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joja [24]
3 years ago
14

Students have been assigned to write reports on cell organelles. Eric’s report is about the organelle that supports and gives st

ructure to plant cells. Which organelle is Eric writing about?
A. Chloroplast
B. Cell wall
C. Mitochondria
D. Nucleus
Physics
2 answers:
zheka24 [161]3 years ago
6 0
B. cell wall is correct
mojhsa [17]3 years ago
4 0
Eric is writing about the cell wall.
Chloroplast is for photosynthesis.
Mitochondria releases energy from respiration.
Nucleus controls the activities of the cell.
But the cell wall supports and gives structure to the cell.
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Tamiku [17]
True, they represent the direction of motion
8 0
3 years ago
An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break
Anuta_ua [19.1K]

Answer:

k = 9876 N/m

Explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

mg(h + x) = \frac{1}{2}kx^2

also we know that maximum acceleration will be 5.3 g

so it is given as

a = \frac{k}{m} x

so we have

x = \frac{ma}{k} = \frac{5.3 mg}{k}

mg(h + \frac{5.3 mg}{k}) = \frac{1}{2}k(\frac{5.3mg}{k})^2

mg(h + \frac{5.3 mg}{k}) = \frac{14.045(mg)^2}{k}

h + \frac{5.3mg}{k} = \frac{14.045 mg}{k}

h = \frac{8.745mg}{k}

k = \frac{8.745 (1439)(9.81)}{12.5}

k = 9876 N/m

6 0
3 years ago
What orientation of velocity and acceleration will cause something to initially slow down
SashulF [63]
Acceleration and velocity in opposite directions
3 0
4 years ago
And I need help with seven and eight only I will appreciate it
Mumz [18]

Answer:

7] Force = mass × acceleration

Force = 2 × 5

<u>Force = 10 N</u>

<u></u>

8] Velocity = acceleration due to gravity × time taken

Velocity = 9.8 × 12

<u>Velocity = 117.6 m/s</u>

8 0
2 years ago
A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar
sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

v^2-u^2 = 2gd

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

y_1 = h - \frac{1}{2}gt^2

While the vertical position of the ball thrown upward is

y_2 = ut - \frac{1}{2}gt^2

The two balls meet when

y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m

So the distance below the top of the cliff is

d=24.0 - 18.0 = 6.0 m

4 0
3 years ago
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