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uranmaximum [27]
3 years ago
15

A kicker punts a football from the very center of the field to the sideline 39 yards downfield.

Physics
1 answer:
MrRa [10]3 years ago
8 0

Answer:

L = 47.15 yards

∅ = 55.80°

Explanation:

Given

y = 39 yards

x = 53/2 yards

L = ?

We apply the formula

L = √(x²+y²)

L = √((53/2 yards)²+(39 yards)²)

L = 47.15 yards

The angle between the direction of the net displacement of the ball and the central line of the feild is obtained as follows

∅ = tg⁻¹(y/x)

∅ = tg⁻¹(39/(53/2)) = 55.80°

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Irina18 [472]

Answer:

the charged particle is electron..

Explanation:

electrons are the only charged particles which can be transferred by rubbing certain materials which generate electrostatic force ...

8 0
3 years ago
An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard
UNO [17]

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,      

velocity of the observer = 17 m/s

speed of the sound = 343 m/s    

velocity of the source = 0 m/s    

frequency emitted from the source  = 2550 Hz              

f = f_0(\dfrac{v-v_0}{v-v_s})              

f = 2550\times (\dfrac{343+17}{343-0})

velocity of observer is negative as it is approaching the source.                   f = 2676.38 Hz ≈ 2677 Hz                    

hence, the frequency heard by the observer is equal to 2677 Hz

7 0
3 years ago
NASA is designing a mission to explore Titan, the largest moon of Saturn. Titan has numerous hydrocarbon lakes containing a mix
Vitek1552 [10]

Answer:

Total pressure in atmosphere  IS 4.36 atm

Explanation:

Given Data:

Depth of lake is 400 mtr

surface tension is 147 kilopascals

surface temperature  94 kelvin

Gravity = 1.35 m/s^2

specific gravity of methane(liquid) = 0.415

specific gravity of ethane(liquid) = 0.546

density of methane can be obtained as

\rho_{methane} = 0.415\times 1000 = 415 kg/m^3

Density of ethane can be obtained as

\rho_{ethane} = 0.546\times 1000 = 546 kg/m^3

Total pressure in atmosphere can be obtained as

P = surface\ tension + \rho_{max}gh

 = 147 +546\times 1.35\times 400\times 10^{-3}

  = 441.84 kPa

= 441.84\times 0.009869 atm\kPa

 = 4.36 atm

7 0
3 years ago
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The statement is true.

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3 years ago
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