Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
Given mass= 1kg
Weight on earth = mg(gravity of earth) = 9.8N
weight on moon = mg(gravity of moon)= 1.62N
weight on outer space mg(gravity outer space = 0) = 0N
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
Answer:
a=2 ok do it and ........
Answer:
20 mangintiude beacuse
Explanation:
mt everst is in mounatin region we eat rice we eat pizza burger sandwich go to thamel for a good reason doing lamo lamo hw so be obtidnet you 4 kaccha fail boy
