Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
<h3>What is the energy of the roller coaster at point E?</h3>
The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.
Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,
Learn more about potential and kinetic energy at: brainly.com/question/18963960
#SPJ1
Answer:
The acceleration is 6 [m/s^2]
Explanation:
We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.
![v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2B%20a%2At%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%20%3D%20final%20velocity%20%3D%2022%20%5Bm%2Fs%5D%5C%5Cv_%7Bi%7D%20%3D%20initial%20velocity%20%3D%204%20%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%3D%203%20%5Bs%5D%5C%5C)
Now replacing the values we have:
![a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_%7Bf%7D%20-%20v_%7Bi%7D%20%7D%7Bt%7D%20%5C%5Ca%3D%5Cfrac%7B22%20-%204%20%7D%7B3%7D%5C%5Ca%20%3D%206%20%5Bm%2Fs%5E%7B2%7D%20%5D)
A) average acceleration = final velocity - initial velocity / time
= 7700 - 0 / 11
= 700ms^-2
B) force = mass x acceleration
= (3.05 x 105) x 700
= 320.25 x 700
= 224,175N
Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
