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daser333 [38]
3 years ago
14

Sugar molecules can be oxidized in the cell of animals . This oxidation releases energy for the animal to use. From what soured

do the cells get these molecules.
Physics
2 answers:
qaws [65]3 years ago
4 0

Answer: through digested food

Explanation: cells get their sugar molecules from the food that is taken in by an organism. The food taken in is digested by the stomach into base materials like sugar, fatty acids and amino acids. These base materials are circulated through the blood till they get to the cells where they are needed.

tamaranim1 [39]3 years ago
3 0

Answer:

The sugar is derived from the breaking down of the carbohydrate molecules in the food we eat during digestion

Explanation:

The cells get the molecules from the food we eat. The food we ingest which consists mostly of carbohydrates, fats and oils and proteins are too complex and must be broken down into smaller molecules before they can be consumed by the body cells.

The process of the breakdown of food starts from the chewing and mixing of the food with saliva after which breakdown by enzymes takes place in the intestine and in the lysosomes.

The carbohydrates are broken down into sugar, the proteins into amino acids and the fats and oils into fatty acids and glycerol.

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An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the
Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec =  78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s   t=1 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s   t=2 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s   t=3 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s   t=4 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s   t=5 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0
2 years ago
What do hydrogen and helium have in common?
Savatey [412]

Atomic disguise makes helium look like hydrogen. ... A helium atom consists of a nucleus containing two positively charged protons and two neutrons, encircled by two orbiting electrons which carry a negative charge. A hydrogen atom has just one proton and one electron

4 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
2 years ago
5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co
Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop.  At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0
2 years ago
Evaluate tan(249).<br> O A. 1.36<br> B. 0.45<br> C. 0.41<br> D. 0.91
vodomira [7]

Answer:

tan 249 = 2.61

tan 249 = tan (249 - 180) = tan 69 = 2.61

3 0
2 years ago
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