A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landin
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Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
1 / p = 0.107375
p = 93.13 cm
Answer:
- 1.3 x 10⁻¹⁵ C/m
Explanation:
Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C
r = Radius of the arc = 5.30 cm = 0.053 m
θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)
L = length of the arc
length of the arc is given as
L = r θ
L = (0.053) (0.84)
L = 0.045 m
λ = Linear charge density
Linear charge density is given as
Inserting the values
λ = - 1.3 x 10⁻¹⁵ C/m