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mestny [16]
3 years ago
13

Three masses A, B & C revolve in the same plane about a centre O in the plane. The

Physics
1 answer:
zhenek [66]3 years ago
4 0

Answer:

Place A along the x-axis, B along the y-axis, and then C is 30 deg below the negative x-axis.

A = Cx to balance forces on the x-axis

Cx = -25 cos 30 = -21.65 kN

A = Ax = 21.61 kN     balances forces in the x-direction

Likewise

Cy = -25 sin 30 = -12.5 kN

B = By = 12.5 kN balances forces in the y-direction

The length of the radii are not important (consider a force table)

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A 1kg mass is thrown to a height of 2cm. what is the potential energy​
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The number of molecules in a container is tripled and the kelvin temperature doubled. the volume remains unchanged. the new pres
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Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c
Neporo4naja [7]

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

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y=-2x+(1/2)  is the first-degree polynomial.

First-degree polynomials are the simplest polynomials. Here, we'll talk about a few qualities and connect the terms polynomial, function, and equation. Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations' answers are the solutions to the derived equations. Factoring cannot always be used to solve polynomial equations. For instance, the polynomial 2x+5 has an exponent of 1. The most typical kinds of polynomials used in algebra and precalculus are zero polynomial functions.

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2 years ago
A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Marrrta [24]

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

The problem can be solved for part A, through the Work Theorem that says the following,

W = \Delta KE

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

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F = \frac {mv ^ 2} {2d}

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