F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s
Answer:
The change in momentum is
Explanation:
From the question we are told that
The mass of the probe is 
The location of the prob at time t = 22.9 s is 
The momentum at time t = 22.9 s is
The net force on the probe is 
Generally the change in momentum is mathematically represented as

The initial time is 22.6 s
The final time is 22.9 s
Substituting values

Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
Explanation:
When,the vehicle has uniform velocity, it's acceleration becomes zero
Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2