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weeeeeb [17]
4 years ago
9

Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A wate

rfall that drops 1,235.2 ft.C. A 21320 ft tall mountain.
D. A canyon with a depth of 6630 ft.
Physics
1 answer:
cricket20 [7]4 years ago
4 0

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

\Rightarrow 5 miles = \times 1.609344 km= 8.04672 km,

\Rightarrow 5 miles =8.04672 \times 1000 m= 8046.72 m [ as 1 km = 1000 m]

\Rightarrow 5 miles = 8046.72 \times 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

\Rightarrow 1235.2 ft.= 1235.2\times 30.48 cm=37648.896 cm,

\Rightarrow 1235.2 ft.=37648.896 \times 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

\Rightarrow 1235.2 ft.=376.48896 \times 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

\Rightarrow 21320 ft.= 21320 \times 30.48 cm=649833.6 cm,

\Rightarrow 21320 ft.=649833.6 \times 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

\Rightarrow 21320 ft.=6498.336 \times 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

\Rightarrow 6630 ft.= 6630 \times 30.48 cm=202082.4 cm,

\Rightarrow 6630 ft.=202082.4  \times 0.01 m =2020.824  m [ as 1 cm = 0.01 m]

\Rightarrow 6630 ft.=2020.824  \times 0.001 km =2.020824  km [ as 1 m = 0.001 km].

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Answer:

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Explanation:

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Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

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(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

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(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

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<h3>Question:</h3>

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Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

<u>FOR STEEL BARS</u>:

Since,

(σallow)st = (Pmax)st/A

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(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

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Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

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