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zhenek [66]
2 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]2 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]2 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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For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear dep
Sladkaya [172]

Answer:

V=-130000x+1080000

Explanation:

<u>Linear Dependence </u>

Some variables are known or assumed to have linear dependence which means the graph of the ordered pairs (x,V) is a straight line.

If we know two points of the line, we can come up with the exact equation and therefore make predictions for other values of x

The linear depreciation gives us these points (2,820000) and (5,430000)

The general equation of the line is

V=mx+b

Where V is the machine value and x is the  number of years after purchase. We need to find the values of m and b.

Replacing the first point

820000=m(2)+b

2m+b=820000

Replacing the second point

5m+b=430000

Subtracting them  

-3m=390000

m=-130000

Replacing in any of the equations, say, the first one

2(-130000)+b=820000

Solving for b

b=820000+260000

b=1080000

The formula for the machine value V is  

\boxed{V=-130000x+1080000}

7 0
2 years ago
A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

4 0
2 years ago
A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i
OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

7 0
2 years ago
Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the
enyata [817]

<u>Answer</u>

27.7


<u>Explanation</u>

The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.

cosx = adjacent/hypotenuse

cos 30 = adjacent / 10

adjacent = 10 cos30

             = 8.66 m/s        ⇒ This is the horizontal speed.

Now  find the horizontal distance.

Distance = speed × time

               = 8.66 × 3.2

                = 27.71

Answer to the nearest tenth = 27.7

4 0
2 years ago
Read 2 more answers
If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (
valkas [14]
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
     = 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
8 0
2 years ago
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