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2 years ago

Why does food need to be broken down (made into chyme) in the stomach (Think about the next step in digestion)

2 answers:

8 0

Answer: Carbohydrates , proteins and lipids are large molecules that are needed by the body for growth, repair and metabolism . They are found in our food. These molecules are too large to pass from the intestine into the blood, so digestive enzymes break them down into smaller molecules.

Explanation:

tatuchka [14]2 years ago

3 0

Answer:

Digestion is the mechanical and chemical break down of food into small organic fragments. It is important to break down macromolecules into smaller fragments that are of suitable size for absorption across the digestive epithelium. Large, complex molecules of proteins, polysaccharides, and lipids must be reduced to simpler particles such as simple sugar before they can be absorbed by the digestive epithelial cells. Different organs play specific roles in the digestive process. The animal diet needs carbohydrates, protein, and fat, as well as vitamins and inorganic components for nutritional balance. How each of these components is digested is discussed in the following sections

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An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

Alexandra [31]

Answer:

a) $\omega _1=7.56rev/s=>47.5rads/s$

b) $a=30.7$

c) $a=35.91$

Explanation:

From the question we are told that:

Initial angular velocity $\omega _1=7.56rev/s=>47.5rads/s$

Initial Length $L_1=0.600m$

Final angular velocity $\omega _2=6.22rev/s=39rad/s$

Final Length $L_2=0.900m$

Generally the rotation with the greater speed is

$\omega _1=7.56rev/s=>47.5rads/s$

Generally the equation for centripetal acceleration at 8.16 is mathematically given by

$a=\omega_1^2*L_1$

$a=8.16 rev/s*0.6$

$a=30.7$

At 6.35 rev/s

$a=6.35 rev/s*0.9$

$a=35.91$

5 0

3 years ago

Ben Walks 2 m from his desk to the teachers Desk. From the teachers Desk, he then walks 3 m in the same direction to the classro

Juliette [100K]

None of the tables that you have attached is a correct one.

7 0

3 years ago

Read 2 more answers

Calculate the frequency of the 3rd normal mode of a guitar string of length 40.0cm and mass 0.5g when stretched with a tension o

Alexandra [31]

Answer:

Explanation:

For third normal mode of vibration

l = $\frac{3\lambda}{2}$ , λ is wavelength , l is length of string .

.4 = $\frac{3\lambda}{2}$

λ = .267 m

velocity = $\sqrt{\frac{T}{m} }$

T is tension and m is mass unit length

m = .5 x 10⁻³ / 40 x 10⁻²

= .00125 kg / m

Putting the values

velocity = $\sqrt{\frac{80}{.00125} }$

= 253 m /s

frequency

= velocity / λ

= 253 / .267

= 947.5 Hz .

7 0

3 years ago

In general, if a sound has intensity of β dB at 1 m from the source, at what distance from the source would the decibel level de

Tpy6a [65]

Answer: The required distance is given by

$r_2=1\text{ m}\cdot 10^\frac{\beta}{20}.$

Explanation: The sound intensity in dB is given by the formula

$\beta \text{ dB}=10\log\frac{I}{I_0}$

where $I_0$ is the hearing threshold in absolute units and $I$ is the absolute intensity of the sound which depends on the distance. In general, for two distances $r_1$ and $r_2$ we have that