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vlada-n [284]
3 years ago
7

Which of the following are true about S waves

Physics
1 answer:
shusha [124]3 years ago
6 0

Answer:

3. they can travel through solids

4. they move rock at right angles to the direction of wave travel

Explanation:

  • S waves are called transverse waves they have the ability to move past the solids. They cannot move through the liquids, these waves are perpendicular to the direction of travel.  
  • They are also called longitudinal waves, the ad is second to record on the seismograph as they slowly pass through the rocks. They have a speed of 3.4 to 7.2 km as per the boundary.
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How might you tell if a food contains an acid
MrRissso [65]

Answer:

it will taste sour

Explanation:

7 0
3 years ago
A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 54.0 s. The
timama [110]

Answer:

a) f=0.1 Hz ; b) T=10s

c)λ= 36m

d)v=3.6m/s

e)amplitude, cannot be determined

Explanation:

Complete question is:

Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.

Given:

number of wave crests 'n'= 5

pass in a time't' 54.0s

distance between two successive crests 'd'= 36m

a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have

f=n/t

f= 5/ 54 => 0.1Hz

b)The time period of wave 'T' is the reciprocal of the frequency

therefore,

T=1/f

T=1/0.1

T=10 sec.

c)wavelength'λ' is the distance between two successive crests i.e 36m

Therefore, λ= 36m

d) speed of the wave 'v' can be determined by the product of frequency and wavelength

v= fλ => 0.1 x 36

v=3.6m/s

e) For amplitude, no data is given in this question. So, it cannot be determined.

5 0
3 years ago
Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e
erik [133]

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

     E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

 we calculate

E_total = 9 10⁹ / (4 10⁻²)²   (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

8 0
3 years ago
A 0.10 g honeybee acquires a charge of +23 pC while flying.
kari74 [83]

Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

Therefore the ratio :

\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

\frac{F}{w} =2.347\times 10^{-6}\ N

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
3 0
3 years ago
A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire
soldi70 [24.7K]

Answer:

a)  v=4.0m/s

b)  B=2.958T

Explanation:

From the question we are told that:

Wire Length l=10cm=>0.10m

Resistance R=0.35

Force F=1.0N

Power P=4.0W

a)

Generally the equation for Power is mathematically given by

P=Fv

Therefore

v=\frac{P}{F}

v=\frac{4.0}{1.0}

v=4.0m/s

b)

Generally the equation for Magnetic Field is mathematically given by

B=\frac{\sqrt{PR}}{vl}

B=\frac{\sqrt{4*0.35}}{4*0.10}

B=2.958T

5 0
3 years ago
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