Answer:
The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.
Explanation:
Answer:
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
Explanation:
The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.
Given are
<em> mass of the student 73 kg</em>
<em> height of water glide 11.8 m</em>
<em> work done as -5.5*10³ J</em>
Have to find speed at which the student goes down the glide.
According to<em> Law of Conservation of energy</em>,
K.E =P.E+Work Done
mv²/2=mgh +W
Rearranging the above eqn for v
v = √2(gh+W/m)
Substituting values,
V = 12.48 m/s.
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
Hello!
We can use the kinematic equation:
a = acceleration (m/s²)
vf = final velocity (45 m/s)
vi = initial velocity (25 m/s)
t = time (5 sec)
Plug in the givens:
A) the periodic time is given by the equation;
T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
= 6.28 × √0.0673
= 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
thus; frequency = 0.614 Hz
b) The vertical distance, the height is given by
h= 0.66 cos 12
h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
mgh = 1/2mv²
v² = 2gΔh = 2×9.81 × 0.01
= 0.1962
v = 0.443 m/s
c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
= 1/2 × 0.310 × 0.443²
= 0.0304 J
