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2 years ago

12

An ultraviolet wave traveling through a vacuum has wavelength of 4.0 x 10^-7 m. The waves frequency, written in scientific notat

ion to two significant figures, is ? X10^14Hz.

1 answer:

Naily [24]2 years ago

5 0

Answer:

λ = c / f or f = c / λ

f = 3.0E8 / 4.0E-7 = .75E15 / sec = 7.5E14 / sec = 7.5 X 10^14 /sec

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Arrange the core steps of the scientific method in sequential order.

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An 7.9 kg crate is pulled 5.5 m up a 30∘ incline by a rope angled 19 ∘ above the incline. The tension in the rope is 140 N and t

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The 1.19 kg head of a hammer has a speed of 8.4 m/s just before it strikes a nail (Fig. 14-13) and is brought to rest. Estimate

KATRIN_1 [288]

Answer:

The temperature rise is $143.9\ K$ .

Explanation:

Given the mass of the hammer $(m)$ is 1.19 kg.

And the speed of the hammer $(v)$ is 8.4 m/s.

We need to find the temperature change of iron nail having 20 hammer blows.

First, we will find the kinetic energy of the each blow of hammer.

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\times 1.19\times 8.4^2=41.99\ J$

For 20 such blows, the kinetic energy will be

$20\times K.E=20\times 41.99=839.8\ J$

Let us assume that the nail absorbs all the kinetic energy. So, this 839.8 J will be converted into heat energy $(Q)$ .

Now,

$Q=839.8\ J$

Also, we know that

$Q=mc\Delta_T$

Where,

$m$ is the mass of the nail, which is $13\ g=\frac{13}{1000}=0.013\ kg$

$c$ is the specific heat capacity of iron, which is 449 J/kg.K

$\Delta_T$ is the change in temperature.

Plug these value we get,

$Q=mc\Delta_T\\839.8\ J=(0.013\ kg)(449\ J/kg.K)\times\Delta_T\\\Delta_T=\frac{839.8}{0.013\times 449}\\\\\Delta_T=143.9\ K$

We can see the temperature rise is 143.9 K.

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4 years ago