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:
The answer to your question is the letter A) F = 9.23 x 10⁻⁷ N
Explanation:
Data
q₁ = -6.25 x 10⁻⁹ C
q₂ = -6.25 x 10⁻⁹ C
d = 0.617 m
k = 9 x 10⁹ Nm²/C²
F = ?
Formula
F = k q₁q₂ /r²
-Substitution
F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²
-Simplification
F = 3.512 x 10⁻⁷ / 0.381
-Result
F = 9.227 x 10⁻⁷ N ≈ 9.23 x 10⁻⁷ N
Answer:
Energy
A wave is a disturbance that carries energy from one place to another through matter and space.
Explanation:
A wave can be defined as a form of disturbance that carries energy from one place to another through matter and space.
The energy of wave depends on the frequency of the wave and the wavelength (lambda) of that particular wave.
Mathematically,
V = f × lambda
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)
