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weeeeeb [17]
3 years ago
14

Gardeners understand that plants have specific pH requirements. Most plants thrive in soil with a pH that is slightly acidic, ne

utral, or slightly basic. Imagine your soil is very acidic, with a pH of 3.5. How can you raise the pH to an appropriate level so your garden plants will survive?
A) Add fertilizer.
B) Water frequently to wash away the acid.
C) Add organic matter like peat moss or humus.
D) Add a base like limestone or calcium carbonate.
Physics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

D) Add a base like limestone or calcium carbonate.

Explanation:

-To reduce the soil's pH, you need to add an alkaline or base solution.

-Calcium carbonate has an alkanizing property and forms a carbonic acid and calcium hydroxide when dissolved in water.

-The solution will reduces the soil's pH.

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create a formula giving the strength of the induced field (B) i terms of current (I) and the distance from the wire to the probe
iren [92.7K]
So we need to find the formula for magnetic field B using the current (I) and the distance from the probe (d). So, We know that the stronger the current I, the stronger the magnetic field B. That tells us that the I and B are proportional. Also we know that the strength of the magnetic field B is weaker as the distance d of the probe increases. That tells us that B and d are inversely proportional. So our formula should have B=(I/d)*c where c is a constant of proportionality. c=μ₀/2π where μ₀ is the permeability of free space. So finally our formula is B=(μ₀I)/(2πd). 
6 0
2 years ago
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
podryga [215]

Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

8 0
3 years ago
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
Earth, has a bigger mass than the Moon .
raketka [301]
Decreased it because you can float a lot
5 0
2 years ago
In the diagram, R1 = 40.0 ,
Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0
2 years ago
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