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Lady bird [3.3K]

2 years ago

7

A 55-kg woman is wearing high heels. If each heel has a circular cross-section 6.0 mm in diameter and she puts all her weight on

one heel, how much pressure does that exert on the floor

1 answer:

bogdanovich [222]2 years ago

5 0

Answer:

The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

Explanation:

Given;

mass of the woman, m = 55 kg

diameter of the circular heel, d = 6.0 mm

radius of the heel, r = 3.0 mm = 0.003 m

Cross-sectional area of the heel is given by;

A = πr²

A = π(0.003)²

A = 2.8278 x 10⁻⁵ m²

The weight of the woman is given by;

W = mg

W = 55 x 9.8

W = 539 N

The pressure exerted by the woman on the floor is given by;

P = F / A

P = W / A

P = 539 / (2.8278 x 10⁻⁵ )

P = 1.9061 x 10⁷ N/m²

Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

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Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the

polet [3.4K]

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

$\rm \theta$ is the angle formed between the path of the incident light and the diffracted light = 9. 2°

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N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

$n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\ d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm$

The number of lines per mm is found as;

$\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm$

Hence the number of lines per mm in the diffraction grating is 326.

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Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

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Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = <em>kdt</em>

In [T-70]= <em>kt </em>+ $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for <em>k </em>using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

<em>K </em>= $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

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3 years ago

If Scoobie could drive a Jetson's flying car at a constant speed of 450.0 km/hr across oceans and space, approximately how long

vladimir2022 [97]

Answer:

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Explanation:

From the question we are told that

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Generally the circumfernce of the earth is mathematically evaluated as

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Generally the time taken is mathematically represented as

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Converting to days

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