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3 years ago

Would a plant placed in an atmosphere of pure oxygen be able to conduct Photosynthesis

Physics

1 answer:

aalyn [17]3 years ago

4 0

NO , because plants require carbon dioxide to preform photosynthesis.

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Which is a correct example of the principle of conservation of momentum? A) reversing of a car at a dead end B) conversion of ra

Hunter-Best [27]

I think the answer would be D.

3 0

3 years ago

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A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an

ANTONII [103]

$a=5000\dfrac{mm}{s}=5\dfrac{m}{s}$

$f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s$

$a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}$

Hope this helps.

r3t40

5 0

2 years ago

An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach

matrenka [14]

Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, $\theta=60^{\circ}$

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m$

So, the maximum height reached by the object is 34.43 m.

6 0

3 years ago

Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t

MA_775_DIABLO [31]

First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200 Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time?
Car A moves 20 meters every second:
20 x 40 = 800 meters

8 0

3 years ago

What is the passenger's apparent weight at t=1.0s?

Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight = m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

<u>Apparent Weight = 1035 N</u>

For downward motion:

Apparent Weight = m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

<u>Apparent Weight = 435 N</u>

4 0

3 years ago

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