Does the fastest car always travel the farthest?

This woman is riding a bicycle down a hill at a constant speed and in a straight line. Which change will increase the speed of t

Marina86 [1]

Answer: Um I would say maybe b

Explanation:

4 0

3 years ago

Read 2 more answers

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0

3 years ago

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.

salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_{1} = u_{g}$ = 2553.6 kJ/kg

$v_{1} = v_{g} = 0.4625 m^{3}/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

$u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$\frac{Q}{m} = \Delta u$

$\frac{Q}{m} = u_{2} - u_{1}$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0

3 years ago

Which of the following is equal to distance divided by time?

givi [52]

To find <em>Speed </em>you divide distance by time.

Reasoning:

Displacement is the change in an object's position from the origin.

To find Acceleration you use the equation a=v^2/r

To find Velocity you divide the distance by the time it takes to travel that distance, then you add your direction to it.

<em>To find speed you use distance is over time</em>

4 0

3 years ago

Read 2 more answers

Q|C Spherical waves of wavelength 45.0 cm propagate outward from a point source. (c) Explain how the phase of the wave at a dist

lbvjy [14]

We are given distance of the center of the source for two waves:

$D_{1} =240 cm$

$D_{2} =60 cm$

C) Calculating how their phases compares:

ΔФ240 $=\frac{D_{2} }{D_{1} } .$ ΔФ60

$\frac{240}{60} .$ ΔФ60 = 4.ΔФ60

How can you determine a wave's phase?

The period, or length of each cycle, is determined by dividing the frequency by 1, so 1/100 corresponds to a period of 0.01 seconds. The phase shift equation is given by ps = 360 * td / p, where td is the time interval between waves, p is the wave period, and ps is the phase shift in degrees.

Phase: The "Phase" of a waveform refers to the location of the moving particle and is expressed in "Radians or degrees."

Phase difference, also known as "Phase angle," is the amount of time that one wave precedes or follows another.

To learn more about Phase of the wave, visit:

brainly.com/question/17095727

#SPJ4

4 0

2 years ago