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poizon [28]
3 years ago
9

Are headlands are formed through wave deposition true orfalse

Physics
2 answers:
KengaRu [80]3 years ago
6 0

Answer: False

Explanation:

 The given statement is false as, the headlands are not form by the wave deposition. The headlands are mainly the part of the shore which sticks out in the oceans. It is basically formed on the harsh coastlines with the help of the bands of the rocks.

The headlands are form with the alternating band of the rocks during the ocean attack in the part of  the coastlines. The headlands are also known as coastal land form and it is characterized as follows:

  • Rocky shores
  • Breaking waves
  • Steep sea cliff
  • Intense erosion

sp2606 [1]3 years ago
5 0

the answer is false hope this helps :]

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The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

5 0
3 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

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6 0
2 years ago
3. How much work is done when you pull a 6 N wagon for 5 meters?
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Answer:

<h2>30 J</h2>

Explanation:

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From the question

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workdone = 6 × 5 = 30

We have the final answer as

<h3>30 J</h3>

Hope this helps you

3 0
2 years ago
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Semmy [17]
8+R=V/I
8+R=6/0.5=12 ohm
then R=4 ohm
4 0
3 years ago
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