<h3>
Answer:</h3>
<h3>
Explanation:</h3>
_______________
S=3 m²
F=900 N
_______________
p - ?
_______________
p=F/S=900 N / 3 m² = 300 Pa
Answer:
efficiancy=40 percent
Explanation:
efficiency=energy output/energy input×100
efficiancy=8J/20J×100
efficiancy=0.4×100
efficiancy=40 percent
Mark brianliest if my answer suit your question..
The
two precipitation peaks in Mbandaka during March to April and September to
November is due to the intertropical convergence zone.
Intertropical
convergence zone is a narrow zone located near the equator. It is where the
northern and southern air masses intersect which results to low atmospheric
pressure. Due to the intertropical convergence zone’s meeting of air masses,
often times the air pressure are lower which will results to colder air, or
even rainfall during the period of March to April, and most especially
September to November in Mbandaka.
<span>Since
Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named
as a Wetland of International importance, there is really a bigger chance that
this place experience above 60mm precipitation in a year, temperatures averaging
from 23 – 26 degrees Celsius.</span>
Answer:
0.339 kgm²
Explanation:
We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.
Since T = 2π√(I/mgh), making I subject of the formula, we have
I = mghT²/4π²
Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.
So, T = 241 s/113 = 2.133 s
So, Substituting the values of the variables into I, we have
I = mghT²/4π²
I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²
I = 15.63/4π² kgm²
I = 0.396 kgm²
Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis
I' = I - mh²
I' = 0.396 kgm² - 2.15 kg × (0.163 m)²
I' = 0.396 kgm² - 0.057 kgm²
I' = 0.339 kgm²
Answer:
Explanation:
Question 1
An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?
mass m = 20g = 20/1000 = 0.02kg
speed v = 50m / s
P.E = K.E = ½mv²
P.E = ½ × 0.02 × 50²
P.E = 25 J
work done = P.E = 25J
Qestion 2
A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.
force of elasticity
F = -kx
x = F / k
in our case F will be the force of pressure or gravity
F = mg
g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg
F = 80kg * 10N / kg = 800 N
x = 800N / -2000N = -0.4
The trampoline will lower, so from the level by 0.4 meters and hence this minus
