Answer:
Explanation:
The two cars are under an uniform linear motion. So, the distance traveled by them is given by:
is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:
We have 
. Thus, solving for t:
Answer:
P = 17.28*10⁶ N
Explanation:
Given
L = 250 mm = 0.25 m
a = 0.54 m
b = 0.40 m
E = 95 GPa = 95*10⁹ Pa
σmax = 80 MPa = 80*10⁶ Pa
ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m
We get A as follows:
A = a*b = (0.54 m)*(0.40 m) = 0.216 m²
then, we apply the formula
ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L
⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)
⇒ P = 24624000 N = 24.624*10⁶ N
Now we can use the equation
σ = P/A
⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa
So σ > σmax we use σmax
⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N
Answer:
The observer sees the space-probe 9.055m long.
Explanation:
Let 
be the length of the space-probe when measured at rest, and 
be its length as observed by an observer moving at velocity 
, then
Now, we know that 
and 
, and putting these into 
we get:
Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.
