<span> Weight = mass x acceleration
Earths acceleration is 9.8 m/s*2
1 kg = 2.2 lbs, so 2.0 lbs x 1 kg/2.2 lbs = 0.91 kg
The bag would have a weight of 9.8 x 0.91 = 8.9 N
1. 8.9 x 1/6 = 1.5 N
2. 8.9 x 2.64 = 23.5 N
The mass of the bag at all three locations is 0.91 kg. Mass does not change, the different locations only change its weight. </span>
Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
Answer:
a
Generally from third equation of motion we have that
![v^2 = u^2 + 2a[s_i - s_f]](https://tex.z-dn.net/?f=v%5E2%20%3D%20%20u%5E2%20%2B%202a%5Bs_i%20-%20s_f%5D%20)
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
![v^2 = 0 + 2g[H - 0]](https://tex.z-dn.net/?f=v%5E2%20%3D%200%20%2B%202g%5BH%20-%200%5D%20)
=> 
b
Explanation:
Generally from third equation of motion we have that
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
=>
When 
we have that
=> 
=>
Answer:
particles
Explanation:
in liquids, particle are close together
