The rocket works on the principal of action and reaction. Newton's third law of motion.
Answer: Newtons First law.
Explanation: Not enough information.
Answer: ![-5013.65\ J](https://tex.z-dn.net/?f=-5013.65%5C%20J)
Explanation:
Given
No of moles ![n=3](https://tex.z-dn.net/?f=n%3D3)
Temperature ![T=290\ K](https://tex.z-dn.net/?f=T%3D290%5C%20K)
Initial volume ![V_1=2\ m^3](https://tex.z-dn.net/?f=V_1%3D2%5C%20m%5E3)
Final volume ![V_2=1\ m^3](https://tex.z-dn.net/?f=V_2%3D1%5C%20m%5E3)
Work done in constant temperature process is
![W=nRT\ln \left(\dfrac{V_2}{V_1}\right)](https://tex.z-dn.net/?f=W%3DnRT%5Cln%20%5Cleft%28%5Cdfrac%7BV_2%7D%7BV_1%7D%5Cright%29)
Insert the values
![\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%3D3%5Ctimes%208.314%5Ctimes%20290%5Cln%20%5Cleft%20%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%5C%5C%5C%5CRightarrow%20W%3D-870%5Ctimes%208.314%5Ctimes%20%5Cln%20%282%29%5C%5C%5CRightarrow%20W%3D-5013.65%5C%20J)
Answer:
25032.47 W
Explanation:
Power is the time rate of doing work, hence,
P = Work done(non conservative) / time
Work done (non conservative) is given as:
W = total K. E. + total P. E.
Total K. E. = 0.5mv²- 0.5mu²
Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s
Total P. E. = mgh(f) - mgh(i)
Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m
=> W = 0.5mv² - mgh(f)
P = [0.5mv² - mgh(f)] / t
P = [(0.5*790*7²) - (790*9.8*7.2)] / 3
P = (19355 + 55742.4) / 3 = 75097.4/3
P = 25032.47 W
Answer:
<em>10500 J</em>
Explanation:
<em>Heat:</em> Heat is defined as a form of energy that brings about the sensation of warmth. The S.I unit of heat is Joules (J)
Q = cm(T₂-T₁)..................................... Equation 1
Where Q = quantity of heat, c = specific heat capacity of water, m = mass of water, T₂ = final temperature, T₁ = initial temperature.
<em>Given: m = 100 g = (100/1000) kg = 0.1 kg, T₁ = 25 °C, T₂ = 50 °C.</em>
<em>Constant: c = 4200 J/kg.°C</em>
<em>Substituting these values into equation 1</em>
<em>Q = 0.1×4200×(50-25)</em>
<em>Q = 420×25</em>
<em>Q = 10500 J</em>
<em>Thus the amount of heat needed to raise the temperature of 100 g of water from 25 °C to 50 °C = 10500 J</em>
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