The average velocity of the car for the whole journey is 69.57 km/h.
The given parameters:
- <em>Length of the road, L = 320 km</em>
- <em>Distance covered = 240 km at 75 km/h</em>
- <em>time spent refueling, t₂ = 0.6 hr</em>
- <em>Final velocity, = 100 km/hr</em>
The time spent by the before refueling is calculated as follows;
The time spent by the car for the remaining journey;
The total time of the journey is calculated as follows;
The average velocity of the car for the whole journey is calculated as follows;
Learn more about average velocity here: brainly.com/question/6504879
Answer:
I believe D
Explanation:
You need to have a more accurate reading and you want to test it multiple times throughout the week though to get a base resting rate.
I hope this is correct good luck!
Answer:
if there is no friction in a simple machine, work output and work input are found equal in that machine
Explanation:
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
Explanation:
5 C = 278 K
25 C = 298 K
V1 / T1 = V2 / T2
1.5L / 278 K = V2 / 298 K
V2 = (1.5L * 298) / 278
V2 = 1.61 L
