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damaskus [11]
3 years ago
11

Fitness can be achieved only through

Physics
2 answers:
Firlakuza [10]3 years ago
7 0
Fitness can only be achieved only through  A. Frequency 
stepladder [879]3 years ago
5 0
A and C can both be correct
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You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
3 years ago
What are the principals of electric arc welding
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Principles<span> of </span>arc welding<span>. </span>Arc welding<span> is a </span>welding<span> process, in which heat is generated by an </span>electric arc<span> struck between an electrode and the work piece. </span>Electric arc<span> is luminous</span>electrical<span> discharge between two electrodes through ionized gas.</span>
7 0
3 years ago
A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration
umka2103 [35]

Answer:

770m/s

Explanation:

caculation using one of the newton law of motion

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2 years ago
why do we not notice a difference in the fall and winter in the amount of oxygen in the air even when leaves fall off
sveticcg [70]
Earths tilt making the sun go haywire lol XD
6 0
3 years ago
What is the force exerted on a charge of 2.5 µC moving perpendicular through a magnetic field of 3.0 × 102 T with a velocity of
OverLord2011 [107]

Given:

B = 3 \times 10^{2} T

V=5 \times 10^{3} \frac{m}{s}

q = 2.5 × 10^{-6} C

α = 90

To find:

Force = ?

Formula used:

Force on the moving charge is given by,

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

Solution:

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

F = 2.5 \times 10^{-6} \times 3 \times 10^{2}  \times 5 \times 10^{3}

F = 37.5 × 10^{-1}

F = 3.75 Newton

Thus, the force acting on the moving charge is 3.75 Newton.

8 0
3 years ago
Read 2 more answers
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