Answer:
The distance of the object placed on the principal axis from the concave mirror.
Explanation:
In a concave mirror, the nature of the image formed formed by the object placed in front of the mirror depends on the position of the object placed in from of the mirror. It all depends on the distance between the mirror and the object placed on the principal axis.
The closer the object is to the lens, the more larger or magnified the image formed will be. For example an object placed between the focal point and the pole of a concave produces a much larger image than an object placed beyond the centre of curvature of such mirror.
The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
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Answer:
a) α = 1.875 
b) t = 8 s
Explanation:
Given:
ω1 = 0 
ω2 = 15
theta (angular displacement) = 60 rad
*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*
a) α = ?
(ω2)^2 = (ω1)^2 + 2α(theta)
= 
+ 2(α)(60)
225 = 120α
α = 1.875
b)
α = (ω2-ω1)/t
t = (ω2-ω1)/α = (15-0)/1.875 = 8
t = 8 s
Answer
A. the work done on the refrigerant in each cycle is 105kJ
B the coefficient of performance of the refrigerator is 4.8
Explanation
Given data
Work done at high temperature T2 Qh=610kJ
Work done at low temperature T1 Ql=505kJ
We know that the net work done by the refrigerator is expressed as
Wnet= Qh-Ql
=610-505
=105kJ
Also we know that the coefficient of performance is expressed as
COP= Ql/Wnet
COP= 505/105
= 4.8
given that snow is projected at an angle of 40 degree
It range is given as a = 19 ft
now we can use the formula of horizontal range
<u>so its initial speed must be 7.6 m/s</u>
