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Elena L [17]
3 years ago
15

Determine whether the interference is constructive or destructive at each location indicated.

Physics
1 answer:
OlgaM077 [116]3 years ago
6 0

Answer:

A. Constructive

B. Destructive

C. Destructive

D. Constructive

Explanation:

Constructive interference takes place at locations along the path of two superposed waves where the waves are in phase such that a high or low point of one of the waves corresponds with a high or low point of the other wave which gives a resulting wave amplitude which is the sum of the amplitudes of the individual waves

Destructive interference takes place at locations along the path of two superposed waves where one wave is out of phase with the other wave such that a high or low point of one of the waves coincides with a low or high point of the other wave respectively thereby cancelling the effect of the other wave and giving a resulting wave that has an amplitude which is the difference in the amplitudes of the individual waves

Therefore;

At point A, the peak of each wave partially coincides resulting in constructive interference

At point B, the peak of the blue wave and the trough of the red wave partially coincides resulting in destructive interference

At point C, the through of the blue wave and the peak of the back wave partially coincides resulting in destructive interference

At point D, the trough of each wave partially coincides resulting in constructive interference.

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postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
3 years ago
Part D
anygoal [31]

Answer: I didn't see a difference because the large ball's vertical displacement and velocity are the same as the small one's.

Explanation:

5 0
3 years ago
How can I convert this?<br>Please answer with solution. Thank you.​
fomenos

Answer:

1 hr 45 min

Explanation:

8 0
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A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
nignag [31]

Answer:

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Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

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3 years ago
Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.
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