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yKpoI14uk [10]
3 years ago
9

In a “minute to win it” game, cards are placed between cups to stack them. The contestant then pulls the card out in hopes that

they will end with a set of stacked cups. Why does this work?
There is no friction between the card and the cup.


Inertia will keep the top cup from moving very far. Gravity will pull one cup on top of another.


The opening of the cup is so much bigger than the bottom, even though the cup moves they will both be stacked together.


Both cups are moved the same amount by the friction between them and the card.
Physics
2 answers:
Luden [163]3 years ago
4 0

Answer:

There is no friction between the card and the cup.

Explanation:

lubasha [3.4K]3 years ago
3 0

Answer:

The answer option to your question is A) - There is no friction between the card and the cup.

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A certain 60.0 Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 11.6 kV/m.
yuradex [85]

Explanation:

Given that,

Frequency of the power line, f = 6 Hz

Value of maximum electric field strength of 11.6 kV/m

(a) The wavelength of this very low frequency electromagnetic wave is given by using relation as :

c=f\lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{60\ Hz}

\lambda=5\times 10^6\ m

(b) As its can be seen that the wavelength of this wave is very high. It shows that it is a radio wave.

(c) The relation between the maximum magnetic field strength and maximum electric field strength is given by :

B_0=\dfrac{E_0}{c}\\\\B_0=\dfrac{11.6\times 10^3}{3\times 10^8}\\\\B_0=3.86\times 10^{-5}\ T

So, the maximum magnetic field strength is 3.86\times 10^{-5}\ T.

7 0
3 years ago
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How do very small objects behave?
nataly862011 [7]
A. very small objects behave like like particles.
7 0
3 years ago
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Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

6 0
3 years ago
GIVING BRAINLIEST PLEASE HELP!!
Akimi4 [234]

Answer:

mechanic and potential?

Explanation:

6 0
2 years ago
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5280 feet to meters
lesya [120]

Answer:

1m= 3.2 feets

? = 5280 feets

let ? be n

cross multiply

3.2n = 1×5280

n= 5280÷3.2

n= 1650

so, 5280 feets is equal to 1650 metres

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