... Extreme temperature changes between seasons.
... Extreme changes in length of daylight and darkness.
... Larger areas around both poles where daylight/darkness
can last more than a whole day.
Altogether, these areas cover half of the planet.
E=274J
h=140cm=1,4m
g≈9,8m/s²
m=?
E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg
"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."
Regards M.Y.
Answer:
a=g(sinθ-μkcosθ)
Explanation:
In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.
wx=m*g*sinθ
wy=m*g*cosθ
As the objects moves down an incline, acceleration in y axis is 0.
Then, by second Newton's Law:
Fy = m*ay
FN - m*g cos θ = 0,
FN=m*g cos θ
In x axis the forces that interacs are the x component of weight and friction force:
Fx = m*ax
mg sen u-FN*μk=m*a
Being friction force, Fr=FN*μk, we replace with its value in below formula:
m*g *sinθ-(m*g*cosθ*μk)=m*a
Then, isolating a:
a=(m*g sinθ-(m*g*cosθ*μk))/m
Solving, we have next equation:
a=g sinθ-(g*cosθ*μk)
Applying distributive property we have:
a=g*(sinθ-μk*cosθ)
Answer:
angular speed = 0.4 rad/s
Explanation:
given data
radius = 5 m
moment of inertia = 2000 kg-m²
angular speed = 1.0 rad/s
mass = 60 kg
to find out
angular speed
solution
Rotational momentum of merry-go-round = I?
we get here momentum that is express as
momentum = 2000 × 1
momentum = 2000 kg-m²/s
and
Inertia of people will be here as
Inertia of people = mr² = 60 × 5²
Inertia of people = 1500 kg-m²
so Inertia of people for two people
1500 × 2 = 3000
and
now conserving angular momentum(ω)
moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum
2000 × 1 = (2000 + 3000 ) ω
solve we get now
ω = 0.4 rad/s