Answer:
The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
hmax = 5740.48 m
Explanation:
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
Explanation:
Hey there!!
Let's simply work with it.
Here,
load = 1200N
Effort = 200N
Load distance = 15cm
We have,
According to the principle of lever.
L×LD = E×ED.
1200×15 = 200× ED.
18000 = 200ED.
Therefore, Effort Distance = 90cm.
<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:14 s
Explanation:
Given
Velocity of ant is 0.1 m/s in a direction 
if it has traveled 1 m perpendicular to the edge of the sidewalk
i.e. from diagram
Answer:
96 m
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²
Δx = 96 m
Answer:
This causes higher average tidal ranges. The gravitational pull of the Sun and moon on Earth combined cause high tides that will be higher and low tides that will be lower than average.
