Answer:
The angle it subtend on the retina is 
Explanation:
From the question we are told that
The length of the warbler is 
The distance from the binoculars is 
The magnification of the binoculars is 
Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as
Now magnification can be represented mathematically as
Where 
is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.
So
=> 
Generally the conversion to degrees can be mathematically evaluated as
Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
From Carnot's theorem, for any engine working between these two temperatures:
efficiency <= (1-tc/th) * 100
Given: tc = 300k (from question assuming it is not 5300 as it seems)
For a, th = 900k, efficiency = (1-300/900) = 70%
For b, th = 500k, efficiency = (1-300/500) = 40%
For c, th = 375k, efficiency = (1-300/375) = 20%
Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid.
Acceleration is the way the motion is changing.
