Given: Mass m = 45 Kg: Density ρ = 2.7 Kg/m³
Required: Volume V = ?
Formula: ρ = m/V
V = m/ρ
V = 45 Kg/2.7 Kg/m³
V = 16.67 m³
Answer:
3.7 m
Explanation:
ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky
The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of
t = √ (2h/g) = √(2(15)/9.8) = 1.75 s
Without the tail wind, the ball travels horizontally
d = vt = 68(1.75) = 119 m
The tailwind exerts a constant acceleration on the ball of
a = F/m = 12/5.0 = 2.4 m/s²
The average horizontal velocity during the flight is
v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s
so the distance with tailwind is
d = v(avg)t = 70.1(1.75) = 122.675 m
The extra distance is 122.675 - 119 = 3.675 = 3.7 m
Answer:
Explanation:
Given that
Length= 2L
Linear charge density=λ
Distance= d
K=1/(4πε)
The electric field at point P
So
Now by integrating above equation
Answer:
As light travels in a straight line at a constant speed, it's acceleration is <u>0 m/s²</u>.
There is no rate of change of speed, so there is no acceleration.
- <u>0 m/s²</u> is the right answer.