I think the answer would be Switch 1 open; switches 2, 3, and 4 closed. It is letter D. This will happen when there is a low resistance and a high circuit so from the choices above, when two bulbs are open and one bulb is close, so there is a low resistance and high voltage.
Answer:
Explanation:
Let the first height be h . second height .75h
third height .75h . fourth height .75²h
fifth height .75²h , sixthth height .75³ and so on
Total distance consists of two geometric series as follows
1 ) first series
h + .75h + .75²h + .75³h......
2 ) second series
.75h +.75²h +.75³h + .75⁴h .......
Sum of first series :
first term a = h , commom ratio r = .75
sum = a / (1 - r )
= h / 1 - .75
= h / .25
4h
sum of second series :--
first term a = .75 h , commom ratio r = .75
sum = a / (1 - r )
= .75h / 1 - .75
= .75h / .25
3h
Total of both the series
= 4h + 3h
= 7h .
h = 1 m
Total distance = 7 m
Answer:
v=u+at
24=0+at
24=a×6
a=4m/s
hence
s=ut+at^2÷2
s=36m
Explanation:
since the car is brought to rest the u=0
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
