Answer: = 5.75 × 10 -6
Explanation:
= 5.75 × 10-6
(scientific notation)
= 5.75e-6
(scientific e notation)
= 5.75 × 10-6
(engineering notation)
(millionth; prefix micro- (u))
= 0.00000575
(real number)
Answer:
The first frequency of audible sound in the speed sound is
f = 662 Hz
Explanation:
vs = 344 m/s
x = 52 cm * 1 / 100m = 0.52m
The wave length is the distance between the peak and peak so
d = 2x
d = 2*0.52 m
d = 1.04 m
So the frequency in the speed velocity is
f = 1 / T
f = vs / x = 344 m/s / 0.52m
f ≅ 662 Hz
Answer:
the distance between the lawyer's home and her office is 124 miles
Explanation:
given information:
first lets assume that
x-axis (west = positive, east = negative)
y-axis (north = positive, south = negative)
thus,
distance of the house = (-88,18)
distance of the office = (13, -54)
thus, the distance between the lawyer's home and her office
R = √(x₂ - x₁)² + (y₂ - y₁)²
= √(13 - (-88))² + (-54 -18)²
= 124 miles
Answer:
λ = 2.7608 x 10⁻⁷ m = 276.08 nm
Explanation:
The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:
Work Function = hc/λ
where,
Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = longest wavelength capable of releasing electron.
Therefore,
7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)
<u>λ = 2.7608 x 10⁻⁷ m = 276.08 nm</u>
Answer:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
