<h3><u>Answer;</u></h3>
volume = 6.3 × 10^-2 L
<h3><u>Explanation</u>;</h3>
Volume = mass/density
Mass = 0.0565 Kg,
Density = 900 kg/m³
= 0.0565 kg/ 900 kg /m³
= 6.3 × 10^-5 M³
but; 1000 L = 1 m³
Hence, <u>volume = 6.3 × 10^-2 L</u>
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
Answer:
1500 milliradians
Explanation:
Data provided in the question:
1.5 radians
Now,
1 radians consists of 1000 milliradians
1 milli = 1000
thus for the 1.5 radians, we have
1.5 radians = 1.5 multiplied by 1000 milliradians
or
1.5 radians = 1500 milliradians
Hence, after the conversion
1.5 radians equals to the value 1500 milliradians
There's so much going on here, in a short period of time.
<u>Before the kick</u>, as the foot swings toward the ball . . .
-- The net force on the ball is zero. That's why it just lays there and
does not accelerate in any direction.
-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.
<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...
-- The net force on the ball is 500N. That's what makes it accelerate from
just laying there to taking off on a high arc.
-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward.
That's how the leg's speed remains constant ... creating a dent in the ball
until the ball accelerates to match the speed of the foot, and then drawing
out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.