Using lens equation;
1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)
Substituting;
1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm
Therefore, the object should be place 99.23 cm from the lens.
Answer:
x = 727.5 km
Explanation:
With the conditions given using trigonometry, we can find the tangent
tan θ = CO / CA
With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun
D =150 10⁶ km (1000m / 1 km)
D = 150 10⁹ m.
We must take the given angle to radians.
1º = 3600 arc s
π rad = 180º
θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =
θ = 4.85 10⁻⁶ rad
That angle is extremely small, so we can approximate the tangent to the angle
θ = x / D
x = θ D
x = 4.85 10-6 150 109
x = 727.5 103 m
x = 727.5 km
Explanation:
c. if the vector is oriented at 0° from the X -axis.
Answer:
The answer is number 2 :)
Speed can never be negative because it does not depend in which direction the car moves whereas, velocity will change if a car turns from due North to East.
Quantities which can be described only by their magnitudes are called scalars and those which are described by both, magnitude and direction are vectors
