Answer:
Explanation:
mass of the coin m = 4.2 x 10⁻³ kg
radius of circular path r in which coin will move = 15 x 10⁻² m
coefficient of static friction μs = .7
at maximum angular velocity of ω , limiting force of friction will provide centripetal force to coin to rotate in circular path .
so centripetal force = frictional force
mω²r = μs x mg
ω²r = μs x g
ω² = μs x g / r
= .7 x 9.8 / 15 x 10⁻²
= 45.73
ω = 6.76 radian / s.
Answer:
Explanation:
acceleration of an object on an inclined plane due to gravitational pull
= g sinθ . This acceleration is irrespective of mass of the object coming down . It only depends on the inclination .
If θ is larger , sinθ will be larger hence gravitational acceleration on inclined surface will be greater . It happens so because a component of weight mg is acting on the body . This component is mgsinθ . The other component mgcosθ acts perpendicular to surface hence it does not contribute to its acceleration .
acceleration produced = force / mass
= mgsinθ / m
= gsinθ .
in the given case , the inclination of plane with mass m₁ is less , hence its downward acceleration will be lesser as compared with the acceleration of m₂ which is put on a surface with greater inclination .
Answer:
"Apparent weight during the "plan's turn" is 519.4 N
Explanation:
The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is 
Given that,
v = 420 m/s
R = 11000 m
Substitute the values in the above equation,
It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector
In magnitude,
Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector
Which is quite heavier than his/her true weigh of 519.4 N
Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m
