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Jet001 [13]
2 years ago
6

calculate the work done to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it.

Physics
1 answer:
Vaselesa [24]2 years ago
8 0

W= F*d which is (10N x 0.4 x 0.4=1.6 J)

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Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D​
erma4kov [3.2K]

Answer:

The car C has KE = 100, PE = 0

Explanation:

The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.

The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.

Kinetic energy is the energy that a body possesses due to its motion.

Potential energy is the energy a body possesses due to its position.

From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.

In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.

Therefore, the car C has KE = 100, PE = 0

6 0
3 years ago
STATE THE LAWS OF CONSERVATION OF MOMENTUM​
dalvyx [7]

\Huge  \mid   \underline {\mathcal {{{\color{purple}{Answer...}}}}} \mid

When two bodies collide with each other in the absence of an external force, then the total final momentum of the bodies is equal to their total initial momentum.

6 0
3 years ago
Explain any 3 alternating ways of producing electricity​
ehidna [41]

Answer:

to produce electricty three ways you have the water wheel the wind mill and solor panels

Explanation:

4 0
2 years 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
A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate
yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

W=Pm\DeltaV

W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}

W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})

Where,R = gas constant

T = temperature

P = pressure

P_{atm}=Atmosphere pressure

m = mass

Put the value into the formula

W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)

W=213\ kJ

Hence, The work done by the steam is 213 kJ.    

6 0
3 years ago
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