Answer:
7.19 * 10^14J
Explanation:
Given that
Density of water Pwater= 1000kg/m3
R=2.1km = 2.1*10^3m
H= 2.3cm. = 2.3*10^-2m
Lv water= 2256 * 10^3J/kg
First, mass of water need to be calculated, using an imaginary cylinder
Density= Mass /Volume
Mass= Density* Volume
Volume of a cylinder= πR2h
Therefore mass= PπR2H
= 1000 * π * (2.1 *10^3)^2 * (2.3 * 10^-2)
= 3.18 *10^8
Heat Released Qv = mLV
= 3.18*10^8 * 2236*10^3
= 7.19 * 10^14J
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
Answer:
Mass doesn't change.
Weight is measured based on gravitational pull.
Explanation:
Answer:
False
Explanation:
Because when you go through east
( +x axis ) then you go to west ( -x axis )
You will subtract -9 from +15
it's become +6
( I talk about the displacement not distance) ( West = - East )
I hope that it's a clear ") .
<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
