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VladimirAG [237]
4 years ago
6

Tita= 55°Answer and proper explanation pls in English​

Physics
1 answer:
Triss [41]4 years ago
6 0

Answer:

the photo is a bit blur

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Anastaziya [24]

Answer:

A. it will move faster than the large object was moving initially

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4 years ago
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If your heart is not strong enough or efficient enough, it is difficult to (3 points)
Lyrx [107]

it's difficult to breathe.

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3 years ago
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A navy diver hears the underwater sound wave from an exploding ship across the harbor. They immediately lift their head out of t
tamaranim1 [39]

Answer:

s = 1800 m = 1.8 km

Explanation:

The distance, the speed, and the time of reach of the sound are related by the following formula:

s = vt

where,

s = distance

v = speed

t = time

FOR WATER:

s = v_wt ---------------------- eq (1)

where,

s = distance between ship and diver = ?

v_w = speed of sound in water = 1440 m/s

t = time taken by sound in water

FOR AIR:

s = v_a(t+4\ s) ---------------------- eq (2)

where,

s = distance between ship and diver = ?

v_a = speed of sound in water = 344 m/s

t + 4 s = time taken by sound in water

Comparing eq (1) and eq (2),because distance remains constant:

v_wt=v_a(t+4\ s)\\\\(1440\ m/s)t = (344\ m/s)(t+4\ s)\\(1440\ m/s - 344\ m/s)t=1376\ m\\t = \frac{1376\ m}{1096\ m/s}

t = 1.25 s

Now using this value in eq (1):

s = (1440\ m/s)(1.25\ s)

<u>s = 1800 m = 1.8 km</u>

4 0
3 years ago
3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a
34kurt

Answer:

The friction force and the x component for the weight should be the reaction forces that are opposite and equal to the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.  

Explanation:

<u>When the locomotive starts to pull the train up, appears two reaction forces opposed to the action force in the direction of the move. </u>

The first one is due to the friction between the wheels and the ground, it will be the friction force (Fr):

Fr = μ*Pₓ =μmg*sin(φ)        

<em>where μ: friction dynamic coefficient, Pₓ: is the weight component in the x-axis, m: total mass = train's mass + locomotive's mass, g: gravity, and sin(φ): is the angle respect to the x-axis.</em>              

And the second one is the x component for the weight (Wₓ):

Wₓ = mg*cos(φ)  

<em>where cos(φ): is the angle respect to the y-axis.    </em>

<em> </em>

These two forces should be the same as the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.          

3 0
4 years ago
Two sites are being considered for wind power generation. In the first site, the wind blows steadily at 7 m/s for 3000 hours per
slavikrds [6]

Answer:

E.year₂ > E.year₁  (Second site is better)

Explanation:

Given data

V_{1}=7 m/s\\Time_{1}=3000 hours/year\\V_{2}=10m/s\\Time_{2}=1500hours/year

The  power generation is the time rate of kinetic energy which can be calculated as:

Power=ΔKE=m×V²/2

Regarding  that m ∝ V.Then

Power ∝ V³ ⇒ Power=constant×V³

Since ρa is constant for both sides and Area is the same as same wind turbine is used

For First site

Power_{1}=const.V^3\\Power_{1}=const.(7m/s)^3\\Power_{1}=const.343Watt\\

For second site

Power_{2}=const.V_{2}^3\\Power_{2}=const.(10m/s)^3\\Power_{2}=const.1000W

Calculating energy generation per year for each of two sites

E.year=Power×Operation time per year

For First site

E.year₁=Power₁×Operation time₁ per year

E_{year1}=const.*343W*3000\\E_{year1}=const.1029000(W.hour/year)

For Second site

E.year₂=Power₂×Operation time₂ per year

  E_{year2}=const.*1000W*1500\\E_{year2}=const.1500000(W.hour/year)

So

E.year₂ > E.year₁  (Second site is better)

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