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postnew [5]
3 years ago
11

Chet plans an experimental investigation to see how well a new fertilizer works on daisies. The fertilizer must be dissolved in

water and delivered to the plants by spraying it onto the leaves and stems.
To set up his control group for the experiment, Chet should leave out what
Physics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

D Fertilizer hope this helps ..

Explanation:

IceJOKER [234]3 years ago
5 0

Answer:

The answer is D or Fertilizer, sorry for the late answer

Explanation:

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1) 4.8
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A 645-turn coil with a 20.250 m ​2 ​​ area is spun in Earth’s 5.00×10 ​−5 ​​ T magnetic field, producing a 1.25-V maximum emf. A
Dmitriy789 [7]

Answer:

\omega = 1.914\ rad/s

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

\varepsilon = NAB\omega

\omega= \dfrac{\varepsilon}{NAB}

\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}

\omega = 1.914\ rad/s

Angular velocity of the coil = \omega = 1.914\ rad/s

5 0
4 years ago
Which of the following is a consequence of the special theory of relativity?
aleksandr82 [10.1K]

Answer:

D.

Explanation:

Specifically, Special Relativity showed us that space and time are not independent of one another but can be mixed into each other and therefore must be considered as the same object, which we shall denote as space-time. The consequences of space/time mixing are: time dilation. and length contraction.

3 0
2 years ago
Where is the sun located in the galaxy?
GarryVolchara [31]
I think it's 3. within an outer arm
6 0
3 years ago
Read 2 more answers
What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J
Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)

and on the surface the current is

I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A

(b)

For Jb = J0(1 - r/R)

I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )

and on the surface

I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi  \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A

and in a simmilar way for Jb

I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi   J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0
3 years ago
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