best. She purchased three of

the same kind of bird feeder and three different kinds of birdseed. Next, she put one kind of birdseed in each of the feeders and placed them near each other outside. Over the next three days, she measured the level of seed in each feeder twice a day. What is the dependent variable in Lana's investigation?
Physics
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

The level of seed in each feeder

Explanation:

The independent variable is the variable which is the suspected cause of an observation, it is the variable that produce the effect observed in the dependent variable. The dependent variable is the variable that is measured

The independent variable is normally the x-value while the dependent variable is the y-value

In the question, Lana wants to find out the kind of birdseed that the neighborhood birds like (to eat) the most by feeding them different types of birdseed and measuring the level of the birdseed in the feeders at regular intervals

the independent variable is the bird seed types which Lana gives to the birds, to determine the type of bird seeds the birds like

The dependent variable is the level of the seed in the feeder which depends on the type of seeds the birds like.

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A stone that is thrown vertically upwards was having a velocity of 15m/s after reaches 2/3 of its maximum height. What is the ma
attashe74 [19]

<em>The correct answer is option</em><em> B.</em> The maximum height that can be reached by the stone is determined as 11.5 m.

<h3>Maximum height attained by the stone </h3>

The maximum height attained by the stone when it is a 2/3 of its total height is calculated as follows;

v² = u² - 2gh

where;

  • v is final velocity at maximum height, v = 0
  • u is initial velocity
  • g is acceleration due to gravity

0 =  u² - 2gh

2gh =  u²

h =  u²/2g

h = (15²)/(2 x 9.8)

h = 11.48 m

h = 11.5 m

Thus, the maximum height that can be reached by the stone is determined as 11.5 m

Learn more about maximum height here: brainly.com/question/12446886

#SPJ1

4 0
2 years ago
2500000000000 in scientific notation
Alex_Xolod [135]
Your answer is 2.5 x 10^(11).

Please take the time to rate and thank me if you feel I helped you! Thanks, it means a lot to me. :)
4 0
3 years ago
Read 2 more answers
What is a plane mirror ? ​
g100num [7]

Answer:

A plain mirrior is a mirrior with flat reflective surface.

hope it is helpful for you.

6 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
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