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2 years ago

What do you call the process of finding the magnitude of fy

Physics

2 answers:

vlada-n [284]2 years ago

4 0

Answer:

The process of finding the magnitude of Fy is called vector resolution.

Explanation:

Given a vector with a magnitude equals to F which forms an angle equals to θ with the x-axis, then it can be decomposed into two smaller vectors (Fx and Fy) as follows:

Fx = F*cosθ (placed on the x-axis)

Fy = F*sinθ (placed on the y-axis)

This process is called vector resolution.

MAVERICK [17]2 years ago

3 0

Answer - vector resolution

Any vector suppose f if lies at some angle, then

fy = F x cos or sin (theta)

and fx = F x cos or sin (theta)

Process of finding the value of component of any vector quantity is called Vector Resolution.

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Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is

Ivahew [28]

Answer:

4 capacitors

Explanation:

Given

$n = 5$ --- conducting plates

Required

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This is calculated as:

$c = n - 1$

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2 years ago

A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the centripetal acceleration of the truck?

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Answer:

B. 25 m/s/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius of curvature.

a = v² / r

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a = (10 m/s)² / 4 m

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2 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

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2 years ago

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