Please explain how you get the answers and what equations you are using.. Willing to pay fg if necessary. Using these equations.





A car accelerates from rest at a constant rate of 2.3m/s2 for 6.0s .

Part A
What is the speed of the car at the end of that time?
Express your answer using two significant figures.

Part B
How far does the car travel in this time?
Express your answer using two significant figures.

Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 m/s2 .
A loaded Boeing 747 jumbo jet has a mass of 2.0×105kg .

Part A
What net force is required to give the plane an acceleration of 4.0m/s2 down the runway for takeoffs?
Express your answer using two significant figures.


Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 m/s2 .
A gun is fired and a 52g bullet is accelerated to a muzzle speed of 100m/s .

Part A
If the length of the gun barrel is 0.60m , what is the magnitude of the accelerating force? (Assume the acceleration to be constant.)
Express your answer using two significant figures.


An ice skater of mass 63kg is skating toward a child of mass 21kg . The adult skater's initial velocity is 0.3m/s . The child skater is initially at rest.

Part A
What is the momentum of the system?

Part B
The grown-up skater picks up the child and keeps sliding forward. To conserve momentum, what must be the skater's final velocity? (Remember, his new mass is adult mass plus child mass.)

child skater isn't moving yet so the only contribution to the system before the collision comes from the adult

p = mv
= 63*0.3
= 18.9 kg *m/s

momentum is conserved
so

before he picks the kid up-------------------------------------after he picks the kid up
they "collide" when he picks the kid up

momentum initial = momentum final

m1v1 + m2v2 = m1v1
63*0.3 + 21(0) = (63+21)*v
v = 18.9 / (63+21)
v = 0.225 m/s

car problem:
A)
Vf= V0 + a*t

Vf = 0 + 2.3*6
Vf = 14 m/s

B.)
Xf = X0 + V0*t + (1/2)*a*t^2

Xf = 0 + 0 + (1/2)*2.3*6^2
Xf=41 m

jet problem:
A)
F=ma

F = 2.0*10^5*4
F = 8.0*10^5 N

This post was edited by Azrad on Nov 12 2013 04:28am

bullet problem:
A)
Vf^2 = V0^2 + 2*a*(Xf-X0)

100^2 = 0^2 + 2*a*(0.6-0)
10000 = 2*a*(3/5)
10000 = (6/5)a
a=(5/6)*10000

F=ma
F=(0.052)*(5/6)*10000
F=(13/250)*(5/6)*10000
F=(1300/3)
F=430N

This post was edited by Azrad on Nov 12 2013 04:28am

skater (solutions consistent with nlin)
A)
p0 = m*v
p0=63*0.3
p0 = 19 N s

B.)
p0 = p1
p1 = (63+21)*v
19 = (63+21)*v
19 = (84)*v
v = 19/84
v = 0.23 m/s

azrad is one smart dewd

whew

This post was edited by nlin on Nov 12 2013 02:57pm

A railway car with mass 2200kg is rolling at an initial velocity of 1.7m/s . The car will run into spring with constant 850N/m . The car will compress the spring by some distance.

Part A
How much kinetic energy does the car have initially?

Part B
How much potential energy does the spring have initially?

Part C
Add parts A and B to get the system total energy. How much total energy does the system have?

Part D
When the spring is compressed 1.5m, how much potential energy does the spring have?

Part E
Subtract D from C to tell how much kinetic energy the car has left.

Part F
When the spring is compressed 1.5m, what is the velocity of the car? Use the car kinetic energy from the previous part.

KE = 1/2 mv^2
= 1/2 (2200)1.7^2
= 3179

PEspring = 1/2kx^2
where k is the spring constant
x is the amount of compression

initially the spring has 0 PE because the car hasn't compressed the spring yet
the spring is in its equilibrium position because its not stretched or compressed yet

C) 3179+0 =

D) PE spring = 1/2kx^2
= 1/2(850)1.5^2
=956.25

E) C-D

3179 - 956.25
2222.75

not sure about F