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shutvik [7]
3 years ago
10

Why is the law of gravity an example of a universal law

Physics
1 answer:
ella [17]3 years ago
8 0
Its an example of a universal law because gravity occurs in every place of the world 
Hoped I helped!!!!!
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Part 1- An ideal gas, initially at a volume of 1.71429 L and pressure of 7 kPa, undergoes isothermal expansion until its volume
Musya8 [376]

Part 1- The work done by the gas during this process will be 5.65 ×10⁻³ kJ.

Part 2-The heat added to the gas during this process will be 5.65 ×10×10⁻³ kJ.

<h3>What is work done by the gas?</h3>

Work is the product of pressure p and volumes V during a volume change for such a gas. The work seems to be the area under the curve that indicates how the state changes.

The work done under the isothermal process is;

\rm W= P_1V_1 log_e(\frac{P_1}{P_2} )\\\\ W= 7 \times 1.71429  \times 10^{-3} log_e(\frac{7 }{2} )\\\\\ W= 0.0044 \  kJ

For the isothermal process;

ΔU=0

\rm \triangle Q = \triangle E + \triangle W \\\\ Q =  0 + 5.65 \times 10^{-3}\\\\ Q = 5.65 \TIMES 10^{-3} \ kJ

Hence, the work done, and the heat added by the gas during this process will be 5.65 ×10⁻³ kJ and 5.65 ×10×10⁻³ kJ respectively.

To learn more about work done by the gas, refer to the link;

brainly.com/question/12539457

#SPJ1

8 0
1 year ago
Describe how switching the desk lamp on and off shows that light waves transfer energy
Ludmilka [50]
The lamp transfers electric energy into both light and thermal energy when it turns on
4 0
2 years ago
Read 2 more answers
Is better to run or walk when it is raining?
lawyer [7]

Answer:

run

Explanation:

run - less wet

walking - more wet

5 0
3 years ago
A 5592 N piano is to be pushed up a(n) 3.79 m frictionless plank that makes an angle of 30.1 ◦ with the horizontal. Calculate th
otez555 [7]

Answer:

10628.87 J

Explanation:

We are given that

Force applied =F=5592 N

\theta=30.1^{\circ}

Displacement=D=3.79 m

We have to find the work done in sliding the piano up the plank at a slow constant rate.

Work done=F\times displacement

The perpendicular component of force=FSin\theta=5592sin(30.1)=2804.45N

Work done =Fsin\theta\times D=2804.45\times 3.79=10628.87 J

Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J

8 0
3 years ago
Read 2 more answers
A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
3 years ago
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