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1 year ago

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals.

Physics

1 answer:

Bess [88]1 year ago

3 0

time should you wait between pushes is 2.83 sec.

the question is incomplete, full statement is-

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals. If you want to increase the amplitude of her motion as quickly as possible, how much time should you wait between pushes?

<h3>What is Amplitude?</h3>

In physics, amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.

regular interval - at similarly spaced intervals: having the same interval of time between occurrences From 4 a.m. to midnight, the buses operate at regular intervals. The boards are positioned at regular intervals, with an equal amount of space between each.

The length of swing, l = 2.1 m

The time between the pushes is nothing but the Time period

and is given by the formula,

$T = 2\pi ( \frac{l}{g} )^{\frac{1}{2} }$

= 2 * 3.14 ( 2.0/ 9.8 ) ^ (1/2)

= 2.83 sec

to learn more about Amplitude go to - brainly.com/question/3613222

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While riding a bike down a steep hill at 9ms you skid to a stop in 3.2 seconds. How far did the bike slide? Assume uniform accel

castortr0y [4]

Answer:

24.30 m

Explanation:

v(final)=v(initial)+at

0 m/s= 9 m/s +a(3.2s)

a=-2.81 m/s^2

delta x= vt+ (1/2)at^2

=(9 m/s * 3.2 s)+ (.5* -2.81 m/s^2 * 3.2s)

=24.30 m

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4 years ago

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electr

Marat540 [252]

Complete Question

$Fe^{2+} + 2e^-----> Fe \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = - 0.441 V$

$Cd^{2+} + 2e^- -----> Cd \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = -0.403V$

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03305 V?

Answer:

The mass is M= 117.37g

Explanation:

The overall reaction is as follows

$Cd^{2+} + Fe Fe^{2+} + Cd$

The reaction is this way because the potential of $Cd^{2+} \ reducing \ to \ Cd$ is higher than the potential of $Fe^{2+} \ reducing \ to \ Fe$ so the the Fe would be oxidized and $Cd^{2+}$ would be reduced

At equilibrium the rate constant of the reaction is

$Q = \frac{concentration \ of \ product }{concentration \ of reactant }$

$= \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}$

The Voltage of the cell $E_{cell} = E_{Cd^{2+}/Cd } + E_{Fe^{2+} /Fe}$

Substituting the given values into the equation

$E_{cell} = -0.403 -(-0.441)$

$= 0.038V$

The voltage of the cell at any point can be calculated using the equation

$E = E_{cell} - \frac{0.059}{n_e} Q$

Where $n_e \ is \ the \ number\ of\ electron$

Substituting for Q

$E = E_{cell} - \frac{0.059}{n_e} \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}$

When E = 0.03305 V

$E = E_{cell} - \frac{0.59}{n_e} \frac{Fe^{2+}}{Cd^{2+}}$

Since we are considering the Cd electrode the equation becomes

$E= E_{cell} - \frac{0.059}{n_e} [\frac{1}{Cd^{2+}} ]$

Substituting values and making [ $Cd^{2+}$ ] the subject

$[Cd^{2+}] =\frac{1}{e^{[\frac{0.03305- 0.038}{\frac{0.059 }{2} }] }}$

$= 0.8455M$

Given from the question that the volume is 1 Liter

The number of mole = concentration * volume

= 0.8455 * 1

= 0.8455 moles

At the standard state the concentration of $Cd^{2+}$ is =1 mole /L

Hence the amount deposited on the Cd electrode would be

= Original amount - The calculated amount

= 1 - 0.8455

= 0.1545 moles

The mass deposited is mathematically represented as

$mass = mole * molar \ mass$

The Molar mass of Cd $= 112.41 g/mol$

Mass $= 0.1545 *112.41$

$= 17.37g$

Hence the total mass of the electrode is = standard mass + calculated mass

M= 100+ 17.37

M= 117.37g

6 0

3 years ago

Read 2 more answers

A simple pendulum consists of a small object with mass m = 1.91 kg and a massless string with length l = 1 m. Halfway down the s

castortr0y [4]

Answer:

around the coop

Explanation:

8 0

3 years ago

Match each term to its definition.

ladessa [460]

Ion is e
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4 0

4 years ago

The "lead" in pencils is a graphite composition with a Young's modulus of about 1 ✕ 10⁹ N/m². Calculate the change in length (in

sweet [91]

Answer:

0.701 mm.

Explanation:

From Hook's law,

Young modulus = Stress/Strain.

Stress = Force/area = F/A

Strain = ΔL/L.

Therefore,

γ = (F/A)/(ΔL/L)

γ = FL/ΔLA ...................... Equation 1

Where γ = Young's modulus, F = Force, L = Length of the pencil, ΔL = change in length of the pencil, A = cross sectional area.

Make ΔL the subject of the equation,

ΔL = FL/γA......................... Equation 2

But,

A = πd²/4......................... Equation 3

Where d = diameter

Substitute equation 3 into equation 2

ΔL = 4FL/γπd²................ Equation 4

Given: F = 4.9 N, L = 55 mm = 0.055 m, d = 0.7 mm = 0.0007 m, γ = 1×10⁹ N/m², π = 3.14.

Substitute into equation 4

ΔL = (4×4.9×0.055)/(1×10⁹×3.14×0.0007²)

ΔL = 1.078/1538.6

ΔL = 7.01×10⁻⁴ m.

ΔL = 0.701 mm.

4 0

3 years ago