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scZoUnD [109]
4 years ago
5

During constructive interference two waves travel through the same medium and the effects of the first wave are ___ the effects

of the second wave
Physics
1 answer:
Ymorist [56]4 years ago
7 0

Answer:  

During constructive interference occurs at any location along the medium  where two inferring waves are in the same direction. In this case both waves have upward displacement. Consequently the medium has an upward displacement has that is greater than the displacement of inferring pulses.


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A ball is thrown horizontally from a platform so that the initial height of the ball is 6.0 m above the level ground below. The
solniwko [45]
We can first obtain time of flight from vertical fall
Initial velocity U=0, d = 6 m, a = 9.8 m/s² 
d = ut + 1/2 at²
6.0 = 0 + (1/2 × 9.80 t²) 
t = √(12/9.8)
 = 1.106 sec
Horizontal velocity = Vh = Dh/t
  = 24.0 /1.106 s
  = 21.69 m/s
The ball was thrown at a speed of 21.69 m/s
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The acceleration due to gravity is
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Answer:

9.8 m/s^2

Explanation:

This is a constant that you should memorize

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A tennis ball is released from a height of 2.0 m above the floor, and then bounces three times. With each bounce, dissipative fo
Leni [432]

Answer:

12 cm

Explanation:

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3 0
4 years ago
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Are all magnetic fields created with electricity
Natasha_Volkova [10]
No not all, because not all electrical field attract and repel
3 0
3 years ago
A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
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3 years ago
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