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zhenek [66]
3 years ago
10

People often break the speed limit and risk getting tickets and injury. On highway 5, people often drive 75mph when the legal sp

eed limit is 70mph. Convert 75mph to m/s.
Physics
1 answer:
lina2011 [118]3 years ago
4 0
33.528 meters per second
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A lawyer drives from her​ home, located 1 mile east and 8 miles north of the town​ courthouse, to her​ office, located 4 miles w
givi [52]

Answer:

d = 13 miles

Explanation:

Lets say the position of court house is origin in this case

her office is located at 4 miles west and 4 miles south of court house

so here we have coordinate of the office with respect to court house is given as

r_1 = (-4\hat i - 4\hat j)

now the position of her home is located at 1 miles east and 8 miles north of the court house

so the coordinates of her home is given as

r_2 = (1\hat i + 8 \hat j)

now the change in the position is given as the distance between office and home

d = r_2 - r_1

d = 5 \hat i + 12 \hat j

d = \sqrt{5^2 + 12^2} = 13 miles

4 0
4 years ago
What type of phase change occurs when water vapor in the atmosphere changes to liquid water in clouds?
Mekhanik [1.2K]

Answer:

condensation, the process of changing from a gas to a liquid

3 0
3 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
What is tan for the given triangle?
Rama09 [41]

Answer:

tan is 15 for that triangle

5 0
2 years ago
Steve and Carl are driving from Scranton to Bridgeport, a distance of 180 miles. If their speed averages 60 miles an hour, how l
Vikki [24]

Time = (distance) / (speed)

Time = (180 miles) / (60 mi/hr)

Time = (180/60) (mi-hr/mi)

<em>Time = 3 hours</em>

4 0
3 years ago
Read 2 more answers
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