Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
Answer:
Velocity = 0.309 m/s
Along negative x axis
Explanation:
A pulse moving to the right along the x axis is represented by the wave function
y(x,t) = 2/ (x - 3t)² + 1
At t =0
y(x,0) = 2/ ((x - 3(0))² + 1)
=2 / (x² + 1)
At t = 1
y(x,t) = 2/ ((x - 3(1))² + 1)
= 2 /(( x - 3)² + 1)
At t = 2
y(x,t) = 2/ ((x - 3(2))² + 1)
= 2 /(( x - 6)² + 1)
For the pulse with expression y(x,t) = 4.5
²
The Velocity is
V = 2.7 / 8.73
= 0.309 m/s
There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.
Answer:
authority, accuracy, objectivity, currency, coverage, and appearance.
Explanation:
There are six (6) criteria that should be applied when evaluating any Web site. or each criterion, there are several questions to be asked. The more questions you can answer "yes", the more likely the Web site is one of quality.
Boyle's law states that when you "shrink"(change its size) a container the pressure increases so the best answer would be D-volume is decreased.
