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Anon25 [30]
3 years ago
8

After a wave passes through a medium, particles in the medium

Physics
1 answer:
sertanlavr [38]3 years ago
3 0

Answer:

If a particle is affected by a wave, then the particles are displaced. They move along the direction of the wave. Hence, After a wave passes through a medium, particles in the medium are moved along with the wave.

Explanation:

hope this helps you

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Wire coils spinning between two magnets will produce …?
klemol [59]

Answer:

B

Explanation:

I don't know but its b

6 0
2 years ago
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A bus is moving at a constant speed of 40 m/s. How many hours will it takes to travel 260 miles?
lawyer [7]

Explanation:

1 mile = 1609m

Distance=260 miles= 418340 m

Speed= 40 m/s

Time = distance/speed

= 418340/40

=10458.5 seconds

= 2.9 hours

8 0
3 years ago
the shock absorbers in a car act as a big spring with k= 21900 N/m. when a 92.5 kg person gets in, how far does the spring stret
r-ruslan [8.4K]

Answer: 0.04139m

Explanation:

First, we need to calculate the weight of the man which will be:

Weight = mass × acceleration due to gravity

Weight = mg

Weight = 92.5 × 9.8

Weight = 906.5N

Then, we calculate the force which will be:

F = kx

mg = kx

x = mg/k

x = 906.5/21900

x = 0.04139m.

The spring stretched for 0.04139m.

4 0
3 years ago
Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)
Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

P_{1}=\rho g h_{1}

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

P_{1}=\rho gh_{1}.....(I)

Pressure for second pipe,

P_{2}=\rho gh_{2}.....(II)

From equation (I) and (II)

P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

Put the value of P₁ and P₂

\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)

2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2....(III)

We know that,

The continuity equation

v_{1}A_{1}=v_{2}A_{2}

v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})

Put the value of v₂ in equation (III)

2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2

2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2

Here, \dfrac{A_{1}}{A_{2}}=\gamma

So, 2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)

v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Hence, (a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

8 0
3 years ago
A 2.72 A current flows through a wire for
agasfer [191]
  • Current=2.72A=I
  • Time=t=9.88s
  • Charge=Q

\\ \rm\longmapsto I=\dfrac{Q}{t}

\\ \rm\longmapsto Q=It

\\ \rm\longmapsto Q=2.72\times 9.88

\\ \rm\longmapsto Q=26.87C

8 0
3 years ago
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