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Fiesta28 [93]
3 years ago
10

Katya is training for an upcoming gymnastics competition and needs to improve her upper body strength. At the moment, she can on

ly support at most 180% of her body weight when she’s hanging off of equipment. Misha, a close friend and circus performer, suggests that she train on a swinging trapeze. Katya grabs onto the trapeze which hangs off a pair of cables of length ` that initially make an angle of 60.0◦ with the vertical. She steps off the platform and swings forward. To what height will she fall/rise as she swings? Will Katya swing back to the platform?
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

she will fall to an height of 21.09° only, beyond such an height ; the spring will break.

Katya will not be able to swing to the platform because the angle will be between -21.09° and 21.09° from vertical

Explanation:

If the distance is from the point of angle A to angle B , partitioned by a perpendicular line in the middle then, The conservation of energy between A and B can be expressed as :

KE_A +PE_B = KE_B + PE_B

0 + mgh = \frac{1}{2}mv_B^2+ 0

where ; the height h = l ( cos \theta - \frac{1}{2})

mgl ( cos \theta - \frac{1}{2}) = \frac{1}{2}mv^2_B

\frac{mv_B^2}{l} = mg (2 cos \theta -1 )

T = \frac{mv_B^2}{l}+ mg cos \theta

T = mg(3 cos θ - 1)

Given that:

T = 180 % = 1.8 mg

Then:

1.8 mg = mg(3 cos θ - 1)

2.8 mg = (3 cos θ - 1)

cos θ = \frac{2.8}{3}

θ = cos ⁻¹ (0.933)

θ = 21.09°

Therefore, she will fall to an height of 21.09° only, beyond such an height ; the spring will break.

Katya will not be able to swing to the platform because the angle will be between -21.09° and 21.09° from vertical

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Zinaida [17]

Answer:

It went about 2 meters away

Explanation:

7 0
3 years ago
A SPARK PLUG IN AN AUTOMOBILE ENGINE CONSISTS OF TWO
kirill [66]

Answer:

Electric potential, V=3.52\times 10^{12}\ volts

Explanation:

It is given that,

The magnitude of electric field, E=4.7\times 10^7\ V/m

Distance between automobile engines that consists of metal conductors, d = 0.75\ mm = 7.5\times 10^4\ m

Let V is the magnitude of the potential difference between the conductors. The relation between the electric field and the electric potential is given by:

V=E\times d

V=4.7\times 10^7\ V/m \times 7.5\times 10^4\ m

V=3.52\times 10^{12}\ volts

So, the magnitude of the potential difference between the conductors is 3.52\times 10^{12}\ volts. Hence, this is the required solution.

6 0
4 years ago
An electron having 500ev energy enters at right angle to a uniform magnetic field of 10^-4 Tesla. If its specific charge is 1.75
gtnhenbr [62]

Answer:

The correct answer might be r = 2.8^{27} meters.

Explanation:

<u>The Answer Given might not be correct, I just did what my brain said.</u>

As the angle is perpendicular so Θ=90.

Putting this in the equation to calculate the magnetic force as:

F = evBsinΘ

F= evBsin90                   *sin90 = 1 so,

F= evB.

Now when the electron will start to move in  a circle, The necessary force that makes the electron rotate in a circle is given by Centripetal force.

So,

    Magneteic Force = Centripetal Force

    evB = \frac{mv^{2} }{r}

    r = \frac{mv}{Be} ......(1)

Now the problem is, We don't know " v " so we need to calculate velocity first,

Calculation of Velocity:

                                   In order to calculate the velocity of electron, We should know the potiential difference with which the electrons are accelerated which in our case is 500ev. If "V" is the potiential difference, the energy gained by electrons during accelreation will be Ve. This appear as kinectic enrgy of electrons as,

         

                        K.E = Ve

                        \frac{1}{2}mv^{2} = Ve

                        v =  \sqrt{\frac{2ve}{m} }................(2)

Putting value of velocity in equation 2 from 1:

r = \frac{mv}{Be}  \sqrt{\frac{2ve}{m} }

r = \sqrt{\frac{2mev}{Be}}

r = \sqrt{\frac{(2)(9.1^{-31})(500) (1.6^{-19} )  }{ (1.75^{11} ) (10^{-4} ) } }

r = 2.8^{27} meters.

                               

8 0
3 years ago
the electric field midway between two equal but opposite point charges is 573 n/c , and the distance between the charges is 16.0
saveliy_v [14]

The electric potential  between the two charges is 91.68 V.

<h3>Electric potential between the two charges</h3>

The electric potential  between the two charges is calculated as follows;

V = Ed

where;

  • V is electric potential
  • E is electric field
  • d is the distance of the charge

Substitute the given parameters and solve for electric potential,

V = 573 N/c x 0.16 m

V = 91.68 V

Thus, the electric potential  between the two charges is 91.68 V.

Learn more about electric potential here: brainly.com/question/26978411

#SPJ4

4 0
1 year ago
A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi
vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel N_1=2\ rev/s

angular speed \omega_1=2\pi N_1=4\pi\  rad/s

mass of wheel m_1=4.5\ kg

diameter of wheel d_1=0.30\ m=30\ cm

radius of wheel r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm

mass of clay m_2=2.8\ kg

the radius of the chunk of clay r_2=8\ cm

Moment of inertia of Wheel

I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2

Combined moment of inertia of wheel and clay chunk

I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2

Conserving angular momentum

\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi

Common frequency of wheel and chunk of clay is

\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s

5 0
3 years ago
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