Answer:
V= 6.974 m/s
Explanation:
Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N
Force of kinetic friction acting parallel and up roof = 18.0 N
Fnet force acting on tool box acting parallel and down roof
Fnet= 55.4 - 18.0
Fnet=37.4 N
acceleration of tool box down roof
a = 37.4(9.81)/88.0
a= 4.169 m/s²
d = 4.90 m
t = √2d/a
t= √2(4.90)/4.169
t= 1.662 s
V = at
V= 4.169(1.662)
V= 6.974 m/s
when object goes under acceleration
c).its velocity always increases
<h3><u>Additional</u><u> </u><u>information</u><u>:</u><u>-</u></h3>
★ Acceleration: Rate of increase in velocity.
★ Velocity: Distance travelled by a body per unit time in given direction is called velocity .
Answer:
The average force on ball by the golf club is 340 N.
Explanation:
Given that,
Mass of the golf ball, m = 0.03 kg
Initial speed of the ball, u = 0
Final speed of the ball, v = 34 m/s
Time of contact, 
We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :
So, the average force on ball by the golf club is 340 N.
When acceleration is constant, the average velocity is given by
where 
and 
are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by
where 
are the final/initial displacements, and 
are the final/initial times, respectively.
Take the car's starting position to be at 
. Then
So we have
You also could have first found the acceleration using the equation
then solve for 
via
but that would have involved a bit more work, and it turns out we didn't need to know the precise value of 
anyway.
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
