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svetoff [14.1K]

3 years ago

5

Which is a front in which cold air is replacing warm air at the surface?

2 answers:

Galina-37 [17]3 years ago

6 0

The answer should be a. cold front! a stationary front is when a cold front and warm front meet and do not move. STATIONARY IS INCORRECT!

Snezhnost [94]3 years ago

5 0

Answer:

C.stationarty front

Explanation:

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A force of 2N will stretch a rubber band 0.02m. Assuming that Hooke's Law applies, answer the following: How far will a 1600N fo

denis-greek [22]

Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant

F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m

Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ

7 0

3 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Which of the following is not a symptom associated with hypertension

Romashka-Z-Leto [24]

Do you have any answer choices?

3 0

3 years ago

An element has 4 protons, 5 neutrons, and 4 electrons. What is the atomic mass of the element? 4 5 8 9

alex41 [277]

at mass = protons+neutrons=9

5 0

3 years ago

Read 2 more answers

What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat

Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water $\left ( m\right )=250gm$

Initial temperature $\left ( T_i\right )=20^{\circ}C$

Final temperature $\left ( T_f\right )=100^{\circ}C$

Specific heat of water $\left ( c \right )$ =4190 J/kg-k

heat of vaporization $\left ( L\right )=22.6\times 10^5 J/kg$

Heat required for process $\left ( Q\right )$ =heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc $\left ( T_f-T_i\right )+mL$

Q= $0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5$

Q= $\left ( 0.838+5.5\right )\times 10^5$

Q= $6.338\times 10^5J=633.8 KJ$

4 0

3 years ago