Answer:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s
Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:
Rewriting the above expression in terms of 'x', we get:
Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:
Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.
the more pressure put on the string, the more frequency and higher pitch.
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
