Car A take a time of 2.55hr and car B take a time of 2.14 hr
We know that distance divide by time is speed
here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr
so speed=distance/time
s=d/t
t=d/s
=189/74
=2.55hr
In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr
s=d/t
t=d/s
=199.8/93
=2.14hr
so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h
learn more about Speed here brainly.com/question/13943409
#SPJ9
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ 
(b)
For m₃,
⇒ 
Or,
⇒ 
Answer:
2.572 m/s²
Explanation:
Convert the given initial velocity and final velocity rates to m/s:
- 65 km/h → 18.0556 m/s
- 35 km/h → 9.72222 m/s
The motorboat's displacement is 45 m during this time.
We are trying to find the acceleration of the boat.
We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.
Substitute the known values into the equation.
- (9.72222)² = (18.0556)² + 2a(45)
- 94.52156173 = 326.0046914 + 90a
- -231.4831296 = 90a
- a = -2.572
The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².
Momentum
= mass x velocity
= 0.2 x 5
= 1 kg m/s
