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Vikentia [17]
3 years ago
13

A rectangular block has dimensions 2.9 cm x 3.5 cm x 10.0 cm. the mass of the block is 615.0 g. what is the volume and density o

f the block? (answer it with correct numbers of significant figures)
Physics
1 answer:
Goshia [24]3 years ago
5 0
Density is the ratio of the substance's mass to volume. It is an extensive property, meaning it does not depend on the amount of the substance. The ratio is always fixed per identity of a substance. The density is equal to:

V = 2.9 cm x 3.5 cm x 10.0 cm = 101.5 cm³
Density = 615 g/101.5 cm³ = 6.06 g/cm³
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Explain how energy balance sets planetary temperature? Imagine a planet colder than expected for energy balance and explain why
RUDIKE [14]

The planetary temperature energy balance is obtained by radiating back the absorbed radiation energy from outer-space, by the planet and thus acquiring thermal equilibrium.

What is the process of attaining thermal equilibrium by Earth?

The Stefan-Boltzmann law states that the more the temperature a planet has, the more it will radiate out to reach thermal equilibrium.

We know that outer space contains large masses of radiative energy freely distributed in its vast expanse. A small fraction of this energy is absorbed by the Earth through the atmosphere, surface land, clouds etc.

Now, radiative balance is achieved when a planet's surface continuously warms up until it reaches its peak at which point the same amount of absorbed energy can then be radiated back to space. The relative amount of energy radiated back by a planet is dependent upon the size of the planet.

A colder planet relatively absorbs lower amount of radiation energy from space. In some time, as the planet heats up enough, the energy is radiated back to the space attaining thermal equilibrium.

Learn more about Stefan-Boltzmann law here:

<u>brainly.com/question/14919749</u>

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6 0
1 year ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0
ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure  = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

T_{H} = 62.9 °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

T_{L} = - 10.1 °c = 262.9 K

COP of the system is given by

COP = \frac{T_{L} }{T_{H} -T_{L}   }

COP = \frac{335.9}{335.9 -262.9}

COP = 4.6

Therefore the COP of the system is = 4.6

8 0
3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
What material is most commonly used to make a photovoltaic solar cell?
zaharov [31]

Solar cells can be classified into first, second and third generation cells. The first generation cells—also called conventional, traditional or wafer-based cells—are made of crystalline silicon, the commercially predominant PV technology, that includes materials such as polysilicon and monocrystalline silicon.

3 0
3 years ago
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