Answer:
(C) four times as far
Explanation:
As we know that the range of the launcher is given as
here we know that all the parameters will remain same but only change is the velocity is doubled
So we will have
so we have
so here correct answer will be
(C) four times as far
Answer:
4213.2 Km/h
Explanation:
Given:
Initial Speed of space vehicle relative to Earth, u = 4150 km/h
Mass of rocket motor = 4m
speed of the rocket motor relative to command module , v' = 79 Km/h
Mass of the command module = m
Now,
let the speed of command module relative to earth be 'v'
From the conservation of momentum, we have
( 4m + m ) × u = m × v + (4m × (v - v'))
or
5m × 4150 = mv + 4mv - 4mv'
or
20750 = 5v - ( 4 × 79 )
or
20750 = 5v - 316
or
v = 4213.2 Km/h
Answer:
2.4583 ± 0.0207 seconds
Explanation:
The time period of a pendulum is approximately given by the formula ...
T = 2π√(L/g)
The maximum period will be achieved when length is longest and gravity is smallest:
Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds
The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:
Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds
If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.
T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2
T ≈ 2.4583 ± 0.0207 . . . seconds
__
We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.
Since q=mct, and q is heat ENERGY, masses and specific heat are equal, Temperature is the determining factor here. So as the equation suggests, liquid water has more energy as it contains more heat energy.