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2 years ago

The movement of an object at a constant speed around a circular radius is known as

Physics

1 answer:

forsale [732]2 years ago

8 0

Answer:

Uniform Circular Motion

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle.

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A fold is a _ in a rock , and a fault is a _in a rock?

777dan777 [17]

A geological fold occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.

So A fold is a Bend? in a rock. Maybe.

A fault is a planar fracture or discontinuity in a volume of rock, across which there has been significant displacement as a result of rock-mass movement.

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2 years ago

Read 2 more answers

Formula for weight on moon compared to weight on earth

Rzqust [24]

Weight on moon = (0.16) • Earth weight

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3 years ago

14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change

klio [65]

Lets se

And

$\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}$

$\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}$

$\\ \rm\Rrightarrow k\propto m$

If spring constant is doubled mass must be doubled

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2 years ago

Multiply the following three numbers and report your answer to the correct number of significant figures: 0.020cm x 50cm x 11.1c

Neko [114]

Answer:11.1

Explanation:

Three significant figures

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3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$