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brilliants [131]
2 years ago
8

What do slinky waves and seismic waves have in common? A. They both always occur as transverse waves. B. They both always occur

as longitudinal waves. C. Both are electromagnetic waves. D. Both exhibit the same particle-to-particle interaction. E. Both are mechanical waves.
Physics
2 answers:
anyanavicka [17]2 years ago
8 0

Answer:

E> Both are Mechanical waves

Explanation:

As we know that Slinky waves are transverse in nature and medium particles will move to and fro while wave propagate through the medium.

While Seismic waves are of two type

(i) P waves   (ii) S Waves

these both moves in form of compression and rarefaction and it can travel through the medium.

So here in case of Seismic wave we can say it is longitudinal waves.

So here we can say that they both have common that they both requires medium to move from one end to other end.

So they both are mechanical waves

Usimov [2.4K]2 years ago
3 0

the answer is E both are mechanical waves

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An ultrasound pulse has a wavelength of 1.0mm. Its speed in water is 1400m. What’s the frequency?
timama [110]

Answer:

1.4\cdot 10^6 Hz

Explanation:

The relationship between the frequency, the wavelength and the speed of a wave is given by the wave's equation:

v=f \lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

For the pulse in this problem,

\lambda = 1.0 mm = 0.001 m\\v = 1400 m/s

Solving for  f, we find the frequency:

f=\frac{v}{\lambda}=\frac{1400}{0.001}=1.4\cdot 10^6 Hz

6 0
2 years ago
A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo
Digiron [165]
Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
8 0
2 years ago
A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

mv_o=m(v_1 + v_2) where m is the mass, v_o is the initial speed before impact, v_1 and v_2 are velocity of the impacting object after collision and velocity after impact of the originally constant object

5m=m(v_1 +v_2)

Therefore v_1+v_2=5

After collision, kinetic energy doubles hence

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting 5 m/s for v_o then

2*(5^{2})= v_1^{2} + v_2^{2}

50= v_1^{2} + v_2^{2}

Also, it’s known that v_1+v_2=5 hence v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then v_2=6.83 m/s

Substituting, v_1=-1.83 m/s

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0
2 years ago
A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the
zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂=  L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0
2 years ago
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vladimir2022 [97]

Answer:

Explanation:

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Mass of block A = 604 g

density of mass A = mass / density

= 604 / 75 = 8.05 g / cm³

Since density of both A and B are less than that of mercury , both will float in mercury.

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