Given:
Time: 3.5 hrs
Velocity: 120 miles/hr
Now Distance= Speed × Time
Now Velocity and speed have the same magnitude. Velocity being a vector quantity has a definite direction. Whereas speed is a scalar quantity,it indicates only the magnitude an doesn't define any direction.
Hence Distance = Velocity x time
Distance = 3.5 × 120 = 420 miles
More energy is released in nuclear reactions than in chemical reactions; this is because in nuclear reactions, mass is converted to energy. Nuclear energy released in nuclear fission and fusion is several 100 million times as large as an ordinary chemical reaction like the combustion process. The reason why nuclear energy release so much energy is because tremendous amounts of energy is released at one time. The nuclei in a nuclear reaction undergo a chain reaction, causing the neutrons to move extremely fast and release high amounts of energy.
Answer:
2156J
Explanation:
Given parameters:
Height of lift = 10m
Mass = 22kg
Unknown:
Work done by the machine = ?
Solution:
Work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Here;
Work done = mass x acceleration due to gravity x height
Work done = 22 x 9.8 x 10 = 2156J
The change in the internal energy of the system is 110 kJ.
<h3>What is internal energy?</h3>
Internal energy is defined as the energy associated with the random, disorder motions of molecules.
calculate the change in internal energy, we apply the formula below.
Formula:
- ΔU = Q-W.................... Equation 1
Where:
- ΔU = Change in internal energy
- Q = Heat absorbed from the surroundings
- W = work done by the system
From the question,
Given:
Substitute these values into equation 1
Hence, The change in the internal energy of the system is 110 kJ.
Learn more about change in internal energy here: brainly.com/question/4654659
Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
