One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.
From the temperature the value is given as
Where,
T_L = Cold focus temperature
T_H = Hot spot temperature
Our values are given as,
T_L = 20\° C = (20+273) K = 293 K
T_H = 440\° C = (440+273) K = 713 K
Replacing we have,
Therefore the maximum possible efficiency the car can have is 58.9%
Answer:
a) 
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo =
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s
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Answer:
Explanation:
d = Diameter of spot = 30 μm
r = Radius of spot = 
P = Power of the laser = 
A = Area = 
Intensity is given by
The light intensity within this spot is 
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.
Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get
