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garik1379 [7]
3 years ago
11

An astronaut has a mass of 74.0 kg. 1) how much would the astronaut weigh on mars where surface gravity is 38.0% of that on eart

h? (express your answer to three significant figures.)
Physics
2 answers:
Step2247 [10]3 years ago
6 0
74 kg * 0.38 * 9.8 m/s/s = 275.576 N
3 significant figures rounds to 276 N
NeX [460]3 years ago
3 0
274.614N or 61.736lbs
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Right when it’s about to go down and stopped.
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3 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
3 years ago
Please solve no.g <br>Anyone!???​
steposvetlana [31]

Answer:

The answer is already given... it's 31.25 cm

5 0
3 years ago
When searching for your word processing file to finish writing your report, you should look for a file with which extension?
ryzh [129]

Answer:

doc

Explanation:

6 0
3 years ago
A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti
ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = 6.31\times 10^{4} T

Solution:

To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}

N = 22.74 = 23 turns (approx)

 

8 0
3 years ago
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