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yan [13]
3 years ago
15

4. A solution that contains a large amount of salt and a small amount of water is said to be a _______ solution.

Physics
1 answer:
n200080 [17]3 years ago
4 0
The answer is diluted solution

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What happens when the dew point and the temperature are the same?
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The value of relative humidity becomes 100% leading to condensation of water vapor in the air into water droplets or water (dew)
4 0
3 years ago
A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen
leva [86]

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

  • Work function is a material property defined as the minimum amount of energy  required to infinitely remove electrons from the surface of a particular solid.
  • The potential difference required to support all emitted electrons is called the stopping potential which is given by v_0=\frac{K.E_m_a_x}{e} .....(1)
  • where v_0 is the stopping potential and e is the charge of the electron given by 1.6\times10^-^1^9 .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation  is given by:

K.E_m_a_x=E-\phi      ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

  ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV

As we know that E=\frac{hc}{\lambda}  ....(3)

where Speed of light,c = 3\times10^8 m/s and Planck's constant , h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs

From equation (3) , we get

\lambda=\frac{hc}{E} \\\\\lambda=\frac{  4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm

Learn about more einstein photoelectric equation  here:

brainly.com/question/11683155

#SPJ4

8 0
1 year ago
A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec
seropon [69]

Answer:

\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

\vec{C} = \vec{B} - \vec{A}

So:

\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )

\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )

\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )

4 0
3 years ago
Which of the following statements about Masters programs is not correct?
uranmaximum [27]

The correct answer is C. The level of competition is not very high in most Masters  programs.

Explanation:

In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters  programs".

7 0
3 years ago
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0
3 years ago
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