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3 years ago

8

As you move from left to right across the electromagnetic spectrum in the image, the wavelength becomes

1 answer:

svp [43]3 years ago

4 0

The shortest wavelength is on the (high energy)left side of this diagram with Gamma rays. The longest wavelength waves are radio, around 3 meters. So, as you progress, the wavelengths get A) <u><em>longer. </em></u>

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A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between

Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

$a = \dfrac{F}{m}$

Where, F = frictional force

m = mass

Put the value into the formula

$a=\dfrac{\mu mg}{m}$

$a=\mu g$

$a=0.229\times9.8$

$a=2.244\ m/s^2$

We need to calculate the speed of the sled

Using equation of motion

$v^2=u^2-2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v ^2=(5.49)^2-2\times2.244\times3.89$

$v=\sqrt{(5.49)^2-2\times2.244\times3.89}$

$v=3.56\ m/s$

Hence, The speed of the sled is 3.56 m/s.

8 0

3 years ago

A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.

DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy $Q_{out}=1000\ Btu$

Work done $W_{cycle}=300\ Btu$

We need to calculate the coefficient of performance

Using formula of the coefficient of performance

$COP=\dftrac{Q_{in}}{W_{cycle}}$

We need to calculate the $Q_{in}$

$W_{cycle}=Q_{out}-Q_{in}$

Put the value into the formula

$300=1000-Q_{in}$

$Q_{in}=300-1000$

$Q_{in}=700\ Btu$

Now put the value of $Q_{in}$ into the formula of COP

$COP=\dfrac{700}{300}$

$COP=\dfrac{7}{3}=2.33$

Hence, The coefficient of performance for the cycle is 2.33.

5 0

4 years ago

What is the relationship between the buildup of electons and the sudden flow of electrons from one charged object to another?

melisa1 [442]

Static electricity

hope this helps :)

5 0

4 years ago

Read 2 more answers

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/

Hunter-Best [27]

Answer:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

$n_{T} = n_{1} + n_{2}$

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$

$n_{T} = 31.68\,rev$

4 0

4 years ago