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Anestetic [448]

3 years ago

9

A student heats a liquid on a burner. What happens to the portion of liquid that first begins to warm? It loses energy. It becom

es less dense. It sinks downward. Its molecules get closer together.
PLEASE HURRY

2 answers:

4vir4ik [10]3 years ago

6 0

Answer:

B

Explanation:

Because i said so

scoray [572]3 years ago

5 0

Answer:

they get closer together

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Svetllana [295]

Researchers found the "cosmic microwave background radiation", which is a heat imprint left over from the big bang.

The redshift of light emitted by most galaxies indicates the universe is expanding.

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3 years ago

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The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo

hjlf

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m= h^2/(2*m*λ^2)

replacing we obtained the above values.

6 0

3 years ago

What are some forces that come in pairs?

Nikolay [14]

Answer:

ACTION REACTION FORCES

Explanation:

When there is an action frce there will be a reaction force

3 0

3 years ago

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c

den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, $L_1=L-1$

New time period of the pendulum is, $T_1=2.81\ s$

We know that, the time period of a simple pendulum of length 'L' is given as:

$T=2\pi\sqrt{\frac{L}{g}}$ -------------- (1)

So, for the new length, the time period is given as:

$T_1=2\pi\sqrt{\frac{L_1}{g}}$ ------------ (2)

Squaring both the equations and then dividing them, we get:

$\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1$

Now, plug in the given values and calculate 'L'. This gives,

$L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m$

Therefore, the original length of the simple pendulum is 2.97 m

4 0

3 years ago

A proton has been accelerated from rest through a potential difference of -1000 v . part a what is the proton's kinetic energy,

wlad13 [49]

We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
$\Delta U + \Delta K=0$
which means
$\Delta K = - \Delta U$
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference ( $\Delta V=-1000 V$ ):
$\Delta U = q \Delta V=(1 eV)(-1000 V)=-1000 eV$
Therefore, the kinetic energy gained by the proton is
$\Delta K = -(-1000 eV)=1000 eV$
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>

4 0

3 years ago

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