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3 years ago

How will you modify the station so that coin travels with the index card

Physics

1 answer:

Marina CMI [18]3 years ago

7 0

Answer:

When the edge of the card passes the coin, the coin has nothing underneath it, so gravity pulls it down. 3) How you modify this station so the coin travels with the index card? You change the card's acceleration by starting to move the card at a lower speed, then slowly increasing the speed.

Explanation:

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Answer the point I wish you would have the greatest potential energy is when you are coming down the swing and getting ready to go up the greatest kinetic energy is whenever you’re falling back down from the height of how far you went up

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3 years ago

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A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

Lilit [14]

Answer:

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

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3 years ago

A transverse traveling wave on a cord is represented by, where D and x are in meters and t is in seconds.

yarga [219]

Answer with Explanation:

We are given that a transverse travelling wave on a cord is represented by

$D=0.51sin(6.1x+76t)$

Where D and x in meters

t(in seconds)

a.General equation of transverse wave

$y=Asin(kx+\omega t)$

By comparing we get

A=0.51

$k=6.1$

$k=\frac{2\pi}{\lambda}$

Wavelength, $\lambda=\frac{2\pi}{k}=\frac{2\pi}{6.1}=1.03$

b. $\omega=76$

Frequency, $f=\frac{\omega}{2\pi}=\frac{76}{2\pi}=12.096Hz$

c.Velocity, $v=f\lambda=12.096\times 1.03=12.46m/s$

Direction:Towards negative x- axis

d.Amplitude,A=0.51 m

e.Maximum speed, $v_{max}=A\omega=0.51\times 76=38.76 m/s$

Minimum speed, $v_{min}=0$

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4 years ago

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Explanation:hope this helps you

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3 years ago

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Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apar

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The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

We have to calculate much work is needed to move the charges apart and double the distance between them.

<h3>What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?</h3>

The potential energy of the system of two charges separated by a distance is given by -

$U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}$

In order to solve this question, it is important to remember the work - energy theorem which states -

"The change in the energy of the body is equal to work done on it"

Hence, using this work -energy theorem in the question given to us we get -

$U_{f} -U_{i} =W_{net}$

In our case -

$U_{f} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}$

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

To solve more question on potential energy, visit the link below -

brainly.com/question/15014856

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2 years ago

How will you modify the station so that coin travels with the index card​

How will you modify the station so that coin travels with the index card