Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m
Where Fe is elastic force, W the weight and the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
Answer:
magma
Explanation:
I wanna think that that's right if it's not in so sorry but I'm pretty sure it's magma
Answer:
Explanation:
For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:
which can also be rewritten as
In our case, we have:
is the initial pressure
is the initial volume
is the final pressure
Solving for V2, we find the final volume:
The orbital with the lowest energy is 3s.
Answer:
Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?
Explanation:
solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).