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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:

You might be interested in
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
3 years ago
If the mass of a material is 42 grams and the volume of the material is 15 cm^3, what would the density of the material be?
Fantom [35]

Density = mass / volume

Density = 42g / 15cm^3

Density = 2.8g/cm^3

3 0
3 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

F_c = F_g

\frac{mv^2}{r} = \frac{GmM}{r^2}

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

v^2 = \frac{GM}{r}

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

v = \frac{2\pi r}{T}

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

v^2 = \frac{GM}{r}

( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

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

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})

T= 1.74hour

Therefore the correct answer is C.

7 0
3 years ago
You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the
lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

\Delta p=p_2-p_1

The momentum is computed as

p=mv

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

p_1=mv

When it hits the wall, it sticks, thus its final speed is 0 and

p_2=0

The change of momentum is

\Delta p=0-mv=-mv

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

p_2=-mv

The change of momentum is

\Delta p=-mv-mv=-2mv

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0
3 years ago
Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016 Na+ ions arrive at the negative electrode and 3.
sladkih [1.3K]

Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = \frac{q_{1}}{t} = \frac{0.004288}{1} = 0.004288 A

i₂ = Current due to Cl⁻ ions = \frac{q_{2}}{t} = \frac{0.006272}{1} = 0.006272 A

Current passing between the electrodes is given as

i = i₁ + i₂

i = 0.004288 + 0.006272

i = 0.01056 A

i = 10.6 x 10⁻³ A

i = 10.6 mA

8 0
3 years ago
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