Answer: The riders are subjected to 11.5 revolutions per minute
Explanation: Please see the attachments below
A is growth!!!!! B is reproduction!!!
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:
4 m/s² down
Explanation:
We'll begin by calculating the net force acting on the object.
The net force acting on the object from the left and right side is zero because the same force is applied on both sides.
Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:
Force up (Fᵤ) = 15 N
Force down (Fₔ) = 25 N
Net force (Fₙ) =?
Fₙ = Fₔ – Fᵤ
Fₙ = 25 – 15
Fₙ = 10 N down
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (ml= 2.5 Kg
Net force (Fₙ) = 10 N down
Acceleration (a) =?
Fₙ = ma
10 = 2.5 × a
Divide both side by 2.5
a = 10 / 2.5
a = 4 m/s² down
Therefore, the acceleration of the object is 4 m/s² down
Answer:
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