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Gala2k [10]
3 years ago
10

A=vf-vi/t solve for vi

Physics
1 answer:
amid [387]3 years ago
7 0
A =  vf - vi/t
A - vf = -vi/t  Multiply both sides by -1
-A + vf = vi/t
vf - A = vi/t    Cross multiply both sides by t.
tvf - At  = vi

vi = tvf - At

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A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t
kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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3 years ago
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find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo
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Vector with direction 18.4° and magnitude 0.250 has x and y components of:

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Answer:

2.61 atm

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Explanation:

P_1 = Presión inicial = 0.96 atm

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V_2 = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}

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