Answer:
The horizontal displacement of the arrow is not larger than the banana split.
Explanation:
Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.
So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.
y - y₀ = ut - 1/2gt²
0 - y₀ = 0 × t - 1/2gt²
-y₀ = -1/2gt²
t² = 2y₀/g
t = √(2y₀/g)
Substituting the values of the variables, we have
t = √(2y₀/g)
= √(2 × 8848.04 m/9.8 m/s²)
= √(17696.08 m/9.8 m/s²)
= √(1805.72 s²)
= 42.5 s
The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s
So, d = vt
= 100 m/s × 42.5 s
= 4250 m
= 4.25 km
Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.
Answer:
recoil velocity of canon is 0.1039 m/s
Explanation:
given,
mass of shell = 97 kg
speed at which the shell is fired = 105 m/s
angle at which the shell is fired (θ) = 60°
mass of cannon plus car = 4.9 × 10⁴ kg
by conservation of momentum
∑ P = 0
recoil velocity of canon is 0.1039 m/s
Electricity is the movement of flow of electrons.
Answer:
6.78 km
Explanation:
Length of path due east = 4.3km
Length of path south = 2.48km
Unknown:
Distance covered = ?
Solution:
The distance covered is the total length of path from start to finish. It takes cognizance of the turns and every direction moved.
Unlike displacement which only considers the net direction from start to finish, distance sums up the total path.
So;
Distance = 4.3km + 2.48km = 6.78km
Answer:
the bearing of the line AC will be 154° 51' 48"
Explanation:
given,
bearing of the line AB = 234° 51' 48"
an anticlockwise measure of an angle to the point C measured 80°
to calculate the bearing of AC.
As the bearing of line AB is calculated clockwise from North direction.
the angle is moved anticlockwise now the bearing of AC will be calculated by
= bearing of line AB - 80°
= 234° 51' 48" - 80°
= 154° 51' 48".
= 154.5148
so, the bearing of the line AC will be 154° 51' 48"