(1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)
= (1.9 x 365.24 x 86,400 x 10⁹) nanosec
= 6.00 x 10¹⁶ nanoseconds
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
A region within a magnetic material in which magnetization is in a uniform direction this means the individual magnetic moments of the atoms are aligned with one another and they point the same direction. when cooled bwlow a temperature called the curie temperature the magnetization of a piece of ferromagnetic material.<span />
Given:-
- Speed of the unicycle = 20 m/s
- Time taken = 15 s
To Find: Distance travelled by the unicycle.
We know,
s = vt
where,
- s = Distance travelled,
- v = Speed &
- t = Time taken.
Therefore,
s = (20 m/s)(15 s)
→ s = (20 m)(15)
→ s = 300 m (Ans.)