The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
where
m=225 kg is the mass of the boat
is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].
The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.
At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.
Therefore we will have the following equation:
![(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]](https://tex.z-dn.net/?f=%286.5%2A9.81%2A120%29%2B%280.5%2A6.5%2A18%5E%7B2%7D%20%29%3D%286.5%2A9.81%2A60%29%2B%280.5%2A6.5%2Av_%7BB%7D%5E%7B2%7D%20%29%5C%5C3.25%2Av_%7BB%7D%5E%7B2%7D%20%3D4878.9%5C%5Cv_%7BB%7D%3D%5Csqrt%7B1501.2%7D%5C%5Cv_%7BB%7D%3D38.75%5Bm%2Fs%5D)
The kinetic energy can be easily calculated by means of the kinetic energy equation.
![KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]](https://tex.z-dn.net/?f=KE_%7BB%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7BB%7D%5E%7B2%7D%5C%5CKE_%7BB%7D%3D0.5%2A6.5%2A%2838.75%29%5E%7B2%7D%5C%5CKE_%7BB%7D%3D4878.9%5BJ%5D)
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
![E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7BC%7D%5C%5C6.5%2A9.81%2A120%2B%280.5%2A9.81%2A18%5E%7B2%7D%20%29%3D0.5%2A6.5%2Av_%7BC%7D%5E%7B2%7D%20%5C%5Cv_%7Bc%7D%5E%7B2%7D%20%3D%5Csqrt%7B2843.39%7D%5C%5Cv_%7Bc%7D%3D53.32%5Bm%2Fs%5D)
Please be determined and being hardworking person do not rely on the other people to make your problems solved
Explanation:
Ok?
Answer:
I THINK it’s A
Explanation:
Because all the other answers don’t make sense.
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² = / r
ω = √( / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
