Monday 22 Nov:
Target: Review
Review Monday 22 Nov
Test L# 19 - 23 Tomorrow Tues 23 Nov
Simple Machine Project:
Requirements:
1. Individual
2. Build and its due Wed 24 Nov
(must be made and not just a store example)
2. Any size & any of the 6 types you want.
(as long as you can carry it by your self through the doorway)
3. One page paper explaining your simple machine and its use in History
. Bridge Contest:
-
Due by Friday Dec 17
|
Tower Contest:
-
Due by Friday Jan 7th
1) A 10g object is placed into 1000C boiling water and brought to equilibrium with it. The object is then placed into 50g of tap temperature water (200C) causing the tap water and the hot object to come to a equilibrium of 250C. What is the specific heat (cObj) of the object?
cwater = cal / 1g 0c
Qwater = mwater Dtwater cwater
= (50g)(50C)(1g/cal0) = 250Cal
Qobj = mobjDtobjcobj
250Cal = (10g)(750C)(cobj)
250Cal / 750g0C = 0.33 Cal/g0C
2. An object is dropped from a height of 50m on planet X (acceleration of gravity is 2.5 m/s2).
a) How long does it take to hit the ground?
b) What is the speed when it hits the ground?
2 a) xf = Vot + 1/2at2
-50m = 0 + 1/2(-2.5m/s2)t2
t = √(50m/1.25m/s2) = 6.33s
2b) Vf = Vo + at
Vf = 0 + (-2.5m/s2)(6.33s)
Vf = -15.8m/s
3. The object is thrown up at a speed of 50m/s on planet X (acceleration of gravity is 2.5 m/s2).
a) How long does it take to reach the apex?
b) What is the height it reaches?
3 a) Vf = V0 + at
0 = 50m/s + (-2.5m/s2)(t)
t = (-50m/s)/(-2.5m/s2)
t = 20.0s
3 b) xf = Vot + 1/2at2
xf = (50m/s)(20.0s) + 1/2(-2.5m/s2)(20.0s)2
xf = (1,000m) - (500m) = 500m
4. The x position in meters of a particle (along the x axis) is given by the equation (function of time):
Find the 1st (velocity)
& 2nd Derivative (acceleration) :
Ex: 5t4 + 4t2 + 7 where t = 3
1st Derivative-->
X(t) = 5t4 + 4t2 + 7
dx/dt = 4(5t3) + 2(4t1) + 0
dx/dt = 20t3 + 8t
= 20(27) + 8(3)
= 540 + 24 = 564 m/s
2nd Derivative (Acceleration) of
1st Derivative --> 20t3 + 8t
2nd: V(t) = 20t3 + 8t1
dV/dt = 3(20t2) + 1(8t0)
dV/dt = 60(32) + 8 = 548 m/s2
5. The velocity of a particle as a function of time is:
V(t) = 2t3 - 3t2 -2t + 40
Find the distance traveled in 4 seconds:
4= (2t3 - 3t2 -2t + 40)dt 0
= [(2t4) / 4] – [(3t3)/ 3] - [2t2 /2] + 40t
= [2(44)/ 4] – [3(43)/ 3] - [2(42)/2] + 40(4)
= [128] – [64] - [16] + 160
= 208m
6. A 10-kg object that was accelerated from 20m/s to 35 m/s in 3 seconds.
a) Find the D Momentum
b) Find the impulse
c) Find the force used
d) What is the final speed if the same force is continued to be applied for 6 additional seconds?
a) D Momentum = (m)(DV)
= (10-kg)(15m/s) = 150-kgm/s
b) Impulse = D Momentum = 150-kgm/s
c) Impulse = D Momentum
(f)(t) = D(mV)
(f) = (150-kgm/s) / 3s = 50N
d) Impulse = D Momentum
(f)(t) = (mDV)
(50N)(6s) = (10-kg)(Vf - Vo)
300Ns / (10-kg) = (Vf - 35m/s)
30 m/s + 35 m/s = Vf
Vf = 65 m/s
7. A ball (mass = 5-kg) and speed of 16 m/s collides head-on with a ball of mass 10-kg and speed of 4 m/s headed in the opposite direction. If the two balls stick together (inelastic collision), their speed after the collision is:
Momentum is conserved & it has direction:
m1V1i - m2V2i = (m1+ m2)V3f
(5-kg)(16m/s) - (10-kg)(4m/s) = (15kg)V3f
(80 kgm/s) - (40 kgm/s) = (15kg)V3f
V3f = 2.67m/s
8. Find center of gravity
Torque Problem (from lower left corner):
(x axis: Center of Gravity is CCWt)
CCWt = CWt + CWt
Fd = Fd + Fd
132x = (100)(5) + (32)(14)
132x = (500) + (448)
x = 948/132 = 7.18
(y axis: Center of Gravity is CWt)
CWt = CCWt + CCWt
Fd = Fd + Fd
132y = (100)(5) + (32)(8)
132y = (500) + (256)
y = 756/132 = 5.73