Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.
Let a = the acceleration of descent., m/s².
Let u = initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u, v, and a are measured as positive upward.
Then
u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a (1)
Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600 (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²
From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s
Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.
Answer:
Measurement is the quantification of attributes of an object or event, which can be used to compare with other objects or events
Answer:
Explanation:
We are given that
Length of antenna,l=1.45 m
Earth's magnetic field,
Speed,v=90 km/h=
Magnitude of maximum induced emf in the antenna=
Substitute the values
Answer:
Explanation:
Given:
- Initial temperature of water,
- final temperature of water,
- energy spent in one hour of walk,
- volumetric capacity of stomach,
<em>Now, let </em><em>m </em><em>be the mass of water at zero degree Celsius to be drank to spend 450 kilo-calories of energy.</em>
We know:
.....................................(1)
where:
m = mass of water
Q = heat energy
temperature difference
= specific heat capacity of water
<u>Putting values in the eq. (1):</u>
Since water has a density of 1 kilogram per liter, therefore the volume of water will be:
