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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.

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At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?
Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$m / s^2

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$m/s^2 is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0
1 year ago
a 3520 kg truck moving north at 18.5 m/s makes an INELASTIC collision with an 1480 kg car moving east after colliding they have
anyanavicka [17]

Answer:

Explanation:

An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is

p = mv and

p = (3520 + 1480)(13.6) so

p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is

p_f=\sqrt{(p_{truck})^2+(p_{car})^2} which, in words, is

the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:

68000=\sqrt{(65100)^2+(1480v)^2} and

(68000)^2=(65100)^2+(1480v)^2 and

4624000000=4238010000+2190400v^2  and

385990000=2190400v^2 and

176.2189554=v^2 so

v = 13.3 m/s at 72.6°

6 0
3 years ago
vector ????⃗ has a magnitude of 17.9 and its direction is 80∘ counter‑clockwise from the x- axis. what are the x- and y- compone
Reika [66]

We have vector (17.9*cos80^{0},17.9*sin80^{0})

Therefore,

x component = 17.9 * cos80 degree = 3.108

y component = 17.9 * sin80 degrees = 17.628

<h3>What is a vector?</h3>

An object with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.

If the magnitude and direction of two vectors match, they are the same vector. This shows that if we move a vector to a different location without rotating it, the final vector will be the same as the initial vector. The vectors that denote force and velocity are two examples. The direction of force and velocity are both fixed. The size of the vector would represent the force's strength or the velocity's corresponding speed.

To know more about vectors, visit:

brainly.com/question/12937011

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5 0
2 years ago
Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5
andrezito [222]

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

             

how angles are measured in radians

          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            \frac{y}{R} = 1.22 \frac{ \lambda}{D}

            R = \frac{ y \ D}{1.22 \lambda}

let's calculate

            R = \frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}

            R = 1.2295 10⁵  m

3 0
2 years ago
The mass of a certain neutron star is 2.5x10^30kg and the radius 7000m. what is the force of gravity on a 1kg object of the surf
Makovka662 [10]

Answer:

3.42N

Explanation:

*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*

Use the equation: F = (GMm)/(r^2)

F = force of gravity

G = gravitational constant (6.7x10^-11)

M = mass1 (2.5x10^30kg)

m = mass2 (1kg)

r = radius (7000m)

Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)

F = (1.675x10^20) / (4.9x10^7)

F = 3.4183673x10^12

F = 3.42N

8 0
3 years ago
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