0.75 km = (0.75×100000) cm = 75000 cm
The scientific notation to find this is ---
Kilo-Hecto-Deca-Meter-Deci-Centi-Milli
Answer:
The frictional torque is
Explanation:
From the question we are told that
The mass attached to one end the string is
The mass attached to the other end of the string is
The radius of the disk is
At equilibrium the tension on the string due to the first mass is mathematically represented as
substituting values
At equilibrium the tension on the string due to the mass is mathematically represented as
The frictional torque that must be exerted is mathematically represented as
substituting values
Answer:
The magnitude of the tension in the cable, T is 1,064.315 N
Explanation:
Here we have
Length of beam = 4.0 m
Weight = 200 N
Center of mass of uniform beam = mid-span = 2.0 m
Point of attachment of cable = Beam end = 4.0 m
Angle of cable = 53° with the horizontal
Tension in cable = T
Point at which person stands = 1.50 m from wall
Weight of person = 350 N
Therefore,
Taking moment about the wall, we have
∑Clockwise moments = ∑Anticlockwise moments
T×sin(53) = 350×1.5 + 200×2
T = 850/sin(53) = 1,064.315 N.
Answer:
707 m/s
Explanation:
Since the root mean temperature is directly proportional to the root mean square velocity.
We have
V1/V2 =√T1/T2
V1= root mean square velocity of nitrogen at 300K
V2= root mean square velocity of nitrogen at 600K
Thus;
517/V2 = √300/600
517/V2 = 0.707
V2 = 500/0.707
V2= 707 m/s