The cyclist who travels 20 kilometers per hour for 15 kilometers
Answer:
Explanation:
The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.
100 mmHg
Givens
V1 = 20 cm^3
V2 = 80 cm^3
P1 = 400 mmHg
P2 = ?
Formula
V1 * P1 = V2 * P2
Solution
20 * 400 = 80 * P2 Divide by 80
20 * 400/80 = P2
P2 = 8000 / 80
P2 = 100 mmHg
<span>Your answer isStratified drift</span>
Answer:
A) 12.57 m
B) 5 RPM
C) 3.142 m/s
Explanation:
A) Distance covered in 1 Revolution:
The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:
s = rθ
where,
s = distance covered = ?
r = radius of circle = 2 m
θ = Angle = 2π radians (For 1 complete Revolution)
Therefore,
s = (2 m)(2π radians)
<u>s = 12.57 m</u>
B) Angular Speed:
The formula for angular speed is given as:
ω = θ/t
where,
ω = angular speed = ?
θ = angular distance covered = 15 revolutions
t = time taken = 3 min
Therefore,
ω = 15 rev/3 min
<u>ω = 5 RPM</u>
C) Linear Speed:
The formula that gives the the linear speed of an object moving in a circular path is given as:
v = rω
where,
v = linear speed = ?
r = radius = 2 m
ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s
Therefore,
v = (2 m)(1.571 rad/s)
<u>v = 3.142 m/s</u>
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2