Answer:
At low pressure-
At high pressure-
Explanation:
Initial speed,
Final speed,
Net horizontal force due to rolling friction
mg where m is mass, g is acceleration due to gravity,
is coefficient of rolling friction
From kinematic relation,
For each tire,
Making
the subject
Under low pressure of 40 Psi, d=18 m
Therefore,
At a pressure of 105 Psi, d=93.7
Therefore,
Answer:
statements <em><u>2, 3, 4, and 7</u></em> are true
Explanation:
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:
Fg=Fcp
m*g=m*(v²/R),
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.
Masses cancel out and we have:
G*(M/r²)=v²/R, R=r so:
G*(M/r)=v²
r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,
r=G*(M/ω²r²),
r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:
r³=G*(M/(4π²/T²)), and finally we take the third root to get r:
r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.