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Ganezh [65]
3 years ago
6

The highest point of a transverse wave is called the _____.

Physics
2 answers:
gladu [14]3 years ago
8 0

The answer is peak. Hope this helps :)

mariarad [96]3 years ago
4 0

Answer:

The answer is Crest just learned it in science.

Explanation:

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How does SCOTUS view information obtained by private citizens and subsequently shared with law enforcement?
scoundrel [369]

Answer:

as a legal search

Explanation:

In the specific case, there was a trespass on private property through the installation of a global positioning system (GPS) to gather sensitive information about the citizen. This is considered a legal search and is not appropriate to conduct such an investigation without proper approval from the relevant authority.

5 0
3 years ago
What is the energy of motion.
AlladinOne [14]

Answer:

Is the sum of potential and kinetic energy in an object that is used to do work.

4 0
3 years ago
Read 2 more answers
In a thunderstorm, the thunder occurs at the same time as lightning. Explain how does a person hears the thunder 5s after he see
sergey [27]

Answer:

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

Explanation:

4 0
2 years ago
A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a
Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

8 0
3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
Read 2 more answers
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