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)
The Kepler's laws predict the planetary motion, so there are three laws for this, namely:
1. The orbit of a planet is an ellipse with the Sun (the sun is a star!) at one of the two focus.
2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
So, let's use second law. The Sun sweeps out equal areas during equal intervals of time means that if A = B, the time the planet takes to travel A1A2 is equal to the time the planet takes to travel B1B2, but given that A = 2B, then takes twice the time to travel A1A2 compared to B1B2.
Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
Answer:
Explanation:
Given that,
The magnitude of magnetic field, B = 0.55 T
The radus of the loop, r = 43 cm = 0.43 m
The current in the loop, I = 5.8 mA = 0.0058 A
We need to find the magnetic moment of the loop. It is given by the relation as follows :
Put all the values,
So, the magnetic moment of the loop is equal to
.
