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a. How much work was done?

b. The same 600-newton person lifts his 100-newton carry-on bag upward a distance of 1 meter. They

travel another 10 meters by riding on the “people mover.” How much work was done in this situation?

2 months ago

Guest #9847817

2 months ago

In a), no work is done to move horizontally 'across" gravity. In b), he does (m) (g) (h) = (100)(9.8)(1) = 980 joules of work to lift the bag, and then no more work is done by the belt to move both of them horizontally.

Question

What is the displacement after 6 seconds when a car is moving at 12m/s and coasts up a hill with a uniform aacceleration of -1.6 m/s^2

Solution 1

The kinematic formula using displacement, time, initial velocity, and acceleration:

x = v t+.5 a t^2

x= (12m/s)(6s) + .5(-1.6m/s^2)(6s)^2

x= 72m - 28.8m

x = 43.2 m

x = v t+.5 a t^2

x= (12m/s)(6s) + .5(-1.6m/s^2)(6s)^2

x= 72m - 28.8m

x = 43.2 m

Question

How do you solve sleep deprivation? And i am not trolling!

Solution 1

The ONLY way to reverse the effects of sleep deprivation is with extra sleep, without drugs. I'm not sure what trolling means.

Question

If a dune Buggy is traveling for 20 seconds at a speed of 8.5m/s how do I calculate the unknown quantities

Solution 1

Right now I can think of about 50 unknown quantities, and the one you need to find is the most unknown of all of them, because you haven't even sait WHAT quantity you need to find.

Question

What are two things that the amount of gravitational force between two object depends on

Solution 1

1). the product of the two masses being gravitationally attracted to each other

2). the distance between their centers of mass

And that's IT. The gravitational force between them depends on

only those two things, nothing else.

2). the distance between their centers of mass

And that's IT. The gravitational force between them depends on

only those two things, nothing else.

Solution 2

The equation behind this question is:

g = (G*m*M) / (r^2)

g is the gravitational force

G is the Universal Gravitational Constant

M is the mass of one object

and m is the mass of the other constant

r is the distance between the two objects

g = (G*m*M) / (r^2)

g is the gravitational force

G is the Universal Gravitational Constant

M is the mass of one object

and m is the mass of the other constant

r is the distance between the two objects

Question

what is the mechanical energy of a falling apple with a kinetic of 5.2 joules and a potential energy of 3.5 joules

Solution 1

Mechanical energy in this problem is the sum of the kinetic energy and the potential energy: 5.2+3.5= 8.7 Joules

Question

How far will a free falling object have fallen from a position of rest when it's instantaneous speed is 10m/s?

Solution 1

Use kinematic equation:

vf² = vo² + 2g(xf - xo)

vf = final velocity = 10 m/s

vo = initial velocity = 0 m/s (i.e. at rest)

g = 9.81 m/s²

xf - xo = distance traveled = d

*rearrange equation to solve for d

(vf² - vo²)/(2g) = d

d = ((10 m/s)² - (0 m/s)²)/(2*9.81 m/s²)

d = (100 m²/s²)/(2*9.81 m/s²)

d = 5.097 m

vf² = vo² + 2g(xf - xo)

vf = final velocity = 10 m/s

vo = initial velocity = 0 m/s (i.e. at rest)

g = 9.81 m/s²

xf - xo = distance traveled = d

*rearrange equation to solve for d

(vf² - vo²)/(2g) = d

d = ((10 m/s)² - (0 m/s)²)/(2*9.81 m/s²)

d = (100 m²/s²)/(2*9.81 m/s²)

d = 5.097 m

Question

Gravity and friction are two kinds of

a.inertia

b.forces

c.masses

d.gravity

a.inertia

b.forces

c.masses

d.gravity

Solution 1

The answer is b, gravity and friction are forces.

Solution 2

**Answer: B forces Hope this helps!!**

Question

A fugitive tries to hop a freight train traveling at a constant speed Of 4.5 m/s. Just as a empty box car passes him, the fugitive starts from rest and accelerates at a= 3.6 m/s squared to his maximum speed of 8.0 m/s. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car?

Solution 1

We use the kinematics equation:

Vf = Vi + a*t

8 = 0 + 3.6 * t

t=2.222s to reach 8.0 m/s

At that time the train has moved

4.5 m/s * 2.222s = 9.999 m

He travelled (another kinematics equation)

Vf^2 = Vi^2 + (2*a*d)

(8.0)^2 = (0)^2 + (2 * 3.6 * d)

d=8.888 m

The train is 9.999m, the fugitive is 8.888m,

He still needs to travel

9.999-8.888= 1.111m

He needs to cover the rest of the distance in a smaller amount of time, however hes at his maximum velocity, so...

8m/s(man) - 4.5m/s(train) = 3.5 m/s more

(1.111m) / (3.5m/s) = .317seconds more to reach the train

So if it takes 2.222 seconds to approach the train at 8.888m, it should take

2.222 + .317 =2.529 seconds to reach the train completely

Last but not least is to figure out the total distance traveled in that time frame:

(Trains velocity) * (total time)

(4.5m/s)*(2.529s)=11.3805m

Vf = Vi + a*t

8 = 0 + 3.6 * t

t=2.222s to reach 8.0 m/s

At that time the train has moved

4.5 m/s * 2.222s = 9.999 m

He travelled (another kinematics equation)

Vf^2 = Vi^2 + (2*a*d)

(8.0)^2 = (0)^2 + (2 * 3.6 * d)

d=8.888 m

The train is 9.999m, the fugitive is 8.888m,

He still needs to travel

9.999-8.888= 1.111m

He needs to cover the rest of the distance in a smaller amount of time, however hes at his maximum velocity, so...

8m/s(man) - 4.5m/s(train) = 3.5 m/s more

(1.111m) / (3.5m/s) = .317seconds more to reach the train

So if it takes 2.222 seconds to approach the train at 8.888m, it should take

2.222 + .317 =2.529 seconds to reach the train completely

Last but not least is to figure out the total distance traveled in that time frame:

(Trains velocity) * (total time)

(4.5m/s)*(2.529s)=11.3805m

Question

How do you find displacement when all you have is time with no velocity, and then in order to find velocity you have to have displacement but you don't have that either?

Solution 1

If you really have nothing else but time, then you can't. There must be some other shred of information. Search around. Look under rocks.

Question

At a highway speeds, a particular automotive is capable of of an acceleration of about 1.6m/s^2. At this rate how long does it take to accelerate from 80km/h to 110km/h.

I know the formula would be v=v0+at. I just don't understand the converting. I would love an explanation on how to do this. I been stuck for a while.

I know the formula would be v=v0+at. I just don't understand the converting. I would love an explanation on how to do this. I been stuck for a while.

Solution 1

Fine. You just need a conversion at some point between seconds and hours.

Let's do it this way:

First, let's review one very useful definition of acceleration:

Acceleration = (change in speed) / (time for the change)

The question is asking for a time, so let's massage this equation to get

time all by itself.

Multiply each side by (time for the change) :

(Acceleration) x (time for the change) = (change in speed)

Divide each side by (acceleration) :

Time for the change = (change in speed) / (acceleration).

There it is. The problem gives you the change in speed and the acceleration.

Change in speed = (110 - 80) = 30 km/hr

Acceleration = 1.6 m/s²

Here's where the only difficulty shows up. One is in hours and the

other one is in seconds.

The way you change the units of a quantity is to multiply it by

a fraction whose value is ' 1 '. That way you don't change the value

of the original quantity.

How does that help ? You just pick the fraction very carefully ! Remember

that the value of a fraction is ' 1 ' if the numerator and denominator are both

the same value. Here are some fractions that are equal to ' 1 ' :

12 inches/1 foot 60 seconds/1 minute 1 meter/100 centimeters

See how that works ? Multiply one of these by a quantity, and it changes

the units of the quantity, but not its value.

Let's change 1.6 m/s² to m/hr² .

1 hour = 3,600 seconds so the fraction is (3600 sec/hr) .

The unit conversion is: (1.6 m/s²) x (3600 sec/hr) x (3600 sec/hr)

I multiplied by the fraction twice because we need sec² on top to cancel

the s² on the bottom.

Multiply this out and you get 20,736,000 m/hr².

Don't let the big number scare you.

Now go back to our equation for the time:

Time for the change = (change in speed) / (acceleration)

Change in speed = 30 km/hr

Acceleration was 1.6 m/sec² but we found that

that's the same as 20,736,000 m/hr² .

And also, since the change in speed is in km,

let's change this one to 20,736 km/hr² .

Time for the change = (30 km/hr) / (20,736 km/hr²)

30 / 20,736 = 0.0001447 hour

= 5.2 seconds.

I realize that this has turned into a nightmare. I probably lost you, and

the answer is likely wrong because the numbers got so ugly. I apologize.

The reason for all the ugly numbers is that I apparently picked the wrong

number to change the units. It probably would have been a lot neater to

convert the change in speed to m/sec instead of changing the acceleration

to m/hr² . But the principle is the same. Make a fraction out of the unit you

have and the new unit you want, wit the same value on top and bottom, and

use it to convert the units of one number to match the units of the others.

========================================

Change in speed = (30 km/hr) x (1000 m/km) x (1 hr / 3600 sec) = 8.333 m/sec

Time = (change in speed) / (acceleration) = (8.333 m/s) / (1.6 m/s²) =__5.2 seconds__

Yes, this way was certainly better and prettier.

But at least I got the same answer going the long tough way!

Let's do it this way:

First, let's review one very useful definition of acceleration:

Acceleration = (change in speed) / (time for the change)

The question is asking for a time, so let's massage this equation to get

time all by itself.

Multiply each side by (time for the change) :

(Acceleration) x (time for the change) = (change in speed)

Divide each side by (acceleration) :

Time for the change = (change in speed) / (acceleration).

There it is. The problem gives you the change in speed and the acceleration.

Change in speed = (110 - 80) = 30 km/hr

Acceleration = 1.6 m/s²

Here's where the only difficulty shows up. One is in hours and the

other one is in seconds.

The way you change the units of a quantity is to multiply it by

a fraction whose value is ' 1 '. That way you don't change the value

of the original quantity.

How does that help ? You just pick the fraction very carefully ! Remember

that the value of a fraction is ' 1 ' if the numerator and denominator are both

the same value. Here are some fractions that are equal to ' 1 ' :

12 inches/1 foot 60 seconds/1 minute 1 meter/100 centimeters

See how that works ? Multiply one of these by a quantity, and it changes

the units of the quantity, but not its value.

Let's change 1.6 m/s² to m/hr² .

1 hour = 3,600 seconds so the fraction is (3600 sec/hr) .

The unit conversion is: (1.6 m/s²) x (3600 sec/hr) x (3600 sec/hr)

I multiplied by the fraction twice because we need sec² on top to cancel

the s² on the bottom.

Multiply this out and you get 20,736,000 m/hr².

Don't let the big number scare you.

Now go back to our equation for the time:

Time for the change = (change in speed) / (acceleration)

Change in speed = 30 km/hr

Acceleration was 1.6 m/sec² but we found that

that's the same as 20,736,000 m/hr² .

And also, since the change in speed is in km,

let's change this one to 20,736 km/hr² .

Time for the change = (30 km/hr) / (20,736 km/hr²)

30 / 20,736 = 0.0001447 hour

= 5.2 seconds.

I realize that this has turned into a nightmare. I probably lost you, and

the answer is likely wrong because the numbers got so ugly. I apologize.

The reason for all the ugly numbers is that I apparently picked the wrong

number to change the units. It probably would have been a lot neater to

convert the change in speed to m/sec instead of changing the acceleration

to m/hr² . But the principle is the same. Make a fraction out of the unit you

have and the new unit you want, wit the same value on top and bottom, and

use it to convert the units of one number to match the units of the others.

========================================

Change in speed = (30 km/hr) x (1000 m/km) x (1 hr / 3600 sec) = 8.333 m/sec

Time = (change in speed) / (acceleration) = (8.333 m/s) / (1.6 m/s²) =

Yes, this way was certainly better and prettier.

But at least I got the same answer going the long tough way!

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