Answer:
The net displacement of the car is 3 km West
Explanation:
Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).
Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.
So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.
Answer:
a) 19440 km/h²
b) 10 sec
Explanation:
v₀ = initial velocity of the car = 45 km/h
v = final velocity achieved by the car = 99 km/h
d = distance traveled by the car while accelerating = 0.2 km
a = acceleration of the car
Using the kinematics equation
v² = v₀² + 2 a d
99² = 45² + 2 a (0.2)
a = 19440 km/h²
b)
t = time required to reach the final velocity
Using the kinematics equation
v = v₀ + a t
99 = 45 + (19440) t
t = 0.00278 h
t = 0.00278 x 3600 sec
t = 10 sec
We r made of atom so v can’t touch anything hehe I just joking
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.
Answer:
Control of air–fuel ratio
Oxygen sensors tell the ECU whether the engine is running rich (too much fuel or too little oxygen) or running lean (too much oxygen or too little fuel) as compared to ideal conditions (known as stoichiometric).
Explanation: