Answer:
The attached diagram explains the system,
Sum of Fy = 0
N=9.81
N - mgCos60 = 0
F= ukN= (0.53)(9.81) =
F= 5.12 N
So
F.d= 1/2(mv.v) - mgdsin60
-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)
(a) v = 2.436 m/s
For deflection
-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)
by solving for with values of v, m, g, F, k
800x^2 - 11.87 x - 5.938 = 0
by solving the quadratic equation
x = 0.093, -0.079
(b) x = 0.093 m
correct Answer is 0.093m
Explanation:
That would be like dropping your cell phone on to the ground by accident. The object (cell phone)'s gravitational potential energy would be converted to kinetic energy or energy in motion more precisely. This is just a hypothetical example though.
m = Q(on moon) * G(on moon) = 200N * 1.63N/kg = 326kg
Q(Earth)= g * m = 10m/s2 * 326kg = 3260N
Answer:
strong winds that blow for a long time over a great distance
weak winds that blow for short periods of time with a short fetch
Explanation:
When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.
Answer:
Given a tube of diameter d, = 3cm = 0.03m
Pressure Balance
Mercury pressure at the tube bottom Pₓ = Pa + ρgh
where
Pa = Atmospheric pressure = 101kpa
ρ = Density of mercury = 13,546kg/m3
g = acceleration due to gravity
h = height of the tube?
Given
Bottom pressure in excess of the atmospheric pressure = 48kPa = Pₓ - Pa
Therefore, 48kPa = ρgh
h = 48(kN/m2)/ρg
h = 48,000kgms⁻²m⁻²/(13546kgm⁻³ x 9.81ms⁻²)
h = 0.36m
the tube is 36cm tall