1) It is because they produce a lot of pollution in the atmosphere
2) They are limited in number and can't be recycled
Hope this helps!
Answer:
715 N
Explanation:
Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.
Let g = 9.8 m/s2
Gravity and equalized normal force is:
N = P = mg = 107*9.8 = 1048.6 N
Kinetic friction force and equalized tension force on the rope is

The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
- It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
- We have the expression for slit width w as,

where, d is the distance between slit and the screen, and a is the slit width.
- Thus, distance between slit and the screen is,

Thus, we can conclude that, the distance between slit and the screen is 1.214m.
Learn more about the width of the central maximum here:
brainly.com/question/13088191
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Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.