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Eva8 [605]
2 years ago
11

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti

cal. String A is attached at a distance d
Physics
1 answer:
NeX [460]2 years ago
5 0

Answer:

a)  T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} ) ,  b) T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c)  x = d - \frac{m_1}{m_2} \  \frac{L}{2d},  d)  m₂ = m₁  ( \frac{ L}{2d} -1)

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )  - g m₂ - g m₁

       T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = d - \frac{m_1}{m_2} \  \frac{L}{2d}

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         \frac{m_1}{m_2} \ (0.5L -d) = d

          \frac{m_1}{m_2} = \frac{ d}{0.5L-d}

          m₂ = m₁  \frac{0.5 L -d}{d}

          m₂ = m₁  ( \frac{ L}{2d} -1)

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An object located near the surface of Earth has a weight of a 245 N
marusya05 [52]

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

4 0
3 years ago
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the
Rufina [12.5K]

Answer:

The  angular speed is w = 5.89 \ rad/s

Explanation:

From the question we are told that

    The time taken is  t = 1.6 s

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         \theta  =  n *  2 *  \pi

substituting values

        \theta  =  1.5 *  2 *  3.142

       \theta  = 9.426 \ rad

 The angular speed is mathematically represented as

         w =  \frac{\theta }{t}

substituting values

         w =  \frac{9.426}{1.6}

          w = 5.89 \ rad/s

     

3 0
3 years ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in
Novay_Z [31]

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

  • μ(s) is coefficient of static friction
  • m is mass of the block
  • g is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

#SPJ1

4 0
2 years ago
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