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3 years ago

15

If an airplane flies at 364 km/h, how long will it take to get from New York to Washington DC (288 km)? (If needed, round to the

nearest thousandths and answer with unit 'h' for hours.)

1 answer:

Kazeer [188]3 years ago

3 0

Answer:

0.791 h

Explanation:

Distance = rate × time

288 km = 364 km/h × t

t = 0.791 h

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In the Compton effect, the wavelength of the scattered photon is __________ the wavelength of the incident photon.

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Answer:

he wavelength is different (greater) than the wavelength of the incident photon

Explanation:

The Compton effect is the scattering of a photon by an electron, this process is analyzed using the conservation of momentum, in which we assume that initially the electron is at rest and after the collision it recedes, therefore the energy of the incident photon decreases and consequently its wavelength changes

To complete the sentence we use the wavelength is different (greater) than the wavelength of the incident photon

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Electromagentic waves can travel through space. they don't require any medium.

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wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needl

Bingel [31]

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

$P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\$

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

$S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}$

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

$P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa$

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3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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3 years ago

Elements in a period have _____________.

ahrayia [7]

Answer:

Elements in a period have wide range of chemical properties.

Explanation:

Period is the row of chemical elements's arrangement in the periodic table. It is numbered from 1 to 7.
In period as we move towards right the atomic number of element constantly increases along with the change in chemical properties.
As we move from left towards right in period, the valence electron number regularly increases in every element.
This valence electron differs the reactivity of the elements.

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3 years ago