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3 years ago

15

OSCilloscope shows wave starts cycle minimum 20 units reaches max of 100 units before completing 5 milliseconds what is relation

ship between time and wave function

1 answer:

Vitek1552 [10]3 years ago

6 0

Answer:

$T=0.02\ s$

$f=50\ Hz$

Explanation:

Given:

minimum amplitude at the start of oscillation cycle, $a_0=20\ unit$

the first maximum amplitude after the start of oscillation cycle, $a_{m1}=100\ units$

Time taken to reach from the first minima to the first maxima, $t=5\times 10^{-3}\ s$

As we know that an oscilloscope executes a wave cycle represented by a sine wave. So we can deduce that it has executed one-fourth of the cycle in going from the amplitude of 20 units to 100 units in 0.005 seconds.

<u>So the time taken to complete one cycle of the oscillation:</u>

$T=4\times 0.005$

$T=0.02\ s$ is the time period of the oscillation

<u>We know frequency:</u>

$f=\frac{1}{T}$

$f=\frac{1}{0.02}$

$f=50\ Hz$

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3 years ago

What is average velocity of a particle?

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3 years ago

Read 2 more answers

A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density

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Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude $mg$ , where m = 5.32 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude $kx$ , where $k=179N/m$ is the spring constant and $x$ is the elongation of the spring

- The buoyant force, upward, of magnitude $\rho V g$ , where $\rho=1000 kg/m^3$ is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

$mg+kx-\rho V g=0$ (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

$\rho_w = 622 kg/m^3$ (wood density)

So, the volume is

$V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3$

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

$x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m$

So, the spring is compressed by 0.18 m.

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3 years ago

7. How much thermal energy is absorbed when 10.0 g of ice at 0.0 degrees C changes phase to liquid water at 0.0 degrees C? The m

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Answer:

3.34 kJ

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Now we know that the heat of fusion is

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$Q=n H_{fus} = (0.555 mol)(6.01 kJ/mol)=3.34 kJ$

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3 years ago