how do you solve this problem. A bat flying at 3.7m/s spots an insect 23.8 meters away. How much must the bat increase its speed
1 answer:
You might be interested in
Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as
By putting the values
ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation
150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.
Answer:
Shawn's speed relative to Susan's speed = 10 mph
Resultant velocity = 82.32 mph
Explanation:
The given data :-
i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.
ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.
iii) The speed of both Susan and Shawn is relative to earth.
iv) The angle between Susan in north and Shawn in east is 90°.
We have to find Shawn's speed relative to Susan's speed.
v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph
Resultant velocity,
v = 82.32 mph
Explanation:
It is given that,
Magnetic field, B = 0.15 T
Charge on a proton,
Mass of a proton,
The cyclotron frequency is given by :
f = 2286785.40 Hz
or
Hence, this is the required solution.