The question is not complete
Answer:
Part 1) Time of travel equals 61 seconds
Part 2) Maximum speed equals 39.66 m/s.
Explanation:
The final speed of the train when it completes half of it's journey is given by third equation of kinematics as
where
'v' is the final speed
'u' is initial speed
'a' is acceleration of the body
's' is the distance covered
Applying the given values we get
Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as
Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance
Thus total time of journey equals
Part b)
the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Answer:
400 trips
Explanation:
Mechanical energy needed to climb 14 m by a man of 68 kg
= mgh
= 68 x 9.8 x 14
= 9330 J
1 Kg of fat releases 3.77 x 10⁷ J of energy
.45 kg of fat releases 1.6965 x 10⁷ J of energy
22% is converted into mechanical energy
so 22% of 1.6965 x 10⁷ J
= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.
one trip of climbing of 14 m requires 9330 J of mechanical energy
no of such trip possible with given mechanical energy
= 3732.3 x 10³ / 9330
= 400 trips
Answer:
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Explanation:
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