The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.
<h3 /><h3>Velocity: </h3>
This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.
⇒ Formula:
- mv²/2 = ke²/2
- mv² = ke².................. Equation 1
⇒ Where:
- m = mass of the stuntman
- v = velocity of the stuntman
- k = force constant of the spring
- e = compression of the spring
⇒ Make v the subject of the equation
- v = √(ke²/m)................. Equation 2
From the question,
⇒ Given:
- m = 48 kg
- k = 75 N/m
- e = 4 m
⇒ Substitute these values into equation 2
- v = √[(75×4²)/48]
- v = √25
- v = 5 m/s.
Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
Learn more about velocity here: brainly.com/question/10962624
Answer:
Explanation:
We have to find electric potential V at a distance r.
a) For r>R,
The electric field in the cylinder is given by
E.A equating it to the other electric field given by
б.A/ε₀
Here the area of cylinder is given by= 2*3.14*r*L
While for the outside, the area= 2*3.14*R*L
Equating both, we get
E= бR/rε₀
Now,
The potential difference is given as:
ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.
Where ΔV= V₀-V
So solving we get
V₀=V-бR/ε₀ln (r/R)
b) For r<R i.e. inside the cylinder
There will be no electric field produced as E=0
So ultimately Vin= V
c) V=0 at r= infinity.
K=0.5 mu×u
K=2200J no matter the direction
By using common factors of physics: weight, gravity, and stability.
Weight would keep them at a constant height.
Gravity helps the weight and how much force it propels the person, or objects, into the air.
Stability helps adjust how much distance the person, or object, needs to be.
Even tho one is stronger then the other... they are both alike because they are still nuclear forces.
