1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor
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 !
Solved your another question same like this with scaling to Cm this time we go with metre(m)
Scale factor
Mercury
Ven us
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Answer:
2697.75N/m
Explanation:
Step one
This problem bothers on energy stored in a spring.
Step two
Given data
Compression x= 2cm
To meter = 2/100= 0.02m
Mass m= 0.01kg
Height h= 5.5m
K=?
Let us assume g= 9.81m/s²
Step three
According to the principle of conservation of energy
We know that the the energy stored in a spring is
E= 1/2kx²
1/2kx²= mgh
Making k subject of formula we have
kx²= 2mgh
k= 2mgh/x²
k= (2*0.01*9.81*5.5)/0.02²
k= 1.0791/0.0004
k= 2697.75N/m
Hence the spring constant k is 2697.75N/m
Answer:
You are given that the mass of the clock M is 95 kg.
This is true whether the clock is in motion or not.
Fs is the frictional force required to keep the clock from moving.
Thus Fk = uk W = uk M g the force required to move clock at constant speed. (the kinetic frictional force)
uk = 560 N / 931 N = .644 since the weight of the clock is 931 N (95 * 9.8)
us is the frictional force requited to start the clock moving
us = static frictional force = 650 / 931 -= .698
