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3 years ago

The ocean helps absorb: hydrogen carbon dioxide nitrogen sulfur dioxide

Physics

2 answers:

Romashka-Z-Leto [24]3 years ago

5 0

Carbon Dioxide (CO2), this leads to a greenhouse effect.

icang [17]3 years ago

3 0

Answer: Carbon dioxide

Explanation:

The ocean helps in the absorption of carbon dioxide as phytoplankton absorbs carbon dioxide and performs the process of photosynthesis.

This process releases oxygen and consumes carbon dioxide by the process of diffusion.

More than normal level of carbon dioxide in water has lead to ocean acidification which is not good for the aquatic animals.

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Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

8 0

3 years ago

Energy is converted from solar to chemical in Process A and then from one form of chemical to another in Process B. Process A is

Pepsi [2]

I think the answer is photosynthis, when plants turn light into food and energy.

9 0

3 years ago

Read 2 more answers

The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8

Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) $\tau = \dfrac{\Delta L}{\Delta t}$

$\tau = \dfrac{0.8-3}{1.5}$

τ = -1.467 N m

b) angle at which fly wheel will turn

$\theta= \omega t +\dfrac{1}{2}\alpha t^2$

$\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2$

$\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2$

θ = 20.35 rad

c) work done on the wheel

W = τ x θ

W = -1.467 x 20.35 rad

W = -29.86 J

d) average power of wheel

$P_{av} =-\dfrac{W}{t}$

$P_{av} =-\dfrac{(-29.86)}{1.5}$

$P_{av} =19.91\ W$

7 0

3 years ago

Read 2 more answers

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

6 0

3 years ago

Read 2 more answers

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

4 0

3 years ago