<span>Answer:
skater x km/h
cyclist 20 faster x + 20 km/h
skater 30 km
cyclist 80 km
skater time = cyclist time
t=d/r
30 / x = 80 /( x + 20
cross multiply
30 ( x + 20 ) = 80 x
30 x + 600 = 80 x
30 x - 80 x = -600
-50 x = -600
/ -50
x = 12 km/h
12 km/h skater</span>
Answer:
<u>We are given:</u>
displacement (s) = 130 m
acceleration (a) = -5 m/s²
final velocity (v) = 0 m/s [the cars 'stops' in 130 m]
initial velocity (u) = u m/s
<u>Solving for initial velocity:</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(0)² - (u)² = 2(-5)(130)
-u² = -1300
u² = 1300
u = √1300
u = 36 m/s
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
Answer:
150153.06122 N
Explanation:
m = Mass of person = 75 kg
h = Height of fall = 1 m
g = Acceleration due to gravity = 9.81 m/s²
F = Force
s = Displacement = 0.49 cm
Potential energy is given by
Work is given by
The average force exerted is 150153.06122 N
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
