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malfutka [58]
3 years ago
8

A certain part of a flat screen TV has a thickness of 150 nanometers. How many meters is this?

Physics
1 answer:
arlik [135]3 years ago
3 0

Answer:

Around 1.5e-7 meters.

.00000015 meters

Explanation:

.000000001 meters = 1 nanometer.

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You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1
drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

  • <em>Mass of the pendulum, = M </em>
  • <em>Length of the pendulum, = L</em>
  • <em>Initial angular speed, </em>\omega _i<em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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3 years ago
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A biologist formulates a hypothesis, performs experiments to test his hypothesis, makes careful observations, and keeps accurate
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3 years ago
A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?
musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where we have:

p_1 = 1.01\cdot 10^5 Pa (initial pressure is stp pressure)

V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3 is the initial volume

T_1 = 0^{\circ}=273 K is the initial temperature (stp temperature)

p_2 = 368 torr = 4.9\cdot 10^4 Pa is the final pressure

V_2 = ? is the final volume

T=155^{\circ}=428 K is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3

which corresponds to 2.41 L.

7 0
3 years ago
Could someone help me with this question please?
BabaBlast [244]
You are correct...amplitude will be the answer
3 0
3 years ago
Read 2 more answers
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