I dont know but i know i dont lnow if this is true but the gravity is slowly going away every 4 year i dont know i think.
He feels a 10 N to the left force moves. Yes ,he moves.
Answer:
44100 N
Explanation:
Each wall will have dimension of 4 m x 1.5 m
Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m
pressure at CM = h d g , h = .75 , d ( density of water = 10³ )
pressure at CM = .75 x 10³ x 9.8
= 7350 N / m²
Total force on each wall
= pressure x area
= 7350 x 4 x 1.5
= 44100 N Ans
b ) If h = 1.5 x 2 = 3
Pressure = hdg
1.5 x 10³ x 9.8
= 14700 N / m²
Force
= pressure x area
14700 x 3 x 4
= 176400 N
Which is 4 times 44100 N
So force will quadruple.
It is so because both area and height have become twice.
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Given that,
Radius of track, r = 50 m
time , t = 9 s
velocity, v = ?
Distance covered by car in one lap around a track is equal to the circumference of the track.
C = 2 π r = 2 * 3.14 * 50
C = 314.159 m
Distance covered by car, s = 314.159 m
Velocity = distance/ time
V = 314.159 / 9
V = 34.9 m/s
The average velocity of car is 34.9 m/s.
