Answer:
L = ¼ k g / m
Explanation:
This is an interesting exercise, in the first case the spring bounces under its own weight and in the second it oscillates under its own weight.
The first case angular velocity, spring mass system is
w₁² = k / m
The second case the angular velocity is
w₂² = L / g
They tell us
w₂ = ½ w₁
Let's replace and calculate
√ (L / g) = ½ √ (k / m)
L / g = ¼ k / m
L = ¼ k g / m
Solution
x(t) = 8 cos t, x(5π/6)= 8 cos(<span>5π/6)
</span>cos(5π/6)=cos(3π/6 + 2π/6 )=cos(π/3 +π/2)= - sin π/3 (cos (x+<span>π/2)= -sinx)
</span>x(t) = -8sin <span>π/3 = - 4 .sqrt3
</span>v(t) = -8sint = -8sin (π/3 +<span>π/2)= -8 cosπ/3 </span>(sin (x+π/2)= cosx)
v(t) =<span> -8 cosπ/3 = -8/2= - 4
</span>a(5π/6) = - 8cost = -(- sin π/3)= 4 .<span>sqrt3
</span>a(5π/6) = 4 .<span>sqrt3</span>
To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,
Here,
v = Lineal velocity
= Angular velocity
r = Radius
Our values are
Replacing to find the angular velocity we have,
Convert the units to RPM we have that
Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm
Because then it could mess up the machine with to much energy
Answer:
Explanation:
Distance between plates d = 2 x 10⁻³m
Potential diff applied = 5 x 10³ V
Electric field = Potential diff applied / d
= 5 x 10³ / 2 x 10⁻³
= 2.5 x 10⁶ V/m
This is less than breakdown strength for air 3.0×10⁶ V/m
b ) Let the plates be at a separation of d .so
5 x 10³ / d = 3.0×10⁶ ( break down voltage )
d = 5 x 10³ / 3.0×10⁶
= 1.67 x 10⁻³ m
= 1.67 mm.
