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3 years ago

13

a spring with an unknown force constant is compressed as shown. If the original length of the spring is 60cm, determine (a)the s

pring force constant (b) the compressive stress in the spring (c) the strain of the spring

1 answer:

aalyn [17]3 years ago

6 0

<span>(b) the compressive stress in the spring </span>

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In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

cluponka [151]

Answer:

The thickness is $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 480 \ nm = 480*10^{-9} \ m$

The first order of the dark fringe is $m_1 = 16$

The second order of dark fringe considered is $m_2 = 6$

Generally the condition for destructive interference is mathematically represented as

$y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=> $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=> $\Delta y = 2.4 *10^{-6} \ m$

8 0

3 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

3 years ago

Middle school physics

Alecsey [184]

Answer:

8 V

Explanation:

There is no resistance between the left legs of voltmeters 2 and 3 and there is no resistance between the right legs of voltmeters 2 and 3. They are measuring the same voltage.

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3 years ago

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co

Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

$v_{d}=4.842\times10^{-5}\ m/s$

We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

$v_{d}$ = drift velocity

Put the value into the formula

$t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}$

$t=6402.31\ sec$

$t=106.7\ minute$

Hence, The time is 106.7 minute.

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3 years ago

What is the velocity of a worm moving 1 meter in 2 hours to the East?

malfutka [58]

Answer:

Velocity = 0.0001389 m/s

Explanation:

Given that the

Distance covered = 1 metre

Time taken = 2 hours

Convert the hour to second

1 hour = 60 × 60 = 3600

2 hours = 2 × 3600 = 7200

What is the velocity of a worm moving 1 meter in 2 hours to the East?

Velocity can be referred as speed.

Velocity = distance/ time

Velocity = 1/7200

Velocity = 0.0001389 m/s

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4 years ago