
Charles asks…
Mechanical Advantage : Science physics; Help pleaseee neeed Asap give you 20 points or more for the best answr?
im doing physics on mechanical advantage really need your help sooooooooo bad!
1. Calculate the mechanical advantage for your small gear wheel.
( you can do it on any of your type of wheels or a toy car)
2.Calculate the mechanical advantage of your medium size gear wheel.
3. Calculate the mechanical advantage of your large gear wheel.
4. Calculate the work done in pushing a car a distance of 10m. When moving slowly the car has a resisting friction force of 200N.
a..2J
b 20J
c. 200J
d. 2000J
5. If there were no resisting frictional force, which answer do you predict would be closest to the amount of work required to move a car 10 m?
a. 2j
b. 20j
c. 200j
d. 2000j
challenge:
6. An incline plane is used to lift a 100 kg barrel onto a drinks bar at a height of 1m. The plane is 5m long. How much force is required to roll the barrel up the plane?
Hints Draw a diagram of the situation. Calculate the force required to lift the barrel straight up without any plane. This info is found in lesson 3, weight force. Use the information in lesson 5 on inclined planes to calculate the mechanical advantage. Divide the force straight up by the mechanical advantage factor to find the force to roll up the plane.
a. 20N
b. 980N
c. 196N
d. 4900N
7. The baby weighs 15kg and the elephant weighs 3T. How many pulleys and ropes are required for the baby to lift the elephant?
a. 2000 ropes & pulleys
b. 200 ropes & pulleys
c. 20 ropes & pulleys
d. 1000 ropes & pulleys
8. strong meter rule is used as a lever and the fulcrum is placed under the 90cm mark. What mass would need to be placed on the long end of the lever to lift a 1kg mass on the short end?
a. 111g
b. 9kg
c. 1kg
d. 12g
( i know this is alot , if you give me a really good answer ill giev you like 20 points.)
please help really appreciated.:).

Abest Energy answers:
G’day!!
You are unable to give 20 points even if you wanted to!
1,2 and 3 are impossible to do… I know nothing about your wheels but MA = Load/effort.
4) Work = force x distance moved = 200N x 10m = 2000J.
5) 2J… It’s like pushing it on ice.
6) Are you sure you mean ‘force’ not work?
100kg = 100 x 9.8N = 980N( that’s the force)
Work = 980N x 1m = 980J ( up against gravity)
7) Need a diagram of the pulley system
90M = 1000 x 10
M = 10000/90
M = 111g

Robert asks…
Why don’t people care about the American Car Industry?
Everywhere I look, I keep hearing people talk about how the government should just let Ford, GM, and Chrysler die. My question is why would anyone ever wish for that to happen?
I’m a big fan of Capitalism, I adhere strongly to the belief that companies should be held accountable, however, in the situation of GM, Ford, and Chrysler, I honestly think the government needs to step in. The government needs to protect itself from foreign influences and it needs to protect its key core businesses from international takeover.
We all know that the Japanese government helps out its car industry like Toyota, Nissan, Mitsubishi, they have very good government backing therefore they are very strong. For instance, you would never see a foreign company being allowed to buyout Toyota Motor Company even if that hypothetical company could afford it, why? Because the Japanese government would never allow it, just like France would never allow Boeing to buyout Airbus.
This lead to an unfair marketplace for the American Auto Industry. Now, had our government stepped in and supported our Auto Industry, I think the industry as a whole would look so much different today. Unfortunately it didn’t and GM, Ford, and Chrysler were forced to focus on SUV’s because that’s what sold during the 90′s, and unlike Toyota, GM, Ford, and Chrysler, did not have the deep pockets to fully support next-gen passenger car program as well as a money-making SUV program. And as you know, money is necessary to maintain future operations, so GM was forced to devote most of its resources to making the cars people wanted to drive at the time.
Had the Big 3 had government support/aid during the 90′s perhaps we would have gotten a vehicle like the 2008 Malibu and CTS much earlier. However, if GM had decided to focus it’s attention on the cars that it needed rather than the cars that people wanted, it would have quickly gone bankrupt.
And right now as we speak rumors of bankruptcy are heard throughout the market place, as well as rumors of the end of the Big 3.
This would be one of the worst things to happen to the American Economy during a good economic period. If it happens now in our current economic state, it will be disastrous.
GM directly employes what, about 200,000+ (maybe less since all the recent layoffs) and I know Ford and Chrysler employ quite a few workers as well, but let’s focus on GM.
So if GM dies, that’s 200,000 for the unemployment office. However, that number doesn’t include those that will become jobless as the result of the domino effect a GM bankruptcy will have on other companies.
GM has the most dealerships with it’s many different brands, if these were all to suddenly close, we can add more people to the unemployment line. In addition to dealerships, many part manufacturers have GM as a primary customer ( think American Axle, Allison, Eaton, Delphi, Bose, AC Delco, etc), then there are those companies that contract with GM for vehicle delivery,etc. And as a result of this sudden unemployment, forclosure rates could go up at an unprecedented rate as those no unemployed can no longer afford to stay in their houses.
A GM bankruptcy wouldn’t just affect GM workers and GM Fans, it would affect the jobs of a lot of people around the country. If you thought our current unemployment rate was high, a GM bankruptcy would make todays numbers look small.
In fact, a GM bankruptcy wouldn’t just affect American workers, but workers in Canada, Britain, China, India, all places where GM has factories.
People blame GM and other companies for building factories in other countries and bringing those cars here, well, who can blame them? With UAW fighting for $25+per hour wages for High School educated people, I’d move my factory to a place where I can pay the people what they truly deserve to be payed rather than what some labor union tells them that they are entitled to be payed.
The government really needs to step back and take a look at the bigger picture regarding the Big 3. A decision regarding a “bailout” of the Big 3 will either make the presidency of whoever wins in November, not to mention it may make or break the future of our country as an economic/world power.
Perhaps the old say really is true, “What’s good for General Motors, is good for America”.

Abest Energy answers:
I’ve read only half of ur question….. But anyway, let me tell you what I think.
GM killed the electric car! The electric was to become a success!! GM gives a sh*t 4 z environment! He’s just concerned about making profits(just like any other company). So, why should I care for a company that had the opportunity to do something for the community, BUT did not do it!?!?!

Thomas asks…
Physics force problem? need help?
In a front-end collision, a 1200 kg car with shock-absorbing bumpers can withstand a maximum force of 65 kN before damage occurs.
If the maximum speed for a nondamaging collision is 9.0 km/h, by how much must the bumper be able to move relative to the car?
distance in meters

Abest Energy answers:
Equation to use:
2*a*x = (vi)² – (vf)²
Solve for x:
x = ((vi)² – (vf)²)/(2*a)
Then we relate F, m, and a:
where F=65kN, m=1200kg, a =?
F = ma
a = F/m
a = 65kN/1200kg
a = 54.17 m/s²
Now plug into:
x = ((vi)² – (vf)²)/(2*a)
where vi = 9km/h = 2.5 m/s; vf = 0 m/s
x = ((2.5)² – (0))/(2*54.17)
x = 0.0577 m

Donald asks…
Need some help with a few Physics questions about force?
You are a crime scene investigator arriving at an automobile accident in a school zone. The scene contains vehicle skid marks that measure 25.0 meters long. Use your knowledge of physics to determine and report the estimated initial speed of the vehicle involved (in m/s) to the court official, so they will have the crucial evidence to charge the driver for speeding above the 20 mph limit (~ 12 m/s). Assume a deceleration rate of 11.0 m/s2 for dry pavement. (I do not even know what formula to use to this problem.)
Frictional forces from brake pads are responsible for stopping an automobile. A car is moving at 35.0 m/s. If the car‘s brakes are capable of generating a deceleration rate of 6.00 m/s2, then how many meters will this vehicle move before coming to a stop?
Name the forces, their directions, and the distances involved in the following problem. Do not solve it mathematically.
A car is traveling 10 m/s when the driver sees a dog in the road. The driver takes 0.7 s to react, then steps on the brakes and slows down at 3.0 m/s2. How far does the car go before it stops?
Thank you so much
:D
The third and forth paragraphs are one question.

Abest Energy answers:
Deceleration = (original velocity – final velocity / time
But in this problem, we are not given time.
Instead, we are given stopping distance.
So we use the formula
time = distance / velocity
or
time = stopping distance / original velocity
This allows us to change
deceleration = (original velocity – final velocity / (time)
into
deceleration = (original velocity – final velocity / (distance / original velocity)
We are given
deceleration = 11 m/s^2
final velocity = 0
stopping distance = 25 m
So,
deceleration = (original velocity – final velocity / (distance / original velocity)
11 m/s^2 = (original velocity – 0) / (25 m / original velocity)
11 = (original velocity) / (25 / original velocity)
11 = (original velocity)(original velocity) / 25
11 * 25 = (original velocity)^2
275 = (original velocity)^2
16.6 m/s = original velocity
Next we convert 16.6 m/s into miles per hour.
16.6 m/s * 60 seconds/minute * 60 minutes/hour * 3.28 feet/meter * 1/5280 mile/feet = 37.1 mph
Since 37.1 – 20 = 17.1, this proves that the car was going more than 17 mph over the speed limit.
===============================================================
deceleration = (original velocity – final velocity / time
But in this problem, we are not given time.
Instead, we are given stopping distance.
So we use the formula
time = distance / velocity
or
time = stopping distance / original velocity
This allows us to change
deceleration = (original velocity – final velocity / (time)
into
deceleration = (original velocity – final velocity / (distance / original velocity)
We are given
deceleration = 6 m/s^2
original velocity = 35 m/s
final velocity = 0
stopping distance = x
So,
deceleration = (original velocity – final velocity / (distance / original velocity)
6 m/s^2 = (35 – 0) / (x / 35)
You should be able to solve for x which is the distance to stop.
========================================================
I’m confused about the last problem (last two paragraphs). It asks for the distances involved, but there are no distances given and it indicates not to solve it mathematically.

George asks…
Physics help needed urgently?
Tomorrow is my physics UT and I am so depressed. Although I have been able to solve maximum number of questions, yet I am unable to solve the following questions from motion in a straight line. Please solve them with explanation and also give me tips on how to solve such questions ( Note: I have written the correct answers that were given at the back as (A)) :-
1. The two ends of a train moving with constant acceleration pass a certain point with velocities u and 3u. The velocity with which the middle point of the train passes the same point is :
Root 5 *u (A)
2. The initial velocity of a particle is u (at t = 0) and the acceleration a is given by Bt^3/2. Which of the following relations is valid?
v = u + 2/5Bt^5/2 (A)
3. A train starts from rest from a station with acceleration 0.2 m/s^2 on a straight track and then comes to rest on another station due to retardation 0.4 m/s^2 after attaining maximum speed in half an hour. The distance between the two stations (neglecting the length of the train) is :-
216 km (A)
4. The position x of a particle moving along x-axis varies with time t as x = Asin (ct) where A and c are positive constants. The acceleration a of the particle varies with its position (x) as
a = -c^2x (A)
5. A particle moves in a straight line and its position x at time t is given by x^2 = 2 + t. Its acceleration is given by
-2/4x^3 (A)
6. A particle starts moving from rest state along a straight line under the action of a constant force and travels distance x in the first 5 seconds. The distance travelled by the particle in the next 5 seconds is
3x (A)
7. A train of 150 m length is going towards north at a speed of 10 m/s. A bird is flying parallel to the track towards south. The time taken by the bird to cross the train is :
10 s (A)
8. Two cars are moving in the same direction with a speed of 30 km/h. They are separated from each other by 5 km. Third car moving in the opposite direction meets the two cars after an interval of 4 minutes. The speed of the 3rd car is :-
45 km/h (A)
9. Two cars A and B are moving in the same direction with velocities 30 m/s and 20 m/s. When car A is at a distance d behind car B, the driver of the car applies brakes producing uniform retardation of 2 m/s^2. There will be no collision when :-
d>25 m (A)
10. Two trains each of length 100 m are moving parallel towards each other at a speed of 72 km/h and 36 km/h respectively. In how much time will they cross each other ?
6.67 seconds (A)

Abest Energy answers:
2. V_0 = u, a = Bt^3/2, v(t) = at + v_0 = Bt^5/2 + u.
But it also has to be true that, dv/dt = a = Bt^3/2, which can only be true if,
v(t) = u + (2/5)Bt^5/2, as dv/dt = Bt^3/2.
4. X(t) = A sin (ct), d^2x/dt^2 = – c^2A sin (ct) = -c^2 x(t).
6. X(t) = ct^2, x(5) = x = 25c. X(10) – x(5) = distance travelled in the next 5 seconds = 100c – 25c = 75c = 3x.
8. This is the same thing as a car moving at 30 +x km/h travelling 5 km in 240 seconds. So,
(30 + x)(1/15 h) = 5 km, 30 + x = 75 km, x = 45.
So the car moves at 45 km/h.
Some tips are to remember to use the functions x(t), v(t) and a(t), and to understand that dx/dt = v, dv/dt = a, etc. Also, when you have a problem that involves two cars moving towards one another t velocties A and B, this is the same thing as one car moving towards a stationary target at velocity A+B.
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