(a) The stress in the post is 1,568,000 N/m²
(b) The strain in the post is 7.61 x 10⁻⁶
(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.
<h3>Area of the steel post</h3>
A = πd²/4
where;
d is the diameter
A = π(0.25²)/4 = 0.05 m²
<h3>Stress on the steel post</h3>
σ = F/A
σ = mg/A
where;
- m is mass supported by the steel
- g is acceleration due to gravity
- A is the area of the steel post
σ = (8000 x 9.8)/(0.05)
σ = 1,568,000 N/m²
<h3>Strain of the post</h3>
E = stress / strain
where;
- E is Young's modulus of steel = 206 Gpa
strain = stress/E
strain = (1,568,000) / (206 x 10⁹)
strain = 7.61 x 10⁻⁶
<h3>Change in length of the steel post</h3>
strain = ΔL/L
where;
- ΔL is change in length
- L is original length
ΔL = 7.61 x 10⁻⁶ x 2.5
ΔL = 1.9 x 10⁻⁵ m
Learn more about Young's modulus of steel here: brainly.com/question/14772333
#SPJ1
Since the bag was at rest, its initial momentum is zero. The velocity of the ball before collision is 500 ms-1.
<h3>Linear momentum</h3>
The term momentum in physics refers the product of mass and velocity. If we know mass of the object and its velocity, then we calculate the momentum.
Momentum before collision for the bullet = 0.01 kg × v
Momentum before collision for the bag = 0
Momentum after collision for the bag and bullet = (0.01 kg + 0.49 kg) 10 = 5 Kgms-1
The velocity of the bullet before collision = 0.01 kg × v + 0 = 5 Kgms-1
v = 5 Kgms-1/0.01 kg
v = 500 ms-1
Learn more about momentum: brainly.com/question/904448
Answer:
R=V/I=6/2=3ohm
time =5minutes =5*60=300seconds
I=2A
Heat =I^2Rt=(2)^2*3*300=4*900=3600J
Explanation:
Given:
v₀ = 0 m/s
v = 49 m/s
a = 9.8 m/s²
Find: t
v = at + v₀
49 m/s = (9.8 m/s²) t + 0 m/s
t = 5 s
KE = 1/2 mv^2 is the relationship betwee mass and kinetic energy
