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ASHA 777 [7]
3 years ago
6

Which piece of evidence did Alfred Wegener use to develop the theory of continental drift?

Physics
1 answer:
Eduardwww [97]3 years ago
3 0
The right answer for the question that is being asked and shown above is that: "A.tectonic activity concentrated in certain areas." A piece of evidence did Alfred Wegener use to develop the theory of continental drift is that <span>A.tectonic activity concentrated in certain areas</span>
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*Materials that regulate the flow of current through them *
4vir4ik [10]

Answer:

electromagnet

Explanation:

mark me as brainlest

3 0
2 years ago
Can someone explain to me #4.
nalin [4]
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
4 0
3 years ago
Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
The refractive index for glycerine is n_g=1.473, while for air it is n_a = 1.00.

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
40 Points!!!!!!!!!!!!!
Firlakuza [10]
The answer is A for sure
3 0
3 years ago
One hundred turns of insulated copper wire are wrapped into a circular coil of crosssectional area 1.20⇥103 m2. The two ends of
arsen [322]

Answer:

236.3  x 10^-^3 C

Explanation:

Given:

B(0)=1.60T and B(t)=-1.60T

No. of turns 'N' =100

cross-sectional area 'A'= 1.2 x 10^-^3m²

Resistance 'R'= 1.3Ω

According to Faraday's law, the induced emf is given by,

ℰ=-NdΦ/dt

The current given by resistance and induced emf as

I = ℰ/R

I= -NdΦ/dtR

By converting the current to differential form(the time derivative of charge), we get

\frac{dq}{dt}=  -NdΦ/dtR

dq= -N dΦ/R

The change in the flux dФ =Ф(t)-Ф(0)

therefore, dq = \frac{N}{R} (Ф(0)-Ф(t))

Also, flux is equal to the magnetic field multiplied with the area of the coil

dq = NA(B(0)-B(t))/R

dq= (100)(1.2 x 10^-^3)(1.6+1.6)/1.3

dq= 236.3  x 10^-^3 C

5 0
3 years ago
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