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diamong [38]
3 years ago
11

Suppose an earthquake occurs on an imaginary planet. Scientists on the other side of the planet detect primary waves but not sec

ondary waves after the quake. This suggests that:________.
a. part of the planet's interior is liquid
b. all of the planet's interior is solid
c. the planet has an iron core
d. the planet's interior consists entirely of rocky materials
e. the planet's mantle is liquid
Physics
1 answer:
MrMuchimi3 years ago
4 0

Answer:

a. part of the planet's interior is liquid

Explanation:

Seismic waves are slowed down in a liquid medium. On earth, molten areas reduces the speed of P waves and completely stops S waves which have shearing motion that can only be transmitted through a solid

The motion has the highest speed in materials, allowing it to be detected first on seismograms. The P wave is has a smaller wavelength and a higher frequency than the surface ans the S waves.

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Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft
vlabodo [156]

Answer:

<u><em>note:</em></u>

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4 0
3 years ago
A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.
Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

\epsilon=NA\dfrac{dB}{dt}

For N = 1

\epsilon=A\dfrac{dB}{dt}

We need to calculate the current

Using formula of current

i=\dfrac{\epsilon}{R}

Put the value of emf

i=\dfrac{A\dfrac{dB}{dt}}{R}

Now, if the number of turn is 22 , then induced emf would be

\epsilon'=NA\dfrac{dB}{dt}

Then the current would be

i'=\dfrac{\epsilon'}{NR}

i'=\dfrac{NA\dfrac{dB}{dt}}{NR}

i'=\dfrac{A\dfrac{dB}{dt}}{R}

i'=i

Hence, The current would be same in both situation.

4 0
3 years ago
A clock radio is rated as 30 w of power output. if the radio also draws 30 w at 120 v, which will the current draw be?
g100num [7]
P = IV

I = P/V =  30 / 120 = 0.25 A.

Current = 0.25A  
4 0
3 years ago
Hey can yall help me out in dis
Karolina [17]

Answer:

Scientific method

Explanation:

5 0
2 years ago
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