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zhenek [66]
3 years ago
15

Lizette works in her school’s vegetable garden. Every Tuesday, she pulls weeds for 15 minutes. Weeding seems like a never-ending

task. Each time Lizette goes to the garden, there are just as many weeds as the week before!
Lizette’s teacher suggests that she use a strong solution of vinegar to kill the weeds. Vinegar is acidic and prevents plants from maintaining homeostasis. Lizette sets up a controlled experiment to test her hypothesis that the solution will kill the weeds without harming nearby vegetables. She plans to spray one group of weeds with the solution and another group of weeds with water as a control.

What variables should Lizette keep the same between the control group of weeds and the sprayed weeds?
Physics
2 answers:
Gnoma [55]3 years ago
8 0

Answer:

Constant or Controlled variables: Same concentration of vinegar solution, same quantity of vinegar, same type of weed etc

Explanation:

In an experiment, certain variables are kept unchanged or constant for both the experimental group and control group in order not to influence the outcome of the experiment. These variables are called CONTROLLED VARIABLES or CONSTANTS.

In the case of this experiment where Lizette is testing the effect of vinegar on weed, the variable that should be kept the same (controlled variables) for the control group of weeds and the sprayed weeds include Same concentration of vinegar solution, Same quantity of vinegar, same type of weed.

lorasvet [3.4K]3 years ago
7 0

Answer:

Constant or Controlled variables: Same concentration of vinegar solution, same quantity of vinegar, same type of weed etc

Explanation:

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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
The wetter the surface, the friction
allochka39001 [22]
Answer: Water can either increase or decrease the friction between surfaces.
7 0
3 years ago
How does the thermoflask prevent heat loss​
riadik2000 [5.3K]

Answer:

The way that the flask is built it has 3 protective layers.... the inside layer to keep the heat in, the outside layer to reflective the cold, and a vacuum layer, which is an empty layer that limits conduction and convection

Explanation:

3 0
3 years ago
Two multiple choice questions.
m_a_m_a [10]
C and c infeijnveirnvinefine
4 0
3 years ago
Read 2 more answers
The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the
Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

r = radius of the orbit of moon around the earth = 3.85\times10^{8} m

M = Mass of earth = 5.98\times10^{24} m

T = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}

inserting the values, we get

T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days

1 month = 30 days

T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month

6 0
3 years ago
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