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LenKa [72]
3 years ago
14

What part of the plant takes in carbon dioxide?

Physics
2 answers:
murzikaleks [220]3 years ago
4 0

The answer is number 2 stomata.

Irina18 [472]3 years ago
3 0
The answer is stomata 2
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What is 12 and 1 half % expressed as a fraction
kiruha [24]
12 1/2 i would assume
6 0
3 years ago
Read 2 more answers
A transform boundary occurs where two tectonic plates _____.?
gladu [14]

Where they slide over each other.

Transform boundaries are formed or occur when two plates slide past each other in a sideways motion.  They do not tear or crunch into each other (but the rock in between them may be ground up) and therefore none of the spectacular features are seen such as occur in divergent and convergent boundaries.

In a transform boundary, neither plate is added to at the boundary nor destroyed.  They are marked in some places by features like  stream beds that have been split in half and the two halves moved in opposite directions.



7 0
3 years ago
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I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees
Alexeev081 [22]
D=rt
when biker A catches biker B, the time they've been riding is the same, so 
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so 
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters

Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters
8 0
3 years ago
The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s
Anastasy [175]
Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now
6 0
3 years ago
Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and
lyudmila [28]

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

<u>K.E₂ = mg(h - 2R)</u>

3 0
3 years ago
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