Answer:
Speed BA = 18 km/hr.
Explanation:
Given the following data;
Speed AB = 30km/hr
Speed BA = x km/hr
Average speed = 24 km/hr
To find the value of x;
Average speed = (Speed AB + Speed BA)/2
Substituting into the equation, we have
24 = (30 + x)/2
48 = 30 + x
x = 48 - 30
x = 18 km/hr
The spring constant of the spring is 240.05 N/m according to the information on the question above. This problem can be solved using the spring potential energy formula which stated as PE = 1/2 * k * x^2 where PE is the spring's potential energy, k is the constant of the spring, and x is the stretched displacement<span>. (Calculation: 240.05=35*2/(0.54^2) </span>
Answer:
Explanation:
This problem is based on conservation of rotational momentum.
Moment of inertia of rod about its center
= 1/12 m l² , m is mass of the rod and l is its length .
= 1 / 12 x 4.6 x .11²
I = .004638 kg m²
The angular momentum of the bullet about the center of rod = mvr
where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .
5 x 10⁻³ x v sin60 x .11 x .5 where v is velocity of bullet
According to law of conservation of angular momentum
5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and ( I + mr²) is moment of inertia of bullet rod system .
.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12
.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12
.238 x 10⁻³ v = 55.8375 x 10⁻³
.238 v = 55.8375
v = 234.6 m /s
Im afraid i cant help u if we can’t see the image u r working with. could u provide an image of the question?
Hi there!
Recall the conservation of momentum:
For this type of inelastic collision:
Thus, the initial momentum equal the final momentum if there are no external forces.
We can begin by writing out this problem: