A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.
1 answer:
Answer:
3.59 m/s
Explanation:
The average velocity is defined as:
The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement is 1082m north (1100m north - 18m south).
Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.
The walking time is 3.5 minutes. Since 1 minute = 60 seconds, then the walking time is 210 seconds.
To calculate the truck time we use the equation:
Time =
Where the distance the truck travelled is 1100m and the speed of the truck is 12m/s.
Thus,
Truck time= = 91.67s
The total time is the sum of the walking time and the truck time.
Total time = 210s + 91.67s = 301.67s.
As mencioned previously, the average velocity is equal to total displacement/ total time, thus:
Average velocity = = 3.59 m/s North
Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.
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Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
In parallel
Solving the above quadratic equation
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Answer:
The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.
Explanation:
Statement is incomplete. Complete description is presented below:
<em>A freight train has a mass of </em><em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>
<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>
If locomotive have a constant net force (), measured in newtons, then acceleration (
), measured in meters per square second, must be constant and can be found by the following expression:
(1)
Where is the mass of the freight train, measured in kilograms.
If we know that and
, then the acceleration experimented by the train is:
Now, the time taken to accelerate the freight train from rest (), measured in seconds, is determined by the following formula:
(2)
Where:
- Final speed of the train, measured in meters per second.
- Initial speed of the train, measured in meters per second.
If we know that ,
and
, the time taken by the freight train is:
The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:
Finally, you replace the values of all parameters in the previous equation for I:
The moment of inertia of the door around the hinges is 2 kgm^2