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cluponka [151]
2 years ago
5

Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr

evious experiment you plotted time vs. temperature, if you observe closely (e.g. for water) you will see that there is some deviation from a straight line. Use your critical thinking to figure out why this inconsistency.
Physics
1 answer:
Nitella [24]2 years ago
7 0

Answer:

hydrogen bridge

Explanation:

Joule's relationship to heat and temperature is true for all materials where we assume that interatomic forces are linear, when atoms separate these forces decrease. There is a point where the separation between atoms is enough that thermal agitation can separate the molecules and there is a change of state, generally from solid to liquid and from liquid to vapor. When these changes of state are occurring all the energy supplied is used to break the links, so the temperature does not change.

In the specific case of water, there is a bond called a hydrogen bridge that breaks around 4ºC, therefore, at this temperature there is a deviation from the curve since this link is being broken, this does not lead to a change of macroscopic state.

For the other temperatures the water behaves like the other bodies.

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
2 years ago
How do air mass conditions ahead of the squall line support the development of new cell?
IRISSAK [1]
<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>
7 0
2 years ago
A cylinder of radius R and height H is floating upright in
emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

F = mg

so we have

\Delta P \pi R^2 = mg

so we have

\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g

so we have

\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

6 0
2 years ago
A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app
Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0
2 years ago
A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0
GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

F=qE

Put the value into the formula

F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

We need to calculate the total charge on the rod

Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}

Put the value into the formula

Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}

Q=47.8\times10^{-9}\ C

Q=47.8\ nC

Hence, The total charge on the rod is 47.8 nC.

6 0
2 years ago
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