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notsponge [240]
1 year ago
14

As a child walks, for a brief moment his entire weight is placed on one heel, which can be approximated as a small circle. Calcu

late the pressure exerted on the floor by the heel if the circle created by the heel has a diameter of 7 cm, and the child’s mass is 25 kg.Group of answer choices

Physics
1 answer:
Brut [27]1 year ago
4 0

The correct answer is option b, 63.6 kPa

Given,

The diameter of the heel, d=7 cm=0.07 m

The mass of the child, m=25 kg

The pressure is given by the ratio of the force to the area through which the force is applied.

The force applied on the floor by the child is equal to its weight.

Thus the pressure applied on the floor by the child is given by,

\begin{gathered} P=\frac{F}{A} \\ =\frac{mg}{\pi(\frac{d}{2})^2} \end{gathered}

Where A is the area of the cross-section of the heel.

On substituting the known values,

\begin{gathered} P=\frac{25\times9.8}{\pi(\frac{0.07}{2})^2} \\ =63.6\times10^3\text{ Pa} \\ =63.6\text{ kPa} \end{gathered}

Thus the pressure applied on the floor by the heel is 63.6 kPa

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What is nuclear energy and light energy by clear explanation
Kazeer [188]

Answer:

nuclear energy is a energy that holds together of atoms

light energy is a kind of kinetic with the ability to make types of light visible to human eyes

4 0
2 years ago
18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)

6 0
3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

\frac{1}{2}mv^2=0.0052

\frac{1}{2}\times 0.00524\times v^2=0.0052

v = 1.40 m/sec

7 0
3 years ago
Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0
e-lub [12.9K]

Answer:

f = 421.8 Hz

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}

so it will be

\Delta L = 12.5 - 12.093

\Delta L = 0.4066

now we know that

\frac{\lambda}{2} = 0.4066

\lambda = 0.813

now frequency of sound is given as

f = \frac{v}{\lambda}

f = \frac{343}{0.813}

f = 421.8 Hz

4 0
3 years ago
The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec
notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

<u>λ = 1.8 x 10⁻⁷ m = 180 nm</u>

5 0
3 years ago
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