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Lina20 [59]
3 years ago
13

What are the products(what comes out of) photosynthesis

Physics
2 answers:
Tcecarenko [31]3 years ago
4 0

Answer:

D

Explanation:

the 3 products of photosynthesis are glucose (sugar), oxygen and water. D just doesn't include water.

8090 [49]3 years ago
4 0

Answer:

It would be C

Explanation:

plants convert carbon dioxide from the air into oxygen.

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Hey if your answering this can you just answer the number 14 I already got the rest<br> thanks
Scrat [10]
Layer a is the answer
7 0
3 years ago
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Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many
dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

v(t)=t^2e^{-2t}\ m/s

We need to find the distance (x)  traveled by the particle during the first t seconds. It is given by :

x=\int\limits {v.dt}

x=\int\limits {t^2e^{-2t}dt}

Using by parts integration, we get the value of x as :

x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters

Hence, this is the required solution.

6 0
3 years ago
4. What is the acceleration of the car in each section?<br> b<br> с<br> d<br> a
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Answer:

0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.

Explanation:

3 0
2 years ago
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What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared​
mrs_skeptik [129]

Answer:

8.5 kg

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F = ma

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6 0
3 years ago
A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

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To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

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w = width

\lambda =wavelength

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\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

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w_2 = \frac{4.4*10^{-6}}{487}496

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Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
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