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Vitek1552 [10]
3 years ago
15

A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j

Physics
1 answer:
elena55 [62]3 years ago
6 0

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

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50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui
soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0
3 years ago
A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and
frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

m=\sqrt{\dfrac{hP}{KA}}

For circular fin

m=\sqrt{\dfrac{4h}{KD}}

m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

m = 37.08

\eta_f=\dfrac{tanhmL}{mL}

0.65=\dfrac{tanh37.08L}{37.08L}

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}

\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}

ε = 23.52

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