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Sergio [31]
3 years ago
15

The following chart provides the pH of some common foods. How many would be considered basic?

Physics
1 answer:
Yuri [45]3 years ago
6 0

The ones with pH greater than 7

Explanation:

A pH scale is a scale for expressing the level of acidity or alkalinity of aqueous solutions.

This scale is called a pH or pOH scale. The pH is the common one.

pH of a solution is the negative logarithm to base 10 o the hydrogen ion concentration of the solution.

  • The scale ranges from 1 through 14.
  • An acidic solution has a pH value less than 7. Neutral solutions have pH of 7 and basic solutions have pH greater than 7.

learn more:

acidity brainly.com/question/5121777

#learnwithBrainly

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Aloiza [94]
10/70×360°
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3 years ago
According to this equation, F=ma, how much force is needed to accelerate an 82-kg runner at 7.5m/s2?
Murljashka [212]
You multiply the mass by the acceleration 82*7.5=615; that's what I would do
5 0
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How did cyanobacteria affect early Earth’s biosphere? Select all that apply. Cyanobacteria allowed organisms that rely on oxygen
FrozenT [24]

Answer:

The correct statements would be

  • Cyanobacteria allowed organisms that rely on oxygen to evolve.
  • Cyanobacteria preceded the first photosynthetic organisms.
  • Cyanobacteria produced excess oxygen.

Cyanobacteria, also termed as blue-green algae are the prokaryotes which are able to perform photosynthesis.

They were the major contributors of oxygen in the atmosphere and thus helped the organisms that rely on oxygen to evolve.

By the process of endosymbiosis, they lead to the origin of plants. The chloroplasts present in green plants is considered as the cyanobacteria living in the plant cell. It helps in photosynthesis and in return plants cell provides shelter to it.

It is believed that the oxygen released from early cyanobacteria reacted with dissolved iron ions to form iron oxide.

7 0
3 years ago
The magnitude of the magnetic field B a distance r from a long straight wire is B = μ 0 I 2 π r where μ 0 is the permeability co
PSYCHO15rus [73]

Answer:

The magnetic field is lowest for largest distance and highest when distance is least.

Explanation:

The magnitude of magnetic field strength at a distance 'r' from a long straight wire carrying current 'I' is given as:

B=\frac{\mu_0 I}{2\pi r}\\Where,\mu_0\to permeability\ constant\ of\ free\ space

Now, as per question, the distance 'r' is varied while keeping the current constant in the wire.

As seen from the above formula, the magnitude of magnetic field strength for a constant current varies inversely with the distance 'r'. This means that, as the value of 'r' increases, the magnitude of magnetic field strength decreases and vice-versa.

Therefore, the magnitude of magnetic field strength is maximum when the distance 'r' is least and the magnetic field is minimum for the largest distance.

Example:

If B_1, B_2,\ and\ B_3 are the magnitudes of magnetic field strengths for distances r_1,r_2, \ and\ r_3 respectively such that r_1. Now, as per the explanation above, the order of magnitudes of magnetic field strength is:

B_1>B_2>B_3

6 0
3 years ago
When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted p
jarptica [38.1K]

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

7 0
3 years ago
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