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Temka [501]
3 years ago
10

What does saliva begin to break down in the mouth? answers QUICK I will make you brainst !

Physics
2 answers:
cestrela7 [59]3 years ago
5 0

Answer:

complex carbohydrates

Explanation:

Andrew [12]3 years ago
3 0

An enzyme called amylase breaks down starches (complex carbohydrates) into sugars, which your body can more easily absorb. Saliva also contains an enzyme called lingual lipase, which breaks down fats.

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When atoms of an element are excited, they emit specific wavelengths of light. How is this similar to a fingerprint when Fraunho
Anika [276]

Answer:

As you may know, each element has a "fixed" number of protons and electrons.

These electrons live in elliptical orbits around the nucleus, called valence levels or energy levels.

We know that as further away are the orbits from the nucleus, the more energy has the electrons in it. (And those energies are fixed)

Now, when an electron jumps from a level to another, there is also a jump in energy, and that jump depends only on the levels, then the jump in energy is fixed.

Particularly, when an electron jumps from a more energetic level to a less energetic one, that change in energy must be compensated in some way, and that way is by radiating a photon whose energy is exactly the same as the energy of the jump.

And the energy of a photon is related to the wavelength of the photon, then we can conclude that for a given element, the possible jumps of energy levels are known, meaning that the possible "jumps in energy" are known, which means that the wavelengths of the radiated photons also are known. Then by looking at the colors of the bands (whose depend on the wavelength of the radiated photons) we can know almost exactly what elements are radiating them.

7 0
3 years ago
Help me, i don't know the answer ​
8_murik_8 [283]

Explanation:

1)

A) Bb BB

B) 50%

2)

A) 50%

B) <u> </u><u> </u><u> </u><u> </u><u> </u><u>b</u><u>.</u><u> </u><u> </u><u> </u><u>b</u>

B. Bb. Bb

b. bb. bb

4 0
2 years ago
A petrol tanker ha 2800kg when empty and hold 30m3 0f petrol when full. The denity of petrol i 740kg/m3. Calculate the ma of the
White raven [17]

The mass of the tanker with petrol is 250000 N.

We are given that,

Mass of tanker= m =2800 kg

Volume of petrol= v =30 m³

Density of petrol= d =740 kg/m³

Thus , mass , density and volume relation can be given as,

density= mass/ Volume

Mass = Density × Volume

Mass = 740× 30

Mass = 22200 kg

The mass of the petrol is 22200 kg.

Total mass of tanker with petrol = Mass of petrol + Mass of tanker

Total mass of tanker with petrol= 22200+ 2800 kg

Total mass of tanker with petrol= 25000 kg

Total weight of the tanker with petrol = Mass of tanker full of petrol× g

Where, weight = m × g ,(g =10m/s²)

Total weight of the tanker with petrol= 25000×10 = 250000 N

Therefore, the mass of petrol, total mass of tanker with petrol and weight of tanker with petrol would be  22200 kg, 25000 kg and 250000 N.

To know more about mass

brainly.com/question/24100921

#SPJ4

4 0
1 year ago
Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200
andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

                                   ΔK.E = ΔP.E

                                  K_2 - K_1 = P_1- P_2

Where,  K_2: Kinetic energy of hammer head as it hits the I-beam

             K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

             P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

             P_1: Initial gravitational potential energy of hammer head      

- The expression simplifies to:

                                K_2 = P_1

Where,                     0.5*m*v2^2 = m*g*s12

                                v2 = √(2*g*s12) = √(2*9.81*3)

                                v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

                               Wnet = ( P_3 - P_1 ) + W_friction

                               Wnet = m*g*s13 + F*s23

                               n*s23 = m*g*s13 + F*s23

Where,    n: average force the hammerhead exerts on the I-beam.

               s13 = s12 + s23

Hence,

                             n = m*g*( s12/s23 + 1) + F

                             n = 200*9.81*(3/0.074 + 1) + 60

                             n = 81562 N

                               

                                                   

6 0
3 years ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
3 years ago
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