Your answer is 632,100J which is Choice D
Explanation:
LD₁ = 10⁵ mm⁻²
LD₂ = 10⁴mm⁻²
V = 1000 mm³
Distance = (LD)(V)
Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m
Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m
Conversion to miles:
Distance₁ = 10×10⁴ m / 1609m = 62 miles
Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.
Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J
Answer:
While the use of the type of transformer in a rectifier depends on the voltage requirement or to meet desired operating conditions, a step-down transformer is used mainly to reduce the voltage. It is used to bring the high AC voltage level to a reasonable value or the desired output voltage.
Explanation:
Hope it helps
Correct me if Im wrong
