I agree with the other comment
Answer:reflection by dust particles in air
Answer: 12.67 cm, 8 cm
Explanation:
Given
Normal distance of separation of eyes, d(n) = 6 cm
Distance of separation is your eyes, d(y) = 9.5 cm
Angle created during the jump, θ = 0.75°
To solve this, we use the formula,
θ = d/r, where
θ = angle created during the jump
d = separation between the eyes
r = distance from the object
θ = d/r
0.75 = 9.5 / r
r = 9.5 / 0.75
r = 12.67 cm
θ = d/r
0.75 = 6 / r
r = 6 / 0.75
r = 8 cm
Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
Explanation:
It is given that,
Displacement of the delivery truck,
(due east)
Then the truck moves,
(due south)
Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :
![d=\sqrt{d_1^2+d_2^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7Bd_1%5E2%2Bd_2%5E2%7D)
![d=\sqrt{3.2^2+2.45^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B3.2%5E2%2B2.45%5E2%7D)
d = 4.03 km
Let
is the direction of the truck’s displacement from the warehouse from south of east.
![\theta=tan^{-1}(\dfrac{2.45}{3.2})](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cdfrac%7B2.45%7D%7B3.2%7D%29)
![\theta=37.43^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D37.43%5E%7B%5Ccirc%7D)
So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.