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3 years ago

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PLZ HELP photosynthesis transforms (blank)into chemical (blank)suger . cellular respration releasses the (blank) from food by (b

lank) suger.

1 answer:

Alex17521 [72]3 years ago

5 0

Answer is:

Photosynthesis transforms light energy into chemical energy. Cellular respration releasses the carbon dioxide from food into the air.

hope this helps :)

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What is a blue moon

Irina18 [472]

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a phenomenon whereby the moon appears bluish owing to smoke or dust particles in the atmosphere

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3 years ago

Which goes faster? *<br> light in space<br> sound in air<br> sound in space<br> light in water

andrezito [222]

Answer:

Hello! Your answer is, sound in the air is faster

Explanation:

The speed of sound through air is about 340 meters per second. It's faster through water and it's even faster through steel. Light will travel through a vacuum at 300 million meters per second. So they're totally different scales.

Hope I helped! Ask me anything if you have any questions! Brainiest plz. Hope you make an 100% and have a nice day! -Amelia♥

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3 years ago

A food processor draws 8.47 A of current when connected to a potential difference of 110 V.

Andreyy89

Answer:

27.95[kW*min]

Explanation:

We must remember that the power can be determined by the product of the current by the voltage.

$P=V*I$

where:

P = power [W]

V = voltage [volt]

I = amperage [Amp]

Now replacing:

$P=110*8.47\\P=931.7[W]$

Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]

$Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min]$

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3 years ago

A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th

mihalych1998 [28]

We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
$a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}$
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.

4 0

3 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago