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lutik1710 [3]
3 years ago
5

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

with a large, stiff 4.1 m rod (assume it has negligible mass) with a heavy object of mass 52 kg and 14 cm from the axle around which the device pivots and a cup on the far end that holds the rock. The war wolf is loaded when it is in a horizontal position, and the rock is projected horizontally when after it has rotated to a vertical position. If the rock has a mass of 123 g, how fast (linear speed) does it move when launched?

Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object m_1=52 kg

distance of m_1 from the axle r_1=14 cm

mass of rock m_2=123 gm

Length of rod =4.1 m

distance of m_2 from axle r_2=4.1-0.14=3.96 m

Net torque acting is

T_{net}=m_1gr_1-m_2gr_2

T_{net}=52\times 0.14\times g-0.123\times 3.96\times g

T_{net}=6.793\times 9.8

T_{net}=66.57 N-m

Work done by T_{net} is converted to rock kinetic Energy

thus

T_{net}\times \theta =\frac{mv^2}{2}

Where \theta =angle\ turned =\frac{\pi }{2}

v= velocity\ at\ launch

66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}

v^2=66.57\times \pi

v=\sqrt{1700.511}

v=41.23 m/s

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2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗
Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

   The tension is  T =  48.255 \ N

b

   The time taken is  t =  0.226 \ s

c

   The position for maximum velocity is  

        S = 0  

d

     The maximum velocity is  V_{max} =0.384 \ m/s

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

    The mass of the bob  is m_b  =  5 \ kg

     The angle is  \theta = 10^o

     The length of the string is  L =  0.5 \ m

The tension on the string is mathematically represented as

              T  = mg cos \theta

substituting values

             T =  5 * 9.8 cos(10)

             T =  48.255 \ N

The motion of the bob is mathematically represented as

             S =  A sin (w t  + \frac{\pi }{2} )

     =>   S =  A sin (wt)

Where  w is the angular speed

and  \frac{\pi}{2} is the phase change

At initial position S =  0

   So   wt  = cos^{-1} (0)

          wt  = 1

Generally w can be mathematically represented as

          w =  \frac{2 \pi }{T}

Where T is the period of oscillation which i mathematically represented as

          T  =  2 \pi \sqrt{\frac{L}{g} }

So  

      t   = \frac{1}{w}

       t   = \frac{T}{2 \pi}

       t =  \sqrt{\frac{L}{g} }

substituting values

       t =  \sqrt{\frac{0.5}{9.8} }

       t =  0.226 \ s

Looking at the equation

         wt =  1

We see that maximum velocity of the bob will be at  S = 0  

i. e     w =  \frac{1}{t}

The maximum velocity is mathematically represented as

          V_{max}   =  w A

Where A is the amplitude which is mathematically represented as

         A = L sin \theta

So

      V_{max} = \frac{2 \pi }{T }  L sin \theta

      V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} }   L sin \theta  Recall   T  =  2 \pi \sqrt{\frac{L}{g} }

     V_{max} = \sqrt{gL} sin \theta

substituting values        

       V_{max} = \sqrt{9.8 * 0.5 } sin (10)

       V_{max} =0.384 \ m/s

6 0
2 years ago
A billiard ball is dropped from a height of 64 feet. Use the position function s(t) = –16???? 2 + ????0???? + ????0 to answer th
Delicious77 [7]

Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

a(t) = -32 ft/s^2

v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

Find:-

a. Determine the position function s(t),

b. the velocity function v(t),

c. the acceleration function a(t).

d. What is the velocity of the ball at impact?

Solution:-

- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

                                  a(t) = d v(t) / dt

                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

6 0
3 years ago
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w
Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

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