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3 years ago

Write down applications of mechanics

Physics

1 answer:

harkovskaia [24]3 years ago

7 0

Answer:

Explanation:

applied Mechanics and its Growing Utilisation of Theoretical Mechanics.\

Structural Engineering.

Hydraulics.

Mechanical Engineering.

External Fluid Dynamics.

Planetary Sciences.

Life Sciences.

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A rock from space in Earth's Atmosphere would be considered?

Gnesinka [82]

Answer:

A meteor mate

Explanation:

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8 0

3 years ago

Differentiate scalar & vector quantity?

Keith_Richards [23]

$\textbf{Hello Friend}$

Scalar Quantity :-

→ These are the quantities with magnitude only . These quantities doesn't have to be mentioned with direction

eg.)=> Mass , Temprature .

Vector Quantity :-

→ These quantities are described with both Magnitude and Direction . These quantities follow special type of algebra called Vector algebra .

eg.)=> Force , Displacement

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5 0

3 years ago

Eugene describes the physical property of a material as “sweet and floral.” What physical property of the material is Eugene mos

shusha [124]

The correct answer would be odor. Because it's sweet. Boiling shape and hardness have nothing to do with sweet and floral :)

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3 years ago

Read 2 more answers

On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the

Margaret [11]

Answer:

The change in temperature, $\Delta T=9.45^{\circ} C$

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from $-4^{\circ} F\ to\ 45^{\circ} F$ in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

$\Delta T=T_f-T_i\\\\\Delta T=45-(-4)\\\\\Delta T=49^{\circ}F$

The relation between degrees Celsius and degrees Fahrenheit is given by :

$F=1.8C+32$

Here, F = 49 degrees

$49=1.8C+32\\\\\Delta T=9.45^{\circ} C$

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.

8 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

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

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$