Answer:
a= 17.69 m/s^2
Explanation:
Step one:
given data
A car accelerates uniformly from rest to 23 m/s
u= 0m/s
v= 23m/s
distance= 30m
Step two:
We know that
acceleration= velocity/time
also,
velocity= distance/time
23= 30/t
t= 30/23
t= 1.30 seconds
hence
acceleration= 23/1.30
accelaration= 17.69 m/s^2
Answers:
a) -2.54 m/s
b) -2351.25 J
Explanation:
This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum
must be equal to the final momentum
:
(1)
Where:
(2)
(3)
is the mass of the first football player
is the velocity of the first football player (to the south)
is the mass of the second football player
is the velocity of the second football player (to the north)
is the final velocity of both football players
With this in mind, let's begin with the answers:
a) Velocity of the players just after the tackle
Substituting (2) and (3) in (1):
(4)
Isolating
:
(5)
(6)
(7) The negative sign indicates the direction of the final velocity, to the south
b) Decrease in kinetic energy of the 110kg player
The change in Kinetic energy
is defined as:
(8)
Simplifying:
(9)
(10)
Finally:
(10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision
Answer:
c = 894.90 m/s
Explanation:
Given data:
Frequency of wave = 471 Hz
Wavelength of wave = 1.9 m
Speed of wave = ?
Solution:
Formula:
Speed of wave = frequency × wavelength
c = f×λ
c = 471 Hz × 1.9 m
Hz = s⁻¹
c = 471s⁻¹ × 1.9 m
c = 894.90 m/s
The speed of wave is 894.90 m/s.
<span>The de-acceleration or negative acceleration of stopping is what damages bones. The ground is rigid and therefore the change in momentum when striking the ground will be large. On the trampoline, the elasticity of the material means that the momentum changes more slowly, resulting in smaller accelerations.</span>
By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)