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2 years ago

What is one way to increase the amplitude of a wave in a medium

1 answer:

3 0

supply it with more energy. one way to do is to produce vibrations in the same frequency as the wave. This would cause resonance leading to higher amplitude

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Timed! I would really appreciate some help! thank you!

GenaCL600 [577]

Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

$t=30[min]*\frac{1h}{60min}= 0.5[h]\\$

We know that speed is defined as the relationship between space and time.

$v=x/t$

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

$x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]$

4 0

2 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

3 years ago

Me izz in canada hehe Xd just in 2 mins wow what is immunization?

Kruka [31]

Answer:

Immunization, or immunisation, is the process by which an individual's immune system becomes fortified against an infectious agent.

Explanation:

HOPE IT HELPS 

HAVE A NICE DAY :) 

XXITZFLIRTYQUEENXX 

4 0

2 years ago

A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?

Slav-nsk [51]

It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.

From the question given above, the following data were obtained:

Height (H) = 206 m

NOTE: Acceleration due to gravity (g) = 10 m/s²

The time taken for the ball to get to the ground can be obtained as follow:

H = ½gt²

206 = ½ × 10 × t²

206 = 5 × t²

Divide both side by 5

$t^{2} = \frac{206}{5}\\\\ t^{2} = 41.2$

Take the square root of both side

$t = \sqrt{41.2}$

Therefore, it will take 6.42 s for the ball to get to the ground.

Learn more: brainly.com/question/24903556

7 0

2 years ago