1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
allsm [11]
3 years ago
9

Extra glucose is _____.

Physics
2 answers:
Deffense [45]3 years ago
5 0

<em>2 weeks late :')</em>

<em>-----------------------------------------------</em>

<em>the person above me is right tho</em>

Murrr4er [49]3 years ago
3 0

Answer: stored in roots, stems, and leaves

Explanation:

As a result of the process of photosynthesis where carbon dioxide and water and used to synthesize glucose, the plant will find itself with extra glucose which is still needed but not at that point.

It will therefore store the glucose in roots, stems and leaves. It will however, convert them to starch first so that they are not affected by osmosis. Starch is not soluble in water which is why osmosis will not affect them and cause they to swell too much which would reduce the space the plant has.

You might be interested in
PLS THIS IS DUE IN 2 MINUTES
Tom [10]

Answer:

The toy car. An object that isn't moving has no momentum

Explanation:

3 0
2 years ago
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0
3 years ago
A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?
Kruka [31]
4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4
4 0
3 years ago
A ray in flint glass (n = 1.61)
Scilla [17]

Answer:0

Explanation:

0

4 0
2 years ago
Read 2 more answers
How does Water Pollution effect the economy
Lelechka [254]
Water pollution can affect the economy because with no clean water, people will be forced to buy water bottles impacting their disposable income. In the search for clean water, potential future conflicts may occur because water is a fundamental need to human survival. :) 
5 0
3 years ago
Other questions:
  • Name the 4 classes of marcolecules and give a funtion for each
    6·1 answer
  • Wich of the following is NOT an example of accelerated motion?
    10·2 answers
  • A mass is connected to a spring on a horizontal frictionless surface. The potential energy of the system is zero when the mass i
    7·1 answer
  • What is emotional strength?​
    11·1 answer
  • A crane lifts a 425 kg steel beam vertically a distance of 64 m. How much work does the crane do on the beam if the beam acceler
    10·1 answer
  • What effect will a turning point have on an individuals life
    15·2 answers
  • You are in a room that has three switches and a closed door. The switches control three light bulbs on the other side of the doo
    15·2 answers
  • A child pushes his younger brother with 54 newtons of force, causing him to accelerate at 3.8 m/s/s. Assuming no friction, what
    13·1 answer
  • HELP PLZZZZZZ
    13·1 answer
  • Stephen is looking through some design diagrams created for a specific application.He spots a diagram which uses a parallelogram
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!