Vector A has magnitude 13.0 m and vector B has magnitude 16.0 m. The scalar product A.B is 116 m2. What is the magnitude of the
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Answer:
a) # buses = 7
Explanation:
For this exercise we use the kinematic equations, let's find the time it takes to reach the same height
y = t - ½ g t²
Let's decompose the speed, with trigonometry
v₀ₓ = v₀ cos θ
= v₀ sin θ
v₀ₓ = 40 cos 32
v₀ₓ = 33.9 m / s
= 40 sin32
= 21.2 m / s
When it arrives it is at the same initial height y = 0
0 = ( - ½ gt) t
That has two solutions
t = 0 when it comes out
t = 2 / g when it arrives
t = 2 21.2 /9.8
t = 4,326 s
We use the horizontal displacement equation
x = vox t
x = 33.9 4.326
x = 146.7 m
To find the number of buses we can use a direct proportions rule
# buses = 146.7 / 20
# buses = 7.3
# buses = 7
The distance of the seven buses is
L = 20 * 7 = 140 m
b) let's look for the scope for this jump
R = vo2 sin2T / g
R = 40 2 without 2 32 /9.8
R = 146.7 m
As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)
Answer:
Second projectile is 1.4 times faster than first projectile.
Explanation:
By linear momentum conservation
Pi = Pf
m x U + M x 0 = (m + M) x V
Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:
KEi + PEi = KEf + PEf
PEi = 0, at reference point
KEf = 0, Speed of system zero at height 'h'
PEf = (m + M) g h
So,
So from above value of V
Initial velocity of projectile =U
Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.
Which means
U₂ = 1.41 U₁
Therefore we can say that ,Second projectile is 1.4 times faster than first projectile.