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2 years ago

6

In a cold winters day, if you left a cup of water sitting outside, it could freeze. Heat is transferred out of the water. Descri

be the behavior of the water molecules and how temperature is affected. Enter your answer in the space provided.

1 answer:

Agata [3.3K]2 years ago

4 0

If you were to sit a hot cup of water out side it would frezze faster

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What is the relationship between the atoms of elements,its isotopes and its ions

klasskru [66]

An element refers to a collection of atoms having the same number of protons and electrons (an atomic number). In each element there is a different atomic number due to a different amount of protons in the nucleus.

An isotope is a variation of an element that contains a different number of neutrons, therefore adding weight to the atom.

An ion is a charged atom, and its charge shows how many electrons it needs to gain or lose in order to become stable.

6 0

3 years ago

Suppose a scientist conducts a series of experiments and the results are so amazing she wants to share them with other experts i

nataly862011 [7]

Answer:

The results have not been through the rigorous process of peer review

Explanation:

When a scientist conducts a study and obtains results, those results ought to be submitted to a reputable journal where the results would go through the rigorous protocol of peer review.

During this process, the reliability of the data presented is ascertained before the results are published for other scientists to see.

If the results are hurriedly published on the internet, many researchers who come in contact with the work may be fed with inaccurate information.

7 0

2 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.

Travka [436]

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

$E = \frac{F}{q} \rightarrow F=Eq$

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

$4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C$

In Second Case,

$KE = q E'd$

$KE = (0.1142)(40)(7)$

$KE = 31.976J$

The total energy change would be subject to,

$\Delta KE = 31.976-4$

$\Delta KE = 27.976J$

Therefore the Kinetic Energy change of the charged object is 27.976J

3 0

2 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

3 years ago