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3 years ago

Definition These are the components of the reservoirs of oxygen that are exchanged in our environment.

Physics

1 answer:

andrew11 [14]3 years ago

4 0

Answer:

Oxygen cycle

Explanation:

The components of the reservoirs of oxygen that are exchange in our environment is the oxygen cycle

It suggests the movement of oxygen between the living and non-living parts.

The cycle does not account for oxygen that is trapped and cannot be exchanged in nature.
Oxygen is important component of the atmosphere.
Gaseous exchange between living organisms and atmosphere involves oxygen to a very large extent.

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A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance. What is the maximum height reached by the

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Answer:

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3 years ago

A 1090 kg car moving +11.0 m/s

stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given:

$m_{1}$ = 1090 kg

$u_{1}$ = 11 m/s

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v = 4.75 m/s

We need to find $m_{2}$

When substituting the given values in the above equation, we get

$(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75$

$11990=5177.5+4.75 m_{2}$

$4.75 m_{2}=11990-5177.5$

$4.75 m_{2}=6812.5$

$m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}$

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3 years ago

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Read 2 more answers

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

3 years ago