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OLga [1]
3 years ago
7

Plsss helpp will give brainliest!!!

Physics
1 answer:
Alexandra [31]3 years ago
6 0

Answer:

1. Flexibility.

2. Muscular Fitness

3. Cardiovascular Fitness

4. Cardiovascular Fitness

5. Flexibility

6. Muscular Fitness

7. Cardiovascular Fitness

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A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid
igor_vitrenko [27]

Answer:

a) attractiva, b) dF = k \frac{Q_1 \ dQ_2}{dx}, c)  F = k Q_1 \frac{Q_2}{d \ (d+L)}, d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = k \frac{Q_1 \ dQ_2}{dx}

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = k \ Q_1 \int\limits^{d+L}_d     {\frac{1}{x^2} } \, dQ_2

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = k \ Q_1 \lambda \int\limits^{d+L}_d  \, \frac{dx}{x^2}

         F = k Q1 λ (-\frac{1}{x})  

we evaluate the integral

        F = k Q₁ λ (- \frac{1}{d+L} + \frac{1}{d} )

        F = k Q₁ λ  ( \frac{L}{d \ (d+L)})

we change the linear density by its value

      λ = Q2 / L

       F = k Q_1 \frac{Q_2}{d \ (d+L)}

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   \frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}

       F = -1.09 N

the sign indicates that the force is attractive

3 0
3 years ago
If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot
Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

I=I_{1}+I_{2}

I=(m+m_{1})\times r^2+(m+m_{2})\times r^2

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2

I=117.45\ kg-m^2

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0
4 years ago
Explain why the light bulb isn't lighting up in the circuit<br> pictured on the right.
FromTheMoon [43]
The wires should be attached to each screw on the light not only one
8 0
3 years ago
The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen
Eddi Din [679]

Answer:

\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }

Explanation:

Additional information:

<em>The ball has charge </em>-q_0<em>, and the ring has  positive charge </em>+Q<em> distributed uniformly along its circumference. </em>

The electric field at distance z along the z-axis due to the charged ring is

E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.

Therefore, the force on the ball with charge -q_0 is

F=-q_oE_z

F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}

and according to Newton's second law

F=ma=m\dfrac{d^2z}{dz^2}

substituting F we get:

- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}

rearranging we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0

Now we use the approximation that

z^2+a^2\approx a^2 <em>(we use this approximation instead of the original </em>d^2+a^2\approx a^2<em> since </em>z<em>, our assumption still holds )</em>

and get

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0

Now the last equation looks like a Simple Harmonic Equation

m\dfrac{d^2z}{dz^2}+kz=0

where

\omega=\sqrt{ \dfrac{k}{m} }

is the frequency of oscillation. Applying this to our equation we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}

\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}

5 0
4 years ago
Question 9 Unsaved A 680-Hz sound wave travels at 340 m/s in air, with a wavelength of 0.5 m. 5 m. 50 m. 500 m. none of the abov
ale4655 [162]

The correct option is 0.5 M

Calculation

Wavelength is defined as the ratio of velocity of a wave to its frequency. It is measure in meters. Mathematically, wavelength is given by the following formula:

Wavelength = wave velocity / frequency

From the details given in the question,

Wavelength =?

Velocity = 340 m/s

Frequency = 680 HZ

Wavelength = 340 /680 = 0.5

Therefore, wavelength = 0.5 M

5 0
3 years ago
Read 2 more answers
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