355465
An electric motor of 100 \(W\) power drives a stirrer in a water bath. Only \(80 \%\) of the energy supplied to the motor is used up in stirring water. The work done on water in 1 second is
355466
A car of mass \(1000\;kg\) accelerates uniformly from rest to a velocity of \(54\,km\,{h^{ - 1}}\) in \(5\;s.\) The average power of the engine during this period in watt is (neglect friction)
1 \(2000\;W\)
2 \(22500\;W\)
3 \(5000\;W\)
4 \(2250\;W\)
Explanation:
Given, \(m = 100\;kg,\,\,v = 54\,km\,{h^{ - 1}}\) \( = \frac{{54 \times 5}}{{18}} = 15\;m{s^{ - 1}}\) and \(t=5 \mathrm{~s}\) From equation of motion, velocity, \(v=u+a t\) \( \Rightarrow a = \frac{{15 - 0}}{5} = 3\;m{s^{ - 2}}\) Hence, \(P_{a v}=\dfrac{1}{2} F v=\dfrac{1}{2} m a v\) \(=\dfrac{1}{2} \times 1000 \times 3 \times 15\) \( = 22500\;W\)
355465
An electric motor of 100 \(W\) power drives a stirrer in a water bath. Only \(80 \%\) of the energy supplied to the motor is used up in stirring water. The work done on water in 1 second is
355466
A car of mass \(1000\;kg\) accelerates uniformly from rest to a velocity of \(54\,km\,{h^{ - 1}}\) in \(5\;s.\) The average power of the engine during this period in watt is (neglect friction)
1 \(2000\;W\)
2 \(22500\;W\)
3 \(5000\;W\)
4 \(2250\;W\)
Explanation:
Given, \(m = 100\;kg,\,\,v = 54\,km\,{h^{ - 1}}\) \( = \frac{{54 \times 5}}{{18}} = 15\;m{s^{ - 1}}\) and \(t=5 \mathrm{~s}\) From equation of motion, velocity, \(v=u+a t\) \( \Rightarrow a = \frac{{15 - 0}}{5} = 3\;m{s^{ - 2}}\) Hence, \(P_{a v}=\dfrac{1}{2} F v=\dfrac{1}{2} m a v\) \(=\dfrac{1}{2} \times 1000 \times 3 \times 15\) \( = 22500\;W\)
355465
An electric motor of 100 \(W\) power drives a stirrer in a water bath. Only \(80 \%\) of the energy supplied to the motor is used up in stirring water. The work done on water in 1 second is
355466
A car of mass \(1000\;kg\) accelerates uniformly from rest to a velocity of \(54\,km\,{h^{ - 1}}\) in \(5\;s.\) The average power of the engine during this period in watt is (neglect friction)
1 \(2000\;W\)
2 \(22500\;W\)
3 \(5000\;W\)
4 \(2250\;W\)
Explanation:
Given, \(m = 100\;kg,\,\,v = 54\,km\,{h^{ - 1}}\) \( = \frac{{54 \times 5}}{{18}} = 15\;m{s^{ - 1}}\) and \(t=5 \mathrm{~s}\) From equation of motion, velocity, \(v=u+a t\) \( \Rightarrow a = \frac{{15 - 0}}{5} = 3\;m{s^{ - 2}}\) Hence, \(P_{a v}=\dfrac{1}{2} F v=\dfrac{1}{2} m a v\) \(=\dfrac{1}{2} \times 1000 \times 3 \times 15\) \( = 22500\;W\)
355465
An electric motor of 100 \(W\) power drives a stirrer in a water bath. Only \(80 \%\) of the energy supplied to the motor is used up in stirring water. The work done on water in 1 second is
355466
A car of mass \(1000\;kg\) accelerates uniformly from rest to a velocity of \(54\,km\,{h^{ - 1}}\) in \(5\;s.\) The average power of the engine during this period in watt is (neglect friction)
1 \(2000\;W\)
2 \(22500\;W\)
3 \(5000\;W\)
4 \(2250\;W\)
Explanation:
Given, \(m = 100\;kg,\,\,v = 54\,km\,{h^{ - 1}}\) \( = \frac{{54 \times 5}}{{18}} = 15\;m{s^{ - 1}}\) and \(t=5 \mathrm{~s}\) From equation of motion, velocity, \(v=u+a t\) \( \Rightarrow a = \frac{{15 - 0}}{5} = 3\;m{s^{ - 2}}\) Hence, \(P_{a v}=\dfrac{1}{2} F v=\dfrac{1}{2} m a v\) \(=\dfrac{1}{2} \times 1000 \times 3 \times 15\) \( = 22500\;W\)