Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
Answer:
: velocity of the crate after the person has pushed the crate a distance of 6 meters
Explanation:
Crate kinetics
Crate moves with uniformly accelerated movement
v f²=v₀²+2a*d (formula 1)
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Known data
v₀=0 The speed of the crate is equal to zero because part of the rest.
a= 2m/s²
d= 6m
Distance calculating
We replace data in the Formula (1)
v f²=0+2*2*d
v f²=2*2*6
v f²=24
As per the question the distance travelled by a car is 28.4 inch.
we are asked to determine the conversion factor in centimeter which when multiplied with 28.4 inch will give a unit.
we know that one inch =2.54 centimeter.
Hence 28.4 inch = 2.54 ×28.4 cm
=72.136 cm.
Now we have to determine the conversion factor .The multiplication factor is calculated as
[p is the multiplication factor.]
Hence the multiplication factor is 72.137 cm which will give unit conversion when multiplied with 28.4 inch.