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inna [77]
2 years ago
6

Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy

starts jumping on the trampoline too. Suddenly, Max goes twice as high even though he did not put any extra effort into jumping. What type of wave interaction is this?
Physics
1 answer:
mylen [45]2 years ago
8 0

Answer:

x_total = (A + B) cos (wt + Ф)

we have the sum of the two waves in a phase movement

Explanation:

In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.

         Max

               x₁ = A cos (wt + Ф)

         Jimmy

              x₂ = B cos (wt + Ф)

         

total movement

             x_total = (A + B) cos (wt + Ф)

 Therefore we have the sum of the two waves in a phase movement

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Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marbl
Lostsunrise [7]
Conservation of linear momentum:

m*v inital = m*v final

0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v

(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)

0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s

Answer 1.8 m/s (positive, to the right).

 

4 0
2 years ago
Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13
Julli [10]
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
                 m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
                    (85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
3 0
2 years ago
The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet
Naily [24]

Answer:

Speed of comet before collision is

v_{2_{i}}=-2.5\times10^{3}\quad km/h

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h

Case is considered as partially inelastic collision, by conservation of momentum

m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h

8 0
2 years ago
Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.
solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination \theta =58.7^{\circ}C

Horizontal distance travel by Particle d=1200 m

Vertical height h=780 m

Let u be the initial velocity

calculating vertical distance

y=u\sin \theta +\frac{at^2}{2}

y=u\sin \theta t-\frac{gt^2}{2}-------1

Calculating horizontal distance

x=u\cos \theta \times t+0

t=\frac{x}{u\cos \theta }

put value of t in equation 1

y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2

y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }

\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y

u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}

u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}

at y=-780\ m\ x=1200 m

u^2=\frac{18.154}{2753.65}\times 1200^2

u=97.41 m/s

6 0
2 years ago
A boy standing on a 19.6 meter tall bridge sees a motorboat approaching the bridge at a constant speed. When the boat is 27 mete
azamat

Answer:

A. 12 m/s

Explanation:

Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.

The equation that describes the motion of the stone is:

y = y_0 + v_0 * t+1/2 * a * t^2

The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:

-19,6 m = -1/2 * 9,8 m/s^2  * t^2

t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2

t=√(4,012 s^2) = 2,003 s

Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:

v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s

Have a nice day! :D

8 0
2 years ago
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