**help Needed**mechanics

Mechanics help?

Hi.
I don't need help solving this problem, there's only one thing I don't understand.

When maiking a free-body diagram of the 10-kg weight, I ended up with 10g - T = 10a.
But accoridng to my teacher this should be 10g - T = 10*2a, since it is 2a for the weight. The only explanation he gives is that a = a₀. Where a₀ is the acceleration of the center of the cylinder.

It did solve the problem. But why is the accerlation twice for the 10-kg weight. Don't they accelerate at the same rate?

Why is it twice as much??

Physics and Mechanics Help Needed?

1. I do not know what is a fortnight

2. with sig digits specially when adding just copy.
so add as usual,1.91 +/- 0.07m. The trick is when you are multiplying or dividing.

3. mass of water is just subtract. 0.094 +/- 0.004 kg.

4. Multiply for A = 0.17*.11=.0187 m(2). now to figure out the uncertainty, use this equation delxy= xy sqrt( (delx/x)(2) + (dely/y)(2)

delxy = uncertainty
x=0.17
y=0.11
delx= 0.2
dely=0.2
substitute
delxy=.004 m(2)

answer for area
area = .0187 +/-.004 m(2)


Volume = 0.17*.011*.00094=.0000173 m(3)

delxyz= xyz sqrt( (delx/x)(2) + (dely/y)(2) +(delz/z)(2))

delxy = uncertainty
x=0.17
y=0.11
z=.0094
delx= 0.2
dely=0.2
delz=.00001

V=.0000173 +/- .0000038 m(3)

1) F=ma

v=v0-at
rearrage this equation gives a=(v-v0)/t substitute into F=ma

F= m(v-v0/t)

m=5kg
v0=7m/s
v=3 m/s
t=3s

F=6.66 N

2. Tricky problem

two equations
y=y0+v0yt-0.5gt(2)
x=x0+v0xt-0.5gt(2)
no vertical velocity means v0x=0
y0=0
x0=0
reduces too
g=9.8 ms(-2)
y=v0yt-0.5gt(2)= 20*3-0.5*9.8*3(2)=15.9m
x=-0.5gt(2)=44.1m

Have to go to work sorry...

Fluid mechanics help needed?

A solid iron cylinder with a diameter of 150 mm, a height of 1 m and a density of 7860 kg/m^3 is suspended in fresh water. Calculate the tension in the supporting cable when two thirds of the volume of the cylinder is submerged

Need Mechanics help why wont my car start?

Muffler bearings cold be worn out.
Check you squelch fluid.
Last but not least, rotate the air in your tires.

Just for fun:
"Won't start" is pretty vague. It could be hundreds of posibilities.
Dead battery, Low battery, bad ground wire to engine, bad power wire to starter, bad power wire to battery, bad connection at battery post or terminals, blown fuse, bad key switch, bad starter, bad fuel pump, bad injectors, out of gas, neutral safety switch, seized engine, broken connecting rod, no compression, no spark, bad oil pressure switch, broken timing belt, plugged fuel filter, bad starter solenoid, broken rotor, cracked distributor cap, Anti-theft device, water in/on engine, vacuum leak, bad ignition coil, blown computer, faulty cam or crank sensor ...

When you turn the key to START it what happens?
1) Nothing happens? (dead battery?, loose battery cables, or it's not in "park" all the way.)
2) You hear a "click", but nothing else? (loose battery cables, bad starter, bad starter solenoid, or broken engine)
3) You hear the starter turning, but the engine doesn't turn? (low battery voltage, bad starter or bad ring gear on flywheel/flex plate)
4) The starter turns the engine, but the engine won't start up? (out of gas, bad fuel pump, plugged fuel filter, broken timing belt, no spark)
5) The engine turns over, but very slowly. (weak battery, worn starter, out of engine oil)
6) the engine turns, but there's an awful banging noise? (something is broken inside the engine)

What happened last time you drove it?

1) Drove home fine last night, just won't start today?
2) Was driving fine, and it just died on you?
3) You haven't driven it in a while. It's been parked in storage?

Need Help with Fluid Mechanics problem?

Fluid Mechanics, Munson, Problem 2-75.

A concrete (S.W = 150lb/ft^3) seawall has a curved surface and restrains seawater at a depth of 24ft. The trace of the surface is a parabola (y = 0.2x^2). Determine the moment of the fluid force (Per unit length) with respect to an axis through the toe A. * The base is 15ft.

_______________|----|
| / |
24' Sea / |
| Water / |
| / |
| / |
|___________/_____ |.A

Physics/ mechanics question, help needed.?

A puck of 0.2kg is placed on a smooth slope at 20 degrees to the horizontal and hit up the slope with an initial velocity of 4ms.

The force that causes the puck to slide DOWN the slope is the force parallel to the slope.
F = m * a
F parallel = mass * g * sin θ
Acceleration = Force ÷ mass = (mass * g * sin θ) ÷ mass
Acceleration = 9.8 * sin 20° = 3.25 m/s^2 DOWN

The puck accelerates at 3.25 m/s^2 DOWN the slope, as it moves UP the slope. So the velocity of the puck decreases 3.25 m/s each second as it moves UP the slope. That’s why the answer is negative.

The initial velocity does not affect the acceleration, but it does affect the time required to stop.

Friction force = μ * m * g * cos θ
The friction force also causes the velocity of the puck to decrease as it slides UP the slope.
Total force decelerating the puck = m * g * sin θ + μ * m * g * cos θ
Total force decelerating the puck = m * g * (sin θ + μ * cos θ)
Total force = mass * acceleration
m * g * (sin θ + μ * cos θ) = m * a
a = g * (sin θ + μ * cos θ) = 9.8 * (sin 20 + 0.5 * cos 20) = 7.96 m/s^2
Of course, the acceleration is negative, since both forces cause the velocity to decrease.
a = -7.96 m/s^2

Forces which are in the opposite direction of the initial velocity should be considered negative.
The direction of Force parallel is always DOWN the slope.
Since the initial velocity is 4 m/s UP the slope, the direction of the friction force is DOWN the slope!

The initial velocity does not affect the acceleration, but it does affect the time required to stop.
Final velocity = Initial velocity + acceleration * time
0 = 4 + (-7.96) * t
-4 = (-7.96) * t

t = -4/-7.96 = 0.5025 seconds
Distance = average velocity * time
average velocity = ½ * (4+ 0) = 2
Distance = 2 * 0.5025 = 1.005 meters

Mechanics Questions. Help Needed Desperately!?

1.
Let α be the angle between the inclined plane and the horizontal (tanα = 3/4)
The motion is decelerated by the component -mgsinα of the gravity force parallel to the plane, and the friction f = -μN = -μmgsinα, where μ is the friction coefficient and N = mgcosα is the component of gravity force normal to the plan.
The acceleration of the motion is A =(f + N)/m = -(sinα + μcosα)g
For uniformly decelerated motion, we have the following relation between velocity and displacement
v² - v0² = 2A.d
d = (v² - v0²) / 2A
d = {v² - [2 sqrt(2ga)]²} / -2(sinα + μcosα)g

distance at which v = 0 is
d = 0 - [2 sqrt(2ga)]² / -2(sinα + μcosα)g
d = - (4*2ga) / -2(sinα + μcosα)g
d = 8a / 2(sinα + μcosα)
d = 4a / (sinα + μcosα)

with tanα = 3/4 and μ = 1/4, we get:
cos²α = 1 / (1 + tan²α) = 1 / [1 + (3/4)²] = 16 / 25 ==> cosα = 4/5
sin²α = tan²α / (1 + tan²α) = (3/4)² / [1 + (3/4)²] = 9 / 25 ==> sinα = 3/5

d = 4a / [3/5 + (1/4)(4/5)] = 4a / (3/5 + 1/5)
d = 4a / (4/5)
d = 5a

2.
Equation for the displacement:
d = v0.t + (1/2)A t² = [v0 + (1/2)At] t
For d = 0, t = 0 or t = -2v0 / A

The particle come back to O(0) at
t = -2v0 / A
t = -4sqrt(2ga) / -(sinα + μcosα)g
t = 4.sqrt(2a) / (sinα + μcosα).sqrt(g)
t = 4.sqrt(2a) / (3/5 + 1/4*4/5).sqrt(g)
t = 4.sqrt(2a)/[(4/5)sqrt(g)]
t = 5.sqrt(2a/g)
(this is different from the given answer by a plus sign, please double check)

Fluid mechanics help needed, please read the question carefully.?

A pump is used to deliver water from a large reservoir to another large reservoir that is 20 m higher. The inlet of the pipe is submerged in the one reservoir and the exit is just above the surface of the other reservoir.

If the friction head loss in the 200-mm-diameter, 4-km-long pipeline is 2.5 m for every 500 m of pipe length, determine the required power output of the pump so the flow is 1.0 m3/s.
Express your answer to three significant figures and include the appropriate units.