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3 years ago

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Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee

n reached, and the temperature is 2°C. The air continues to rise another 1,400 m. What is the final temperature?

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

Explanation:

Given

saturated air temperature by $0.5^{\circ}C/100 m$

Dew point temperature is given by $t=2^{\circ}C$

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be $\Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C$

Final temperature $t_f$

$t_f+\Delta t=t$

$t_f=2-7=-5^{\circ}C$

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A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

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Answer:

The ball and the wall experience the same force.

Explanation:

According to the third law of Newton, which states that "for every action, there is an equal and opposite reaction", this means that when an object 1 acts on object 2 with a certain force, object 2 also acts on object 1 with the same magnitude of force but in an opposite direction.

According to this question, a ball hits a wall with a certain force. This means that the wall will react on the ball with the same force magnitude, but in an opposite manner. Hence, the ball and the wall experience the same force.

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3 years ago

Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo

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Answer:

C) 64lb

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use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

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our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

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Vince weigh on the other planet is C) 64lb

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3 years ago

Read 2 more answers

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

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The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$

$f_2 = n+f_1$

Replacing $3/s$ for n and 202Hz for $f_1,$

$f_2 = 3/s + 202Hz$

$f_2 = 3/s(\frac{1Hz}{1/s})+202Hz$

$f_2 = 206Hz$

The frequency of the tightened is 205Hz

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3 years ago

When this pendulum swings, at which point is its kinetic energy the<br> highest? *

Eddi Din [679]

Answer:

The bottom/center of the pendulum

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As it comes back towards the center that potential energy will convert into kinetic energy until it reaches the middle of its swing (when the pendulum is fully vertical) where all potential energy has been converted into kinetic energy.

This is when the kinetic energy is the highest

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3 years ago

parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet

Rom4ik [11]

Answer:

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

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Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

u = 0.5*ε*E^2

- Where the Electric field strength E between capacitor plates is given by:

E = V / d

Hence,

u = 0.5*ε*(V/d)^2

Where, ε = 8.854 * 10^-12

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V = d*sqrt ( 2*u / ε )

Plug in values:

V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )

V = 576 V

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3 years ago