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yaroslaw [1]
3 years ago
8

Un objeto que es empujado sobre una superficie plana, tiene una masa de 80 kg y dicha fuerza tiene un valor de 300 N. Calcule la

fuerza Normal y la fuerza de Fricción para que este en equilibrio. A. Ff=300 N N=784.8 N B. Ff=300 N N=780 N C. Ff=200 N N=750 N AYUDENME POR FA LA NECESITO
Physics
1 answer:
Gennadij [26K]3 years ago
3 0

Responder:

A. Ff = 300 N N = 784,8 N

Explicación:

Dado

Masa del cuerpo = 80 kg

Fuerza de movimiento Fm = 300N

Dado que el cuerpo no está acelerando, la fuerza de fricción (Ff) es igual a la fuerza de movimiento que actúa sobre el cuerpo, ya que la fuerza de fricción es una fuerza de oposición, es decir, Fm = Ff

Dado que Fm = 300N, Ff = 300N

La reacción normal que actúa en el cuerpo es igual al peso.

N = W = mg

g es la aceleración debida a la gravedad

g = 9,8 m / s

N = mg

N = 80 (9,81)

N = 784,8N

Por tanto, la fuerza normal que actúa sobre el cuerpo es 784,8 N

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cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0
3 years ago
Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each cha
Sloan [31]

Answer:

<em>The force on each charge = 16 N</em>

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

<em>When d = 1 m, F = 1 N,</em>

<em>Constant: 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 2,</em>

<em>q₁q₂ = 1×1²/9×10⁹ </em>

<em>q₁q₂ = 1/9×10⁹  C²</em>

<em>When d = 1/4 m, q₁q₂ = 1/9×10⁹  C² and 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1</em>

<em>F =  9×10⁹×1/9×10⁹ /(1/4)²</em>

<em>F = 1/(1/16)</em>

<em>F = 16 N</em>

<em>Therefore the force on each charge = 16 N</em>

<em />

4 0
3 years ago
How many electrons do the elements of the most stable group have in their outer energy levels
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I think it might be 8 electrons
6 0
3 years ago
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a lamp or bulb is marked 12 volt 240 Watts how many joules does it consume in one hour and what is the current that passes throu
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Answer:

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5 0
3 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
3 years ago
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