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BARSIC [14]
3 years ago
15

Julie wants to know if adding calcium fertilizer to tomato plants will help prevent a disease of tomato plants called blossom-en

d rot. She uses 10 tomato plants. She put 5 plants in a sunny part of the garden and gives each one calcium fertilizer once a week. She places the other 5 plants in a shady part of the garden and gives them calcium fertilizer twice a week. She observes the 10 plants for 3 months to see if the fertilizer helped prevent blossom-end rot. Based on her experiment design, Julie's results are probably reliable. True False
Physics
1 answer:
Ksivusya [100]3 years ago
3 0
It's not true.

The crucial principle for a scientific experiment is to keep only ONE variable at a time.

In this case, the variable of this experiment is actually the tomato is in sunny part or in shady part, instead of whether applying Ca fertilizer. 
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A +0.2 µC charge is in an electric field. What happens if that charge is replaced by a +0.4 µC charge?
AleksandrR [38]

Explanation:

It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.

So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.

                 U = qV

Since, the electric potential is directly proportional to the charge. Hence, when 0.2 \mu C tends to replaced by 0.4 \mu C then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.

Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.

4 0
3 years ago
Forces that are opposite and equal are
Kisachek [45]

Answer:

space

Explanation:

6 0
3 years ago
Acceleration changes and objects...
brilliants [131]

Answer:

speed and direction

Explanation:

Acceleration is the rate of change of speed and direction.

3 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8
Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) \tau = \dfrac{\Delta L}{\Delta t}

   \tau = \dfrac{0.8-3}{1.5}

        τ = -1.467 N m

b) angle at which fly wheel will turn

   \theta= \omega t +\dfrac{1}{2}\alpha t^2

   \theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2

   \theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2

        θ = 20.35 rad

c) work done on the wheel

     W = τ x θ

     W = -1.467 x 20.35 rad

    W = -29.86 J

d) average power of wheel

    P_{av} =-\dfrac{W}{t}

    P_{av} =-\dfrac{(-29.86)}{1.5}

     P_{av} =19.91\ W          

7 0
3 years ago
Read 2 more answers
Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle
Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

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

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

T = 1.812x10^{7} Km

\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU

T = 0.120 AU

T = \frac{0.120 AU}{1 AU} . 1 year

T = 0.120 year

T = \frac{0.120 year}{1 year} . 365.25 days  

T = 43.83 days

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

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

v = \frac{2 \pi (69000 Km)}{(43.83 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

43.83 days . \frac{86400 s}{1 day} ⇒ 3786912 s

69000 Km . \frac{1000 m}{ 1 Km} ⇒ 69000000 m

v = \frac{2 \pi (69000000 m)}{(3786912 s)}

v = 114.483 m/s

 

\frac{114.483 m/s}{81.778 m/s} = 1.399            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

7 0
3 years ago
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