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Nezavi [6.7K]
4 years ago
9

What is the structural difference between plant and animal cells in terms of energy productio?

Physics
1 answer:
MatroZZZ [7]4 years ago
5 0
Most of the energy production in a plant cell happens by "Chloroplasts" through the process of photosynthesis whereas in animal cells, "Mitochondria" does that through the process of Cellular respiration

Hope this helps!
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Vygotsky’s theory of cognitive development states that an individual constructs knowledge primarily through interaction with imp
defon

Answer:

false

Explanation:

trust me i j took the test

4 0
4 years ago
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Vector e is 0.111m long in a 90 deg direction. Vector f is 0.233 m long in 400 deg direction. Find the direction of their vector
-BARSIC- [3]

Answer:

50.6

Explanation:

7 0
3 years ago
You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc
pickupchik [31]

a) 15.4^{\circ}

b) 5.2 m/s

c) 151.2 N

Explanation:

a)

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

mg sin \theta

where

m =16 kg is the mass of the box

g=9.8 m/s^2 is the acceleration due to gravity

\theta is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

F=ma\\mgsin \theta = ma

where a is the acceleration.

From the equation above we get

a=g sin \theta

And we are told that the acceleration must not exceed

a=2.6 m/s^2

Substituting this value and solving for \theta, we find the maximum angle of the ramp:

\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}

b)

Here we are told that the vertical distance of the ramp is

h=1.4 m

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

GPE=KE\\mgh=\frac{1}{2}mv^2

where:

m = 16 kg is the mass of the box

g=9.8 m/s^2

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s

c)

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

mg cos \theta

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

N-mg cos \theta

and by substituting:

m = 16 kg

g=9.8 m/s^2

\theta=15.4^{\circ}

We can find the normal force:

N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N

8 0
3 years ago
A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine
nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop v=26\ ft/s

height of top h=16\ ft

Initial energy at bottom is

E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2

Where m=mass of hoop

I=moment of inertia of hoop

\omega=angular velocity

for pure rolling v=\omega R

I=mR^2

E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2

E_b=mv^2=m(26)^2=676m

Energy required to reach at top

E_T=mgh=m\times 32.2\times 16

E_T=512.2m

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

\Delta E=676m-512.2m=163.8m

Therefore

163.8 m=mv^2

v=\sqrt{163.8}

v=12.798\approx 12.8\ ft/s

7 0
3 years ago
Can some help me with this question​
Luba_88 [7]
The answer is 2000 I need 20 characters do djrjrkrir
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4 years ago
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