Physics

Newtons 3rd law

For every action there is an equal and opposite re-action.

You're given components of velocity Vx=2.6m/s and Vy=-1.8m/s at t1=10.0s. For the interval t1=10s to t2=20s the average acceleration of the dog has a magnitude of 0.45m/s^2 and direction of 31* measured from the x-axis toward the y-axis. At t2=20s: What are the x and y components of the dogs velocity? What are the magnitude and direction of the dogs velocity? Sketch the velocity vectors at t1 and t2. how do they differ?

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A car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v=[5.00m/s-(0.0180m/s^3)t^2]i+[2.00m/s +(0.550m/s^2)t]j What are Ax(t) and Ay(t), the x and y components of the velocity of the car as a function of time? What are the magnitude and direction of the velocity of the car at t=8.00s? What are the magnitude and direction of the acceleration of the car at t=8.00s?

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A book slides off a horizontal table with a speed of 1.10m/s. It strikes the floor in 0.350s. Ignore air. Find: The height of the table above the floor. The horizontal distance from the edge of the table to the point where the book strikes the floor The horizontal and vertical components of the books velocity, and the magnitude and direction of its velocity just before it reaches the floor.

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Two crickets jump from the top of a vertical cliff. Chirpy just drops and reaches the ground in 3.50s, while Milada jumps horizontally with an initial speed of 95.0cm/s. How far from the base of the cliff will Milada hit the ground?

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The radius of the earths orbit around the sun ( assumed to be circular) is 1.50x10^8km, and the earth travels around this orbit in 365 days. What is the magnitude of the orbital velocity of the earth, in m/s? What is the radial acceleration of the earth toward the sun, in m/s^2? Repeat for mercury

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Given ug or mg find how many Newtons it weighs. Given N find out how many pounds it weighs and its mass in kg.

A) 1ug=1.0e^-9kg, 1mg=1.0e^-6kg, multiply these by g=9.81 to find out how many Newtons it weighs. B) 1N=0.224808943 lbF and 1N=0.10197162129

Given three force vectors and angles, find the x and y components of each of the three pulls. Use the components to find the magnitude and direction of the resultant of the three pulls.

A) Find Fx and Fy for each of the forces by using Sin&=Fx/F1 Cos&=Fy/F1, Sin&=Fx/F2 Cos& Cos&=Fy/F2, and so on. B) Combine all Fx factors for each pull to get Fnetx and the same for Fnety. Use those to find magnitude R=SqrRt(Rx^2 +Ry^2). Use &=Tan-1Ry/Rx to find direction.

A rhino is at the origin at t1=0. For the time interval from t1=0 to t2=12.0s the rhinos Average Velocity (Vav) has x component Vavx=-3.8m/s and y component Vavy=4.9m/s. At t2=12.0s: What are the x and y coordinates of rhino? How far is the rhino from the origin?

A) Given Vav-x= -3.8m/s and Vav-y=4.9m/s and t1=0 and t2=12.0s. Knowing Vav-x=x2-x1/t2-t1 and Vav-y=y2-y1/t2-t1, therefore /\x=Vav-x(t2-t1) and /\y=Vav-y(t2-t1) Plug in amounts to find /\x and /\y those are x and y coordinates B) Use r=SqrRt(/\x^2 +/\y^2) to find how far

An astronauts pack on earth is 17.5N but only 3.24N on an astroids surface. What is the acceleration due to gravity on this astroid? What is the mass of the pack on the astroid?

A) Given that W(earth)=17.5N and W(astroid)=3.24N. Plug in the W on earth and gravity on earth to W=mg then divide by gravity to find mass =W/g. Then plug in W on astroid and Mass you just found to find g=W/m. B) Same mass as on earth.

On jupiter the acceleration due to gravity is g=1.81m/s^2. A watermelon weighs 44N on earth. What is the watermelons mass on earths surface? What are its mass and weight on the surface of jupiter?

A) Know that W=mg. Given the W=44N and that g=9.81m/s^2. Plug in W and g for M=W/g to find the mass of the watermelon. B) Since the mass never changes its the same on both planets but when finding the weight you use the g=1.81m/s^2 of jupiter and mass to plug into W=mg.

Sprinters can accelerate nearly horizontal with a magnitude of 15m/s^2. How much horizontal force (F) must a 55-kg sprinter put on the starting blocks to produce this acceleration? What exerts the force that propels the sprinter: the blocks or sprinter herself?

A) Knowing that F=ma and giving A=15m/s^2 and m=55kg. Plug in to find the horizontal force (F). B) The blocks exert the force.

You are given x and y coordinates (1.1m, 3.4m) at t1=0 and (5.3m, -0.5m) at t2=3.0s. For the interval find: The components of the Average Velocity(Vavx and Vavy) The magnitude and the direction of the Average Velocity (Vav).

A) Plot and connect the points on a graph, connect in rectangle shape to form Vavx and Vavy. Find the components of the average velocity by finding Vavx and Vavy Vav-x=x2-x1/t2-t1 Vav-y=y2-y1/t2-t1 B) Plug in Vav-x and Vav-y to SqrRt function to find magnitude of Vav. Vav=SqrRt(Vavx^2 +Vavy^2) Use &=Tan-1Vavy/Vav-x to find the direction.

A dot has a position of r=[4.0cm+(2.5cm/s^2)t^2]i + (5.0cm/s)tj Find the magnitude and direction of the dots Average Velocity(Vav) between t1=0 and t2=2.0s Find the magnitude and direction of the dots instantaneous velocity at t=0, t=1, t=2 Sketch the dots and show trajectory.

A) Plugging into r: At t1=0 X1=4.ocm and y=o. At t2=2.0s X2=14cm and Y2=10cm. Knowing that Vav-x=X2-X1/T2-T1 and Vav-y=Y2-Y1/T2-T1, find both Vav-x and Vav-y. Then plug these into R=SqrRt(Vav-x^2 +Vav-y^2) to find the magnitude and use &=Tan-1Vav-y/Vav-x for direction B) Find derivative of r: r=(5.0cm/s^2)ti+(5.0cm/s)j Plug in for t to find x and y coordinates. Plug in x and y coordinates to R=SqrRt(x^2 +y^2) for each time to find magnitude and use tan-1 to find direction. C) Plot the x and y coordinates on a graph and label their times and velocities.

A dot has a position of r= [(0.280m/s)t+(0.0360m/s^2)t^2]i +(0.0190m/s^3)t^3j What are the Vx(t) and Vy(t), x and y components of the velocity of the dot, as a function of time? At t=5.00s how far is the dot from initial position? At t=5.00s what are the magnitude and direction of the dots velocity?

A) The Vx(t) And Vy(t) are simply the x and y components of the derivative of r. B) Plugging in t=5.00s for the original equation r gives you an x and a y which are then plugged in to R=SqrRt(x^2+y^2) to find magnitude or how far from initial position. C)Plugging t=5.00s to both Vx(t) and Vy(t) will give you a Vx and and Vy which are then plugged into V=SqrRt(Vx^2+Vy^2) to find the magnitude and Tan-1 to find the direction of velocity.

Newtons 2nd law

Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).

Newtons 1st law

An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

A chair has a mass of 12.0kg sitting on a floor. You push on the chair with a F=40N that is directed 37* below the horizontal. Draw a clearly labeled free-body diagram for the chair Use this and Newtons laws to calculate the Normal Force the floor exerts on the chair.

Draw on a normal x and y axis Normal Force NF=mg+FSin& NF=12.0kg(9.81)+(40N)Sin37* NF=142N

A small car( 380kg) is pushing a truck(900kg). The car exerts a horizontal force of 12o0N on the truck. What is the magnitude of the force that the truck exerts on the car?

Due to Newtons 3rd law F1=F2, therefore the magnitude of the force the truck exerts on the car is the same (1200N).

Given a box at rest on a frozen pond. A guy applies a horizontal force with a magnitude of 48N and an acceleration of magnitude of 3.00m/s^2, what is the mass of the box?

Given F=48N and a=3.00m/s^2. Know that F=ma. Divide A in order to solve for m=F/a. Giving you the mass.

A block of ice is given a horizontal force of 80N. The block starts from rest and moves 11.0m in 5.00s. Find the mass of the block of ice. If the worker stops pushing after 5.00s how far does the block move in next 5.00s?

Given F=80N, X1=11.0m, T1=5.00s, T2=5.00s. Knowing the formulas F=ma, /\x=V0t+1/2at^2, and V=V0+at. Knowing that the distance x traveled /\x=11.0m and the V0 (initial velocity)=0 since it starts at rest. Plug in 11.0m=(0)t+1/2at^2 to give you 11.0m=1/2at^2. Given t=5.00s, get a by its self and plug in for t a=2(11.0m)/(5.00s)^2. Once you find A plug in m=F/a to find the mass of box. Given that the box stops accelerating after 5.00s A2=0. Know that /\x=V2t+1/2at^2 and plug in A=0. Giving you /\x=V2t. Knowing V2=V0+a, plug in V0=0. Giving you V2=at, using the acceleration above plug in to find V2. Once V2 is found plug into /\x=V2t to find distance traveled by x (/\x).

A plane is flying at a constant altitude. At t1=0 it has components of velocity Vx=90m/s and Vy=110m/s. At t2=30.0s the components are Vx=-170 and Vy=40m/s. Sketch the velocity vectors at t1 an t2, how do they differ? Calculate the components of the average acceleration and the magnitude and direction of the average acceleration.

Given t1=0 t2=30s and Vx1=90m/s, Vy1=110m/s and Vx2=-170m/s,Vy2=40m/s A) Sketch the vectors on a graph in their quadrants based on their x and y values. B) The components of Average Acceleration are Aav-x and Aav-y, where Aav-x=vx2-vx1/t2-t1 and Aav-y=v2y-v2x/t2-t1. C) Use Aav=SqrRt(Aav-x^2+Aav-y^2) to find magnitude and tan-1 to find degrees. REMEMBER to put Aav vector in right quadrant then set the degrees. 15 +180

Given a net upward force of 5.00N on a patients chin, the angle between forces, and that the tension is the same throughout the strap. To what tension must the strap be adjusted to provide the necessary upward force?

Know that F1=F2 because both forces must be equal. Plot on graph and discover that the sum of Fx=0 due to the fact that the forces are the same both ways. Notice that the sum of forces upward (y direction) was given as 5.00N. Use Sin&=Fy/F1 using 5.00N as F1y to find F1. Once F1 is found divide by two to split forces.

A 2400N boulder is thrown at an enemy. What horizontal force must the person apply to the boulder to give it a horizontal acceleration of 12.0.m/s^2?

Knowing that F=ma and that g=9.81m/s^2 since the bolder is being pulled down by gravity. Given A=12.0m/s^2 and M=2400N/9.81m/s^2=244.65kg. Plug in M and A into F=ma to find the horizontal Force.

Two forces pull horizontally on ropes attached to post. Given the angle between the forces is 60* and that Force A=270N and Force B=300N. Find the magnitude of the resultant force and he angle it makes with dog A's rope.

Plot F1 on the X axis and F2 60* above it in the 1stQaudrant. Know that F1 only has an F1x not an F1y since its on the x axis, F1x given as 270N and F1y=0. Use Cos60 and Sin60 to find F2x and F2y. Find the sum of Fx and the sum of Fy to put into SqrRt formula to find magnitude of the resultant force. Use &=Tan-1Ry/Rx to find direction(angle).

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Given a 68.5kg skater moving 2.40m/s on ice comes to rest in 3.52s due to friction. What force does friction exert on the skater?

You're given M=68.5kg, V0=2.4m/s, T=3.52s. Know formula F=ma and that Vf=Vi+at therefore a=Vf-Vi/t. Once you find A, plug both A and M into the F=ma formula to find the force. Notice its opposite to the motion of the skater. AbsltValue