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Maksim231197 [3]
3 years ago
7

A height of 6 ft 3 in is equal to _______. (Round to 3 significant digits)

Physics
2 answers:
aksik [14]3 years ago
7 0

Answer:

option c

Explanation:

1 feet = 12 inch

1 inch = 2.54 cm

o, 6 ft 3 in = 6 x 12 + 3 = 75 inch

75 inch = 75 x 2.54 cm = 190.5 cm

So, the answer is 191 cm

Olin [163]3 years ago
6 0

A height of 6 ft 3 in is equal to _______. (Round to 3 significant digits)  

A - 0.191 cm

B - 19.1 cm

<u>C - 191 cm</u>

6 ft 3 in = 190.5 cm

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A boat produced water waves of frequency 4 Hz and wavelength 2m, if it reached the shore after 2 seconds how far is the boat fro
kap26 [50]

Answer:

S = 16 m

Explanation:

Given that

The frequency of the water waves, f = 4 Hz

The wavelength of the water waves, λ = 2 m

The time the waves reached the shore, t = 2 s

The relation between the velocity, wavelength, and the frequency of the wave is given by the relation,

                                           v = f λ    m/s

Substituting the given values in the above equation,

                                           v = 4 x 2

                                              = 8 m/s

The velocity of the water waves is v = 8 m/s

The distance between the shore and boat is given by

                                            s = v x t

                                               = 8 x 2

                                               = 16 m

Hence, the distance between the boat and the shore is, s = 16 m

7 0
3 years ago
A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too
maks197457 [2]

Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of black hole = 90\times 1.989\times 10^{30}\ kg

R_{f} = 10000+100 m

R_{sb} = Distance between the nose and the center of the black hole = 10000 m

The difference in the gravitational field in this system is given by

\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2

The acceleration is 2352645198509.9604 m/s²

4 0
2 years ago
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