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statuscvo [17]
3 years ago
13

A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equi- librium posit

ion. Assume it undergoes simple harmonic motion. Determine (a) its period, (b) its total energy, and (c) its maximum angular displacement.
Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

a)   T = 2.997 s

b)   K = 14.3 J

c)   φ = 0.444 rad

Explanation:

a) Determine its period

  The pendulum simple’s period is:

 

  T = 2π\sqrt{\frac{l}{g} }

        Where l: Pendulum’s length

                    g = 9.8 m/s2

  T = 2π\sqrt{\frac{2.23}{9.8} }

  T = 2.997 s

b) Total energy

  Initially his total energy is kinetic

  K = \frac{mv^{2} }{2}

  K = \frac{(6.74)(2.06)^{2} }{2}

  K = 14.3 J

c) Maximum angular displacement

  φ = cos^{-1}(1-\frac{E}{mgl} )

  φ = cos^{-1}(1-\frac{14.3}{(6.74)(9.8)(2.23)} )

  φ = 0.444 rad

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A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
deff fn [24]

Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

         F ≈ - k Q qo / r²

6 0
3 years ago
Determine the value of the current in the solenoid so that the magnetic field at the center of the loop is zero tesla. Justify y
sleet_krkn [62]

Answer:

I will explain the concept of magnetic field and how it can be calculated.

Explanation:

The formula for magnetic field at the center of a loop is given as

B = μ_{o}I / 2R

where B is the magnetic field

R is the radius of the loop

I is the current

and μ_{o} is the magnetic permeability of free space which is a constant 4π × 10^{-7} newtons/ampere²

If the magnetic field at the center of the loop is 0, then μ_{o}I = 0

I = 0 which means there will be no current flow in the loop.

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While looking at bacteria under a microscope you decide to take a break when u return the amount of bacteria has doubled in size
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