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3 years ago

1. ¿Qué presión se ejerce sobre cada una de las cuatro patas de una mesa si su masa es de 20 kg y

1 answer:

8 0

Tenemos.

Masa de la mesa = 20kg
Masa encima = 10kg
Masa total = 30kg
Area de cada pata = 20cm² = 20cm² * 1/10000cm² * 1m² =0,002m²

Presió(P)n que se ejerce sobre cada pata.

P =F/A
P = Masa por gravedad/A Masa = 30kg Gravedad =9,8m/s²
P =(30kg * 9,8m/s²)/0,002m²
P = 294kg * m/s²/0,002m² Pero kg *m/s² = Nw
P = 294Nw0,002m²
P = 147000 Nw/m² Pero Nw/m² = pa
p =1,47 * 10⁵ pa

Respuesta.
La presión que se ejerce sobre cada pata es de 1,47 * 10⁵pa

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Andru [333]

Answer:

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Explanation:

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3 years ago

Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that

stepan [7]

Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm

4 0

4 years ago

Calculate the total resistance for a 650ohm , a 350 ohm , and a 1000 ohm resistor connected in series

Mekhanik [1.2K]

Answer:

2000 ohms

Explanation:

Resisters in series just add.

Rt = R1 + R2 + R3

R1 = 650 ohm

R2 = 350 ohm

R3 = 1000 ohm

Rt = 650 + 350 + 1000

Rt = 2000 ohms.

5 0

3 years ago

Read 2 more answers

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$