1. A train is traveling on a straight section of track at constant speed. In 60
2 answers:
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Answer:
The average velocity of the car is, V = 74.04 m/s
Explanation:
Given data,
The initial velocity of the car, u = 0 m/s
The displacement of the ca, S = 1100 m
The time period of travel, t = 14 s
The velocity of the car at point k, v = 65 m/s
Using the II equation of motion,
S = ut + ½ at²
Substituting the given values,
1100 = 0 + ½ x a x 14²
a = 11.22 m/s²
Using the III equation of motion
v² = u² + 2 as
v = √(2as) (∵ u = 0)
Substituting,
v = √(2 x 11.22 x 1100)
= 157.11 m/s
The average speed of the car,
V = 74.04 m/s
Hence, the average velocity of the car is, V = 74.04 m/s
Answer:
a. The thickness of the wire is 2.5 mm.
b. The wire is 0.25 cm thick.
Explanation:
Number of turns of the wire = 10
The length of total turns = 25 mm
a. The thickness of the wire can be determined by;
thickness of the wire =
=
= 2.5 mm
Therefore, the wire is 2.5 mm thick.
b. To determine the thickness of the wire in centimetre;
10 mm = 1 cm
So that,
2.5 mm = x
x =
= 0.25 cm
The wire is 0.25 cm thick.
Answer:
t = 5 s
Explanation:
Data:
- Initial Velocity (Vo) = 7 m/s
- Acceleration (a) = 3 m/s²
- Final Velocity (Vf) = 22 m/s
- Time (t) = ?
Use formula:
Replace:
Solve the subtraction of the numerator:
It divides:
How much time did it take the car to reach this final velocity?
It took a time of <u>5 seconds.</u>