Explanation:
It is known that the relation between speed and distance is as follows.
velocity =
As it is given that velocity is 6 m/s and distance traveled by the bear is (d + 29). Therefore, time taken by the bear is calculated as follows.
............. (1)
As the tourist is running in a car at a velocity of 4.2 m/s. Hence, time taken by the tourist is as follows.
............. (2)
Now, equation both equations (1) and (2) equal to each other we will calculate the value of d as follows.
=
4.2d + 121.8 = 6d
d =
= 67.66
Thus, we can conclude that the maximum possible value for d is 67.66.
The input voltage is 120 V and the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.
<h3>
Input voltage </h3>
The input voltage of the transformer is the voltage of the primary coil and it is calculated as follows;
Ns/Np = Es/Ep
where;
- Ns is the number of turn in the secondary coil
- Np is the number of turn in the primary coil
- Es is the secondary voltage
- Ep is the primary voltage
2X/X = 240/Ep
2 = 240/Ep
Ep = 240/2
Ep = 120 V
Thus, the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.
Learn more about transformer here: brainly.com/question/25886292
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Answer:
Final velocity of the elevator will be 4.453 m/sec
Explanation:
Let mass is m
Acceleration due to gravity is g m/sec^2
Distance s = 2.2 m
As the elevator is moving upward so net force on elevator
So according to question
0.46 mg = ma
a = 0.46 g
a = 0.46×9.8 = 4.508
Initial velocity of elevator is 0 m/sec
From third equation of motion
So final velocity of the elevator will be 4.453 m/sec
Answer:
the maximum frequency observed is 2.0044 10⁶ Hz
Explanation:
This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is
f ’=
the + sign is used when the observer approaches the source
typical speeds of a baby's heart stop are around 200 m / min
let's reduce to SI units
v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s
let's calculate
f ’= 2 10⁶ ()
f ’= 2.0044 10⁶ Hz
f ’= 1,9956 10⁶ Hz
therefore the maximum frequency observed is 2.0044 10⁶ Hz