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mihalych1998 [28]
3 years ago
8

Una niña está empujando un baúl. El PESO del baúl es de 230 N y el roce es de 50 N, la niña sólo logra ejercer una fuerza de 30

N, por lo que recurre a su papá para que la ayude, ¿Cuánta fuerza debe realizar el papá para que entre ambos logren mover el baúl?
Physics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

Su padre necesita aplicar una fuerza de 20 newtons en la misma dirección que la fuerza aplicada por su hija.

Explanation:

Asúmase que el baúl se mueve en una superficie horizontal. La fuerza de rozamiento dada por el problema es la fuerza de rozamiento estático máximo, se requiere una fuerza externa antiparalela a la fuerza de rozamiento estático máximo para que el baúl se empiece a mover. La ecuación de equilibrio de fuerzas horizontales sobre el baúl es:

\Sigma F = P - f = 0

Donde:

P - Fuerza externa aplicada sobre el baúl, medida en newtons.

f - Fuerza de rozamiento estático máximo, medida en newtons.

Se despeja la fuerza externa:

P = f

Si f = 50\,N, entonces:

P = 50\,N

Si la niña solo logra ejercer una fuerza de 30 newtons, su padre necesita aplicar una fuerza de 20 newtons paralela a aquella fuerza.

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Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b
zmey [24]

Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

    The volume of container A is  V_A = 2 m^3

    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

So substituting for P_A V_A in the equation above

          W =  nRT ln [\frac{V_B}{V_A} ]

Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

          W = 1563.289 \ J

8 0
3 years ago
Select the statement that correctly lists the forces from strongest to weakest. (2 points)
Vilka [71]

Strong Nuclear force: it is the short range force and strongest fundamental force in all type of forces.

Electromagnetism: this is the force due to magnetic and electric behavior of the particles. It is moderate type of force and its range is more than Nuclear force.

Weak Nuclear Force: This force is also short range force which act between the nucleoside. But this force is also moderate type of force

Gravitational force: this force is between two point masses and least order of force. also the range of this force is upto infinite.

so the correct order of this fundamental force is

<em>strong nuclear, electromagnetism, weak nuclear, gravitational</em>

5 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0
3 years ago
At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh
Anna35 [415]

We have the equation for electric field E = kQ/d^{2}

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to d^{2}

So, \frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}

E_{1} = 485 N/C

d_{1} = 0.208 cm

d_{2} = 0.620 cm

E_{2} = ?

\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}

E_{2} = \frac{485*0.208^{2}}{0.628^{2}}

E_{2} = 53.20 N/C

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