PHYSICS COLLEGE

A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’. Please explain.

Answers

Answer 1

Answer:

Answer:

The spring constant k is $1.115* 10^(9) N/m$

Solution:

As per the question:

Length of the solid cylinder, L = 500 mm = $500* 10^(- 3) = 0.5 m$

Diameter pf the cylinder, D = 2 cm = 0.02 m

As the radius is half the diameter,

Radius, R = 1 cm = 0.01 m

Young's Modulus, E = 17.4 GPa = $17.4* 10^(9) Pa$

Now,

The relation between spring constant, k and Young's modulus:

$kL = EA$

where

A = Area

Area of solid cylinder, A = $2\piR(L + R)$

$0.5k = 17.4* 10^(9)* 2\piR(L + R)$

$k = (17.4* 10^(9)* 2\pi* 0.01(0.01 + 0.5))/(0.5)$

k = $1.115* 10^(9) N/m$

Young's modulus, E is the ratio of stress and strain

And

Stress = $(Force or thrust)/(Area)$

Strain = $(length, L)/(elongated or change in length, \Delta L)$

Also

Force on a spring is - kL

Therefore, we utilized these relations in calculating the spring constant.

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Nichrome wire, often used for heating elements, has resistivity of 1.0 × 10-6 Ω ∙ m at room temperature. What length of No. 30 wire (of diameter 0.250 mm) is needed to wind a resistor that has 50 ohms at room temperature?

Answers

Answer:

Length = 2.453 m

Explanation:

Given:

Resistivity of the wire (ρ) = 1 × 10⁻⁶ Ω-m

Diameter of the wire (d) = 0.250 mm = 0.250 × 10⁻³ m

Resistance of the wire (R) = 50 Ω

Length of the wire (L) = ?

The area of cross section is given as:

$A=(1)/(4)\pi d^2\n\nA=(1)/(4)*\ 3.14* (0.250* 10^(-3))^2\n\nA=0.785* 6.25* 10^(-8)\n\nA=4.906* 10^(-8)\ m^2$

We know that, for a constant temperature, the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section. The constant of proportionality is called the resistivity of the wire. Therefore,

$R=\rho (L)/(A)$

Expressing the above in terms of length 'L', we get:

$L=(RA)/(\rho)$

Plug in the given values and solve for 'L'. This gives,

$L=(50* 4.906* 10^(-8))/(1* 10^(-6))\ m\n\nL=(2.453)/(1)=2.453\ m$

Therefore, length of No. 30 wire (of diameter 0.250 mm) is 2.453 m.

Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?

Answers

Answer:

$0.0035\ \text{m}$

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

$\lambda$ = Wavelength = 700 nm

m = Order = 1

We have the relation

$d(y)/(D)=m\lambda\n\Rightarrow y=(Dm\lambda)/(d)\n\Rightarrow y=(1.5* 1* 700* 10^(-9))/(0.3* 10^(-3))\n\Rightarrow y=0.0035\ \text{m}$

The distance from the screen at which the first bright fringe beyond the center fringe appear is $0.0035\ \text{m}$ .

An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.

Answers

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer: i think you should place it on the red line

Explanation:

hope this helps

and need brainliest

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Answers

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Explanation:

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Answers

c is the correct answer

B is correct because I know it is :)

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