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3 years ago

if a metal object was cut on half how much heat will be required to raise both half to a specific temperature

1 answer:

3 0

suppose the mass of a metal object be m and its specific heat capacity be s, and, H joules of heat is required to raise its temperature by t degrees Celsius

Than H is given by relation, $H = mst$

Now if this object is cut in two half's, than mass of half part will also be half i.e. M/2

So heat required to heat the half part will be $= (\frac{m}{2}) st =\frac{mst}{2} = H/2$

Hence, the heat required to raise the half object to a specific temperature will also become half.

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Calculate the minimum thickness (in nm) of an oil slick on water that appears blue when illuminated by white light perpendicular

mr_godi [17]

Answer:

The minimum thickness = 83.92 nm

Explanation:

The relation between the wavelength in a particular medium and refractive index $\lambda_n = \frac{ \lambda }{n}$

where ;

$\lambda$ = wavelength of the light in vacuum

n = refractive index of medium with respect to vacuum

For one phase change :

$2t = \frac{\lambda_n}{2}\\\\where \ \lambda_n = \frac{\lambda}{n}\\\\Then \ \\\\2t = \frac{\lambda}{2n}\\\\t = \frac{\lambda_n}{4n}$

Replacing 1.43 for n and 480 nm for λ; we have:

$t = \frac{480}{4(1.43)}$

t = 83.92 nm

Thus; the minimum thickness = 83.92 nm

4 0

3 years ago

Bettina spoke into a microphone during the school play to increase the sound of her voice so the audience could hear her speak.

grandymaker [24]

The loudness was increased by the amplifier which converted electrical energy into sound energy.

7 0

3 years ago

Read 2 more answers

Which of the following statements best describe destructive forces? A. Forces that build up, create, landmasses. B. Forces that

MissTica

An event that breaks objects into smaller objects or pieces is called destructive force
Tornadoes, Hurricanes, Earthquakes, Volcanoes, Tsunamis and more are some of examples
Forces that wear down, destroy is right answer

4 0

4 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0

3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

4 years ago