Answer:
a) 0.00996 m
b) 109090909 Pa
Explanation:
Unit conversions:
1.2 mm = 0.0012 m
8.5 kN = 8500 N
If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is
With elastic modulus being E = 200 GPa, then its maximum stress must be
Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area
And its corresponding diameter is
To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
<span>a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. </span>
<span>b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. </span>
<span>f = (ma), = .226 x 105.88, = 23.92N. </span>
Answer:
5.95 ml
Explanation:
Given info
P1=2.9 atm
V1=8.21 ml
P2=4 atm
From Boyle's law we know that p1v1=p2v2 where p and v are pressure and volume respectively. This is at a constant temperature. Making v2 the subject of formula then
V2=p1v1/p2
V2= 2.9*8.21/4=5.95 ml
Answer:
friction
Explanation:
the road is the pathway for the vehicle so the tures create friction to stay sturdie
