Given: If a child is at least 4 feet tall, then
1 answer:
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Answer and explanation:
a) How many milliseconds (ms) are in 1 second?
1 second = 1000 milliseconds
1 second = milliseconds
b) How many microseconds (μs) are in 1 second?
1 second = 1000000 microseconds
1 second = microseconds
c) How many nanoseconds (ns) are in 1 millisecond?
1 millisecond = 1000000 nanoseconds
1 millisecond = nanoseconds
d) How many microseconds are in 1 millisecond?
1 millisecond = 1000 microseconds
1 millisecond = microseconds
e) How many nanoseconds are in 1 microsecond?
1 microsecond = 1000 nanoseconds
1 microsecond = nanoseconds
f) How many kilobytes (KB) are in 1 gigabyte (GB)?
1 gigabyte = 1000000 kilobytes
1 gigabyte = kilobytes
g) How many kilobytes are in 1 megabyte (MB)?
1 megabyte = 1000 kilobytes
1 megabyte = kilobytes
h) How many megabytes are in 1 gigabyte?
1 gigabyte = 1000 megabytes
1 gigabyte = megabytes
i) How many bytes are in 20 megabytes?
1 megabytes = 1000000 bytes
20 megabytes = 20000000 bytes
20 megabytes = bytes
j) How many kilobytes are in 2 gigabytes?
1 megabyte = 1000 kilobytes
2 megabyte = 2000 kilobytes
2 megabyte = kilobytes
The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50