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3 years ago

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Physics

1 answer:

Elan Coil [88]3 years ago

8 0

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

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

T is the time period of the crystal's motion.

Time period is given by :

$T=\dfrac{1}{32768}$

$T=3.05\times 10^{-5}\ s$

So, the time period of the crystal's motion is $3.05\times 10^{-5}\ s$ . Hence, this is the required solution.

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An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what

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Answer:

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3 years ago

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0

3 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

3 years ago

A 0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of co

Alex_Xolod [135]

To solve this problem, we should recall the law of conservation of energy. That is, the heat lost by the aluminium must be equal to the heat gained by the cold water. This is expressed in change in enthalpies therefore:

- ΔH aluminium = ΔH water

where ΔH = m Cp (T2 – T1)

The negative sign simply means heat is lost. Therefore we calculate for the mass of water (m):

- 0.5 (900) (20 – 200) = m (4186) (20 – 0)

m = 0.9675 kg

Using same mass of water and initial temperature, the final temperature T of a 1.0 kg aluminium block is:

- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)

- 900 T + 180,000 = 4050 T

4950 T = 180,000

T = 36.36°C

The final temperature of the water and block is 36.36°C

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4 years ago

Read 2 more answers

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Arte-miy333 [17]

<h3>Answer;</h3>

100 times

<h3>Explanation;</h3>

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3 years ago