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2 years ago

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Question 5: A heavy rope is flicked upwards, creating a single pulse in the rope. Make a drawing of the rope and indicate the fo

llowing in your drawing
a. the direction of motion of the pulse
b. Amplitude
c. Pulse length
d. Position of the rest

1 answer:

Anastaziya [24]2 years ago

8 0

Answer:

the direction of motion of the pulse

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A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t

iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

$T = 2\pi \sqrt{\frac{l}{g} }$

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

$T' = 2\pi \sqrt{\frac{2l}{g} }$

$T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }$

$T' = \sqrt{2} * T$

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

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1 year ago

CAN SOMEONE HELP ME PLEASE ASAP

Lady_Fox [76]

The third choice is correct

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3 years ago

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Clarify this statement acceleration due to gravity of the Earth is 9.8m/s square

muminat

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it means that velocity of a body rises by 9.8m/s each second if the air resistance is nrelated

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3 years ago

A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

<em>current in the wire, I = 2.7 A</em>
<em>length of the wire, L = (3.2 i + 4.3j) cm</em>
<em>magnetic filed, B = 1.24 i</em>

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

<em>θ is the angle wire and magnetic field</em>

<em />

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$

The magnitude of the wire length is calculated as follows;

$|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m$

The component of the force in negative z-direction is calculated as;

$F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N$

Thus, the component of the force in negative z-direction is -0.144 N.

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2 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time <em>t</em> (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time <em>t</em> (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at <em>t</em> = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

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2 years ago