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Tema [17]
2 years ago
15

A girl with a mass of 27 kg is playing on a swing. There are three main forces

Physics
1 answer:
N76 [4]2 years ago
3 0

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

  • Mass of the girl, m = 27 kg
  • Speed of the girl, v = 3 m/s
  • Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

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A mover does 422 J of work pushing a crate 8.39 m. How much force did he exert?
kicyunya [14]

Answer:

50.3N

Explanation:

Work done = force x distance

422J. = force x 8.39m

÷8.39 both side to get force

Force is 50.3N to 1 d.p.

Check:

50.3 x 8.39=422.017J

Same as 422J to 1 d.p

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Wave crests in a pond are separated by 0.2m . The period of these waves is 0.2 s . What is the speed at which these waves travel
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7 0
3 years ago
A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the
Aleksandr-060686 [28]

Answer:

(a):  The magnitude of the electric force on the small sphere = \dfrac{q\sigma}{2\epsilon_o}.

(b): Shown below.

Explanation:

<u>Given:</u>

  • m = mass of the small sphere.
  • q = charge on the small sphere.
  • L = length of the silk fiber.
  • \sigma = surface charge density of the large vertical insulating sheet.

<h2>(a):</h2>

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

\rm E = \dfrac{\sigma}{2\epsilon_o}.

<em>where,</em>

\epsilon_o is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

\rm F_e=qE.

Thus, the magnitude of the electric force on the small sphere is given by

\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

<h2>(b):</h2>

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

\rm \tan \theta = \dfrac{F_e}{W}.

<em>where,</em>

  • \rm F_e = electric force on the sphere, acting along left.
  • \rm W = weight of the sphere, acting vertically downwards.

<em />

\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .

g is the acceleration due to gravity.

6 0
3 years ago
an airplanes propeller completes one revolution in 65ms. a)what is the angular speed in rpm? in rad/s?
miss Akunina [59]

-- We already know the rate of revolutions per time ...
it's 1 revolution per 0.065 sec.  We just have to
unit-convert that to 'per minute'.

(1 rev / 0.065 sec) x (60 sec / min) = (1 x 60) / (0.065) = <em>923 RPM</em>  (rounded)
_______________________________

-- 1 revolution = 2π radians

  (2π rad) / (0.065 sec) = (2π / 0.065) = <em>96.66 rad/sec</em>  (rounded)


7 0
3 years ago
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