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Karo-lina-s [1.5K]
3 years ago
14

Which of the following properties of water help to explain why icebergs float in the ocean?

Physics
1 answer:
crimeas [40]3 years ago
7 0
The reason why icebergs float in the ocean has to do with temperature. Icebergs are colder than the ocean water and therefore the cold water is less dense than the warm water and this causes the Iceberg to float.
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What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
Mashcka [7]

<h2>\bf{ \underline{Given:- }}</h2>

\sf• \: The \:  current \:  in \:  a \:  circuit \:  is  \: 5 \: amps.  \: and  \: resistance \:  is \:  30 \:  Ohms.

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf{• \:  The  \: Potential  \: Difference. }

\\

\huge\bf{ \underline{ Solution:- }}

\sf According  \: to  \: the  \: question,

\sf•  \: Current \:  (I) = 5  \: Amps.

\sf• \:  Resistance  \: (R) = 30 \:  Ω

\sf{Potential \:  difference  \: means  \: Voltage \: ( V).}

\sf{We \:  know \: that, }

\bf \red{ \bigstar{ \: V = IR }}

\rightarrow \sf V =5 \times 30

\rightarrow \sf V =150

\\

\sf \purple{Therefore, \:  the \:  potential  \: difference  \: is  \: 150  \: v \: .}

4 0
3 years ago
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.

7 0
3 years ago
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
3 years ago
An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
dsp73

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

3 0
3 years ago
Read 2 more answers
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i
ELEN [110]

Answer:

Pls see attached file

Explanation:

5 0
3 years ago
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