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Lubov Fominskaja [6]
3 years ago
9

How do I ship a pringle in the mail without it breaking? This is a physics project.

Physics
1 answer:
vekshin13 years ago
5 0
Assuming you want it to be as small and lightweight as possible :

Cut a solid box roughly twice as big as the pringle. Put the pringle inside the box, and fill the remaining space with cotton, that will cushion the impacts. Be sure to apply the mention <em>FRAGILE</em> to the box, so that they'll take care of it properly.
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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
8 0
3 years ago
A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running
MrRissso [65]

The runner has initial velocity vector

\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath

and acceleration vector

\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)

so that her velocity at time t is

\vec v=\vec v_0+\vec at

She runs directly east when the vertical component of \vec v is 0:

2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time t would be

\vec x=\vec v_0t+\dfrac12\vec at^2

so that after 10.4 s, her position would be

\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath

which is 19.9 m away from her starting position.

8 0
3 years ago
Read 2 more answers
Samuel hits a ball, which is initially at rest, off of a tee. What is the velocity of the ball after it is hit, if Samuel applie
aniked [119]

Answer:

7.39 m/s

Explanation:

Applying

K.E = 1/2mv²..................... Equation 1

Where K.E = Kinetic Energy, m = mass of the ball, v = velocity of the ball.

Make v the subject of the equation

v = √(2K.E/m)................. Euqation 2

From the question,

Given: K.E = 30 J, m = 1.1kg

Substitute these values into equation 2

v = √(2×30/1.1)

v = √54.54

v = 7.39 m/s

8 0
3 years ago
A 2.45 cm tall object is placed in 33.7 cm in front of a convex lens. The focal length
timofeeve [1]

Answer:

-1.65

Explanation:

First of all, we find the position of the image by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

For the lens in this problem:

f = 21.0 cm (the focal length of a convex lens is positive)

p = 33.7 cm

Solving for q, we find the position of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{21.0}-\frac{1}{33.7}=0.0179 cm^{-1}\\q=\frac{1}{0.0179}=55.7 cm

Then, the magnification of the image is given by:

M=-\frac{q}{p}

And substituting,

M=-\frac{55.7}{33.7}=-1.65

Which means that the image is inverted (negative sign) and enlarged (because M is larger than 1).

3 0
3 years ago
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