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2 years ago

Jack tries to place magnets on his refrigerator at home, but they won’t stick. What could be the reason?

2 answers:

3 0

The most probable reason why the magnets won't stick on the refrigerator is that the body of the refrigerator and the magnets have like poles. If both have negative or both have positive poles facing each other, they will repel. In principle, magnets are attracted to opposite poles and like poles repel.

lukranit [14]2 years ago

3 0

Answer:

The door is made up of aluminum or stainless steel

Explanation:

The outer cabinet and door of the fridge are made of aluminum, steel, or stainless steel. If the door is made of steel, magnets can be attached easily. If its aluminum, magnet won't stick as aluminum does not attract magnets. In stainless steel nickel is added for more strength and thus reducing the magnetic strength of the metal.

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Which term describes an advantage of wind energy?

atroni [7]

The answer is "D". This is because wind is very abundant on earth...

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2 years ago

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What does it mean that are bodies are in osmotic equilibrium but chemical and electrical disequilibrium? Why? What molecules are

Whitepunk [10]

Explanation:

The point is that water is moving smoothly but that the solutes are not.Even though the containers are chemically different (chemical disequilibrium), once all the solutes in one container are contrasted to all the solutes in another container, both have the same total solutes concentrations (this means that they are in osmotic balance).

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2 years ago

The work function for magnesium is 3.70 ev. what is its cutoff frequency?

alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is 3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

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1 year ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

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2 years ago

The graph shows the range of frequencies commonly heard by some animals.

Vsevolod [243]

Concept:

Frequency- It is defined as the number of oscillations occur in one second.

Its SI unit is Hertz (Hz)

Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds

∵ In 0.75 second, produced sound has oscillations = 18,500 cycles

∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz

The frequency of the sound will be ≈ 24,667 Hz

From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.

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3 years ago

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