Answer:
maximum speed of the block is 3.14 m/s
Explanation:
given data
mass = 1.70 kg
stretch in the spring = 0.2 m
time take by block to come to zero t = 0.2 s
solution
we know that Time period of oscillation (T) that is express as
T = 2t ......................1
put here value
T = 2 (0.2)
T = 0.4 s
so here time period is express as
T =
................2
here k is spring constant of the spring so put here value
0.4 =
here k will be
k = 419.02 N/m
so we use here conservation of energy that is
Maximum kinetic energy = Maximum spring potential energy ............3
(0.5) m v² = (0.5) k x²
here v is maximum speed block
so put here value and we get
(1.70) v² = (419.02) (0.2)²
v = 3.14 m/s
so maximum speed of the block is 3.14 m/s
For a 568B crossover cable that already has wht-org and org on pins 1 and 2 of the connector at one end, the connector at the OTHER end should have wht-grn and grn on pins 1 and 2 respectively.
The wht-org and org at that end should drop to pins 3 and 6 respectively.
You're welcome, and good luck.
<u>Answer</u>
The acceleration is
to the nearest tenth
<u>Explanation</u>
Since the car was travelling at
before it starts to decelerate, the initial velocity is
.
The final velocity is
, because the car came to a stop.
The time taken is
.
Using the Newton's equation of linear motion,
, we find the acceleration by substituting the known values.
This implies that,

This gives us,


We divide both sides by 15 to get,

or

D make one revolution around the sun hope this helps