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Section 1

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Magnitude of Force with Vo

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6 years ago

Date created

Mar 1, 2020

Cards (33)

Section 1

(33 cards)

Magnitude of Force with Vo

Front

KE=Fd (KE=0.5mVf^2+0.5Vo^2)

Back

Static Friciton

Front

5 x mass

Back

Tension at bottom of a V

Front

T=(Fw/2sin(theta))

Back

Power solving for resistance

Front

R=(P/I^2)

Back

Object thrown downward

Front

xf=vot+(1/2)at^2

Back

Max height when thrown upward

Front

(Vo^2/a) + (1/2)Vo^2

Back

Fricitonless Track

Front

vf= √(vo^2+2ad)

Back

equilibrium temperature

Front

m1Ti+m2Ti=(m1+m2)Tf

Back

Distance across a River

Front

Dacross=(d/Vobject) x Vpush

Back

Parell Circuit looking for current

Front

(#total resistors/#total ohms)V=I

Back

Density Submerged

Front

Density times submerged percent

Back

Planet "g" density

Front

Fgp= 10 x #(earths radius)

Back

Top of building to ground

Front

building height + 5

Back

Force/Spring Constant

Front

(mg/k)=x

Back

Specific heat

Front

E=cm(change in temp)

Back

Acceleration with no final velcoity

Front

a=2xf/t^2

Back

Resistance (R)

Front

measured in ohms

Back

Escape Velocity

Front

(√(# x earths diameter))(Vescape)

Back

Current (I)

Front

measured in amps

Back

Elastic

Front

m1v1f=m2v2fh

Back

Weight on a Planet

Front

(# times earths mass/ # times earths radius ^2)Fg

Back

Object Dropped (with no final velocity and initial distance)

Front

xf = 5t^2

Back

Finding total power disputed by resistors in parallel circut

Front

P=(V^2/R)

Back

Meterstick

Front

x=((Mms50+MwXw)/(Mms+Mw)) drop zeros on weight

Back

Object fired up

Front

Vi/5

Back

Object accelerated (F)

Front

F=m(vf^2/2d)

Back

inelastic

Front

P1i=(m1+m2)vf

Back

Pulley

Front

a=((m1-m2)g/(m1+m2))

Back

power (P)

Front

Meausred in watts

Back

Solving for vf when object is dropped

Front

vf= √(2ad)

Back

Eleavator Moving Up (scale reading)

Front

Fw+F=Fs (F=ma Fw=mg )

Back

Finding speed with a spring

Front

√((kx^2)/m)

Back

Projectile Motion w/ theta=45

Front

#-10

Back