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Arada [10]
2 years ago
15

A compound has a melting point of 812°C. what type of bonding is in the compound?

Physics
1 answer:
AURORKA [14]2 years ago
7 0

Answer:

Ionic compound.

Explanation:

812° C is a very high melting point. Such high melting points are generally ionic compound. Ionic compounds are have very strong bond between the elements ( electrostatic bond). In order to break this bond, large amount of heat energies are needed. So, they have high melting point. Also, Ionic compound are very good conductors of electricity.

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a wave in the ocean has an amplitude of 3m. Winds pick up suddenly, increasing the wave's amplitude to 6m. How does this change
Paha777 [63]

Answer

Hi,

An increase in amplitude from 3m to 6 m increases the energy it transports. The frequency of the wave is not affected

Explanation

Amplitude is the height of a wave where as frequency is the number of waves that pass by each second. A wave with bigger amplitude has more energy than a wave with smaller amplitude. A point where more waves pass contains more energy that is transferred every second. The change in the amplitude of a wave does not change its frequency. However, frequency is inversely related to the wavelength of a wave.

Best Wishes!

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Which of the following is not a constant or control for this experiment?
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The core of a star must be at temperature of _____ degrees Celsius for hydrogen fusion to take place. 10,000 100,000 1,000,000 1
disa [49]
The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin. 
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3 years ago
True or false? Magnetic reversals are recorded in the newly formed oceanic crust on BOTH sides of a mid-ocean ridge spreading ce
andrey2020 [161]

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4 0
3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
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