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2 years ago

The compass uses a pendulous system to improve?

Physics

1 answer:

ivanzaharov [21]2 years ago

6 0

The compass is designed and developed to use a pendulous system to improve horizontality.

<h3>What is a compass?</h3>

A compass can be defined as a scientific instrument that contains a magnetized pointer, which is used to show and indicate the following four (4) main cardinal directions:

North (N)

South (S)

West (W)

East (E)

Basically, the compass is a scientific instrument that's designed and developed to use a pendulous system to improve horizontality.

Read more on compass here: brainly.com/question/11165627

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Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is $5 N$

Tray makes an angle of $\theta =12^{\circ}$

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force $W_p=W\cos \theta$

$W_p=5\times \cos 12=4.89 N$

Along the Tray

$W_{along}=W\sin \theta$

$W_{along}=5\times \sin 12=1.04 N$

Thus 1.04 N is the magnitude of force that will cause Platter to slide down

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3 years ago

The loudness of a sound is the waves amplitude <br><br><br><br> True or false

kakasveta [241]

Answer:

true i think

Explanation:

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2 years ago

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Name and explain two sleeping disorders

vfiekz [6]

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3 years ago

4) Which of the following gases are typically used for colorful lighting when an electric current is applied ?

PIT_PIT [208]

The answer is B. Hope this Helps!

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How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

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As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

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