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snow_lady [41]
3 years ago
13

The resultant of three vectors is 90.00 cm 34o N of W. If two of these three vectors are 17.89 cm 27o W of S, and 36.00 cm NW, w

hat is the magnitude and direction of the third vector? (Ans: 57.85 cm, 44.76o N of W)
Physics
1 answer:
Art [367]3 years ago
8 0

Answer:

Magnitude of the third vector: 57.85 cm

The direction of the third vector: 44.76 N of W

Explanation:

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<h2>MARK BRAINLIEST</h2>

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<h2><u>Background Information</u></h2>

A wave is any disturbance that carries energy from one place to another. There are two different types of waves: mechanical and electromagnetic. A mechanical wave carries energy through matter. Energy is transferred through vibrating particles of matter. Examples of mechanical waves include ocean waves, sound waves, and seismic waves. Like a mechanical wave, an electromagnetic wave can also carry energy through matter. However, unlike a mechanical wave, an electromagnetic wave does not need particles of matter to carry energy. Examples of electromagnetic waves include microwaves, visible light, X-rays, and radiation from the Sun.

7 0
3 years ago
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A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

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Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

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A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
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q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

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t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



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Explanation:

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