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3 years ago

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

Physics

1 answer:

sweet-ann [11.9K]3 years ago

5 0

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

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What are the products of linear electron flow during the light reactions of photosynthesis?

Katena32 [7]

Answer:

NADPH and ATP

Explanation:

In the clear stage the light that "hits" chlorophyll excites an electron to a higher energy level. In a series of reactions, energy is converted (throughout an electron transport process) into ATP and NADPH. Water breaks down in the process releasing oxygen as a secondary product of the reaction. ATP and NADPH are used to make the C-C bonds in the dark stage.

Photophosphorylation is the process of converting the energy of the electron excited by light into a pyrophosphate bond of an ADP molecule. This occurs when water electrons are excited by light in the presence of P680. The energy transfer is similar to the chemosmotic electron transport that occurs in the mitochondria.

Light energy causes the removal of an electron from a P680 molecule that is part of Photosystem II, the electron is transferred to an acceptor molecule (primary acceptor), and then passes downhill to Photosystem I through a conveyor chain of electrons The P680 requires an electron that is taken from the water by breaking it into H + ions and O-2 ions. These O-2 ions combine to form O2 that is released into the atmosphere.

The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

There is therefore a continuous flow of electrons (non-cyclic) from water to NADPH, which is used for carbon fixation.

Cyclic electron flow occurs in some eukaryotes and in photosynthetic bacteria. NADPH does not occur, only ATP. This also occurs when the cell requires additional ATP, or when there is no NADP + to reduce it to NADPH.

In Photosystem II, the "pumping" of H ions into the thylakoids (from the stroma of the chloroplast) and the conversion of ADP + P to ATP is motorized by an electron gradient established in the thylakoid membrane.

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3 years ago

A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 2.9 m/s when it re

Rufina [12.5K]

This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
$E_m=E_k+E_p$
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
$E_m=Ek=\frac{1}{2}mv^2$
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
$E_m=E_p$
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
$E_M=\frac{1}{2}mv^2$
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
$E_p=mgh=E_m=\frac{1}{2}mv^2$
Solving for h gives us:
$h=\frac{v^2}{2g}.$
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4 0

3 years ago

Read 2 more answers

The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?

Aleks04 [339]

The formula for the period of wave is: wave period is equals to 1 over the frequency. $waveperiod=\frac{1}{frequency}$
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s.

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3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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3 years ago

Thermodynamics

Marizza181 [45]

Answer:

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Explanation:

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2 years ago