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mojhsa
3 months ago
6

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

s modulus for steel is 2.0 x 1011 Pa.
Please show work.
Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
5 0

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half the diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Cross-sectional area of the wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the change in length of the wire be ΔL.

The equation for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the wire stretches by 3.5 cm.

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
3 months ago
A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he
inna [3103]
Let T be the force exerted on the rope by her. This force induces tension in the rope, which exerts an upward force on the crates, while the weight of the crate pulls downward. Thus, the net force acting on the crate can be expressed as mg - T, acting in the downward direction. According to Newton's law, we can set up the equation: mg - T = ma. Given that a = 0 (the speed remains constant), this simplifies our equation to mg - T = 0, which leads to T = mg. Therefore, T = 25 x 9.8 = 245 N, indicating that the force she needs to apply is 245 N.
5 0
2 months ago
A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w
ValentinkaMS [3465]

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

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The ball was in the air for 3.896 seconds

8 0
2 months ago
A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it
Keith_Richards [3271]
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = \sqrt{rg} = \sqrt{2.80 \times 9.8} = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

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where u is the velocity at the lowest point. Consequently, the modified equation is

= \sqrt{v^{2} + 4gr}

= \sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = \frac{(m_{1} + m_{2})u}{m_{2}}

= \frac{(20 + 5) \times 11.71}{5}

= \frac{292.75}{5}

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.

3 0
2 months ago
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