<span>You should deflect the
ball in order to maximize your speed on the skateboard.
Since this creates a larger impulse, you want to deflect the ball. Splitting it
up into catching and throwing the ball may by something you can think of deflecting
the ball. First, you need to catch the ball, which in turn would push you
forward with some speed. (The speed we are talking about should obviously be
equal to option A, where you catch the ball). Now, throw the ball back to him
since these two processes are equal to deflecting the ball. Throwing a mass away
from you would cause or enable you to move even fast.</span>
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the object is 
The unstressed length of the string is 
The length of the spring when it is at equilibrium is 
The initial speed (maximum speed)of the spring when given a downward blow 
Generally the maximum speed of the spring is mathematically represented as

Here A is maximum height above the floor (i.e the maximum amplitude)
and
is the angular frequency which is mathematically represented as

So

=> 
Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

=> 
=> 
Generally at equilibrium position the net force acting on the spring is

=> 
=> 
So

=> 
Answer:
1.7 m/s²
Explanation:
d = length of the ramp = 13.5 m
v₀ = initial speed of the skateboarder = 0 m/s
v = final speed of the skateboarder = 7.37 m/s
a = acceleration
Using the equation
v² = v₀² + 2 a d
7.37² = 0² + 2 a (13.5)
a = 2.01 m/s²
θ = angle of the incline relative to ground = 29.9
a' = Component of acceleration parallel to the ground
Component of acceleration parallel to the ground is given as
a' = a Cosθ
a' = 2.01 Cos29.9
a' = 1.7 m/s²
Answer:
common types of topologies, and we're going to break each of them down in the guide below.
Bus topology. As the simplest design, a bus topology requires nodes to be in a linear order. ...
Ring topology. Another simple design is the ring topology. ...
Star topology. ...
Mesh topology. ...
Tree topology.
Explanation:
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s