If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
<span>(4) 19.6 N</span>
On Earth's surface, a mass of 1kg <span>exerts a force(weight) of 9.81 N
So the weight= 9.81x2.00
=19.62N
=19.6N</span>
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
Answer:
B) Remains constant
Explanation:
The motion of the ball is the motion of a projectile, which consists of two separate motions:
- A horizontal motion with constant velocity, since there are no forces acting along the horizontal direction (if we neglect air resistance)
- A vertical motion with constant acceleration, due to the presence of the force of gravity (pointing downward), that accelerates the ball towards the ground with acceleration g=9.8 m/s^2.
We are only interested in the horizontal motion of the ball: we said that there are no forces acting along the horizontal direction, therefore the horizontal component of the the velocity of the ball is constant.