# A woman lifts a 300 newton child a distance of 1.5 meters in 0.75 seconds. What is her power output in lifting the child?

QUESTION POSTED AT 13/11/2019 - 05:21 PM

QUESTION POSTED AT 13/11/2019 - 05:21 PM

Power = Work done / time

Work done = Force * Distance

= 300 N * 1.5 m = 450 J

Power = 450 / 0.75 = 600 Watts.

Work done = Force * Distance

= 300 N * 1.5 m = 450 J

Power = 450 / 0.75 = 600 Watts.

Force = (mass) x (acceleration)

In order to

(2,300 kg) x (16 m/s²) =

(about 8,270 pounds ... 4.14 tons !)

has to be coming from somewhere ... presumably from the straining,

screaming, smoking engine under the hood.

The force it might apply to an object it runs into will depend on the

shape and composition of that object.

ANSWERED AT 06/12/2019 - 03:37 PM

QUESTION POSTED AT 06/12/2019 - 03:37 PM

It would be 5.94 hours but you can round it to just 6 hours.

ANSWERED AT 06/12/2019 - 03:01 PM

QUESTION POSTED AT 06/12/2019 - 03:01 PM

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

Change in speed = (15 - 30) = -15 km/hr

Time for the change = 4 sec

Acceleration = (-15 km/hr) / (4 sec) =

Is that a lot ? Not much ?

Let's convert it to a unit that we can think about:

(-15 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) =

(-15 x 1,000) / (3,600) = -(4 and 1/6) m/sec .

So the acceleration of the bus is -(4 and 1/6) m/sec² .

The negative sign means that it slowed down.

(4 and 1/6) m/sec² is about 42% of the acceleration of gravity ...

the acceleration the bus would have if it drove off of a cliff.

When the car or the bus you're riding in slows down at that rate,

you feel 42% of your weight pulling you forward against your

seat belt. That's quite a drastic acceleration !

ANSWERED AT 06/12/2019 - 02:36 PM

QUESTION POSTED AT 06/12/2019 - 02:36 PM

There isn;t enough given values to define and determine what is asked. But since

distance = speed / time, time = distance / speed

Calculate the acceleration of a boat that has been cruising at 15 m/s but speeds up to 45 m/s in about 6 seconds. Don’t forget your units, m/s2. In order to calculate the acceleration of the boat we first have to identify the suited formula for acceleration in this scenario.

Hence, a = vf – vi / t. Vi = 15 m/s Vf = 45 m/s T = 6 seconds

Solution:

A = 45 m/s – 15m/s / 6s

**A = 5m/s^2 **

distance = speed / time, time = distance / speed

Calculate the acceleration of a boat that has been cruising at 15 m/s but speeds up to 45 m/s in about 6 seconds. Don’t forget your units, m/s2. In order to calculate the acceleration of the boat we first have to identify the suited formula for acceleration in this scenario.

Hence, a = vf – vi / t. Vi = 15 m/s Vf = 45 m/s T = 6 seconds

Solution:

A = 45 m/s – 15m/s / 6s

ANSWERED AT 06/12/2019 - 12:11 PM

QUESTION POSTED AT 06/12/2019 - 12:11 PM

He covered an amount of 2 km to east and one to south. the displacement, yeah i dont have that sorry :[

ANSWERED AT 06/12/2019 - 12:04 PM

QUESTION POSTED AT 06/12/2019 - 12:04 PM

Here it is. *WARNING* VERY LONG ANSWER

________________________________________...

11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h)

The change in PE =mgh=5*9.8*12=588 J

______________________________________...

12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s.

Commendable and jockey Pat Day had a combined mass =M= 550.0 kg,

Their KE as they crossed the line=(1/2)Mv^2

Their KE as they crossed the line=0.5*550*(15.98)^2

Their KE as they crossed the line is 70224.11 J

______________________________________...

13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m

She trips and drops the spare tire of mass = m = 10.0 kg,

The tire rolls down the hill with an intial speed = u = 2.00 m/s.

The height of top of the next hill = h = 5.00 m

Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2

Initial total mechanical energy =mgH+(1/2)mu^2

Suppose the final speed at the top of second hill is v

Final total mechanical energy =PE+KE=mgh+(1/2)mv^2

As mechanical energy is conserved,

Final total mechanical energy =Initial total mechanical energy

mgh+(1/2)mv^2=mgH+(1/2)mu^2

v = sq rt [u^2+2g(H-h)]

v = sq rt [4+2*9.8(20-5)]

v = sq rt 298

v =17.2627 m/s

The speed of the tire at the top of the next hill is 17.2627 m/s

______________________________________...

14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean.

a.)The mass of bean = m = 2.0 g

Height up to which the been jumps = h = 1.0 cm from hand

Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg

b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s

_____________________________

15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill.

The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s

__________________________________

16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m,

The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2

______________________________________...

EDIT

1.) A train is accelerating at a rate = a = 2.0 km/hr/s.

Acceleration

Initial velocity = u = 20 km/hr,

Velocity after 30 seconds = v = u + at

Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s =

Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr

Velocity after 30 seconds = v = 80 km/hr=22.22 m/s

_______________________________-

2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins.

His acceleration = a =11.1/9=1.233 m/s^2

Distance he covered = s = (1/2)at^2=49.95 m

________________________________________...

11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h)

The change in PE =mgh=5*9.8*12=588 J

______________________________________...

12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s.

Commendable and jockey Pat Day had a combined mass =M= 550.0 kg,

Their KE as they crossed the line=(1/2)Mv^2

Their KE as they crossed the line=0.5*550*(15.98)^2

Their KE as they crossed the line is 70224.11 J

______________________________________...

13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m

She trips and drops the spare tire of mass = m = 10.0 kg,

The tire rolls down the hill with an intial speed = u = 2.00 m/s.

The height of top of the next hill = h = 5.00 m

Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2

Initial total mechanical energy =mgH+(1/2)mu^2

Suppose the final speed at the top of second hill is v

Final total mechanical energy =PE+KE=mgh+(1/2)mv^2

As mechanical energy is conserved,

Final total mechanical energy =Initial total mechanical energy

mgh+(1/2)mv^2=mgH+(1/2)mu^2

v = sq rt [u^2+2g(H-h)]

v = sq rt [4+2*9.8(20-5)]

v = sq rt 298

v =17.2627 m/s

The speed of the tire at the top of the next hill is 17.2627 m/s

______________________________________...

14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean.

a.)The mass of bean = m = 2.0 g

Height up to which the been jumps = h = 1.0 cm from hand

Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg

b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s

_____________________________

15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill.

The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s

__________________________________

16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m,

The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2

______________________________________...

EDIT

1.) A train is accelerating at a rate = a = 2.0 km/hr/s.

Acceleration

Initial velocity = u = 20 km/hr,

Velocity after 30 seconds = v = u + at

Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s =

Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr

Velocity after 30 seconds = v = 80 km/hr=22.22 m/s

_______________________________-

2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins.

His acceleration = a =11.1/9=1.233 m/s^2

Distance he covered = s = (1/2)at^2=49.95 m

ANSWERED AT 06/12/2019 - 11:27 AM

QUESTION POSTED AT 06/12/2019 - 11:27 AM

If you're moving at 70 mi/hr, then you cover 33.88 feet in 0.33 sec. If you're moving at 1,000 ft/sec, then you cover 33 feet in 0.33 sec. 70 mph and 1,000 fps are not equivalent. 1,000 fps is about 682 mph, whereas 70 mph is about 103 fps.

ANSWERED AT 06/12/2019 - 10:54 AM

QUESTION POSTED AT 06/12/2019 - 10:54 AM

I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².

Taking downwards as positive, use v²=u²+2as.

v²=(-2)²+2(9.81)(14)

v=16.7 m/s

Taking downwards as positive, use v²=u²+2as.

v²=(-2)²+2(9.81)(14)

v=16.7 m/s

ANSWERED AT 06/12/2019 - 10:27 AM

QUESTION POSTED AT 06/12/2019 - 10:27 AM

A- 240

B- 10

C- 2

D- ?

B- 10

C- 2

D- ?

ANSWERED AT 06/12/2019 - 10:24 AM

QUESTION POSTED AT 06/12/2019 - 10:24 AM

Let's calculate the angle between two spots, 0.5 cm apart,

that are 10m away from you.

1 m = 100 cm

10m = 1,000 cm

The angle between the two spots is very very close to the angle

whose tangent is

0.5cm/1,000cm = 0.0005 .

That angle is tan⁻¹(0.0005) =

Matilda is looking at two stars that appear to be 0.001 radian apart.

That's (0.001 / 0.0005) = 2 =

As long as Arthur and Matilda are on the same planet while looking at

these two stars ... or as long as they're even in the same solar system ...

Arthur will have no trouble separating them.

ANSWERED AT 06/12/2019 - 09:43 AM

QUESTION POSTED AT 06/12/2019 - 09:43 AM

Portugal was not an axis power

ANSWERED AT 06/12/2019 - 09:34 AM

QUESTION POSTED AT 06/12/2019 - 09:34 AM

The speed of the tortoise can be calculate as follows:

where

S is the distance travelled by the tortoise

t is the time taken

In this problem, S=6.4 m, while we should convert the time into seconds:

Substituting the numbers into the equation, we find the speed:

ANSWERED AT 06/12/2019 - 08:28 AM

QUESTION POSTED AT 06/12/2019 - 08:28 AM

"Acceleration" does NOT mean 'speeding up'. It means any change

in the speed or direction of motion ... speeding up, slowing down, or

moving in a curve.

An object moving in a circle at a constant speed is accelerating, because

when you're moving along a circle, your direction is constantly changing.

ANSWERED AT 06/12/2019 - 08:28 AM

QUESTION POSTED AT 06/12/2019 - 08:28 AM

Hello there!

Explanation:

↓↓↓↓↓↓↓↓↓↓↓↓↓↓

Newton's second law of motion states that the total net force acting on an object is equal to mass times acceleration.

Newton's second law of motion formula:

↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓

F=force=newtons=

M=mass (kg)

A-acceleration

F=ma

M=f/a

A=f/m

For example:

↓↓↓↓↓↓↓↓↓↓↓

Calculate the force needed to accelerate an object to 4.3 m/s². The object has a mass of 2.2 kg. Round the answer to the nearest tenth.

↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓

A= 4.3 m/s²

M= 2.2 kg

F=??

F=ma

2.2 kg (4.3 m/s²)=9.46=9.5N

**Answer⇒⇒⇒⇒⇒9.5N**

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

ANSWERED AT 06/12/2019 - 07:59 AM

QUESTION POSTED AT 06/12/2019 - 07:59 AM

Because acceleration is not speed, and speed is not acceleration.

I'm sure you would not ask "Why isn't temperature given in acres,

as it is for area ?" Speed and acceleration are different things, so

it's only natural that they have different units.

The magnitude (size) part of acceleration is:

(how much speed changes) per second.

Are you speeding up ?

Are you traveling (2 meters per second) faster every second ?

Then your acceleration is

(2 meters per second) per second .

When you write that phrase as an algebraic expression, it's

(2 m/sec) / sec

and when you simplify that fraction, you get 2 m/sec² .

ANSWERED AT 06/12/2019 - 07:40 AM

QUESTION POSTED AT 06/12/2019 - 07:40 AM

From the information as given in the question, the transit time

for light from the sun would be

(6.1 x 10⁹ km) / (3.84 x 10³ km/sec) = 15,885 seconds (rounded).

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

-- A better figure for Pluto's average distance from the sun is 5.874 x 10⁹ km.

-- " 384,000 km/sec " is about 28% too high for the speed of light.

A much better figure is 300,000 km/sec.

Using these numbers, the transit time is

(5.874 x 10⁹ km) / (3 x 10⁵ km/sec) = 19,580 seconds

ANSWERED AT 06/12/2019 - 07:12 AM

QUESTION POSTED AT 06/12/2019 - 07:12 AM

C. push against you with equal force

ANSWERED AT 06/12/2019 - 07:12 AM

QUESTION POSTED AT 06/12/2019 - 07:12 AM

Im pretty sure that it is allied forces

ANSWERED AT 06/12/2019 - 07:11 AM

QUESTION POSTED AT 06/12/2019 - 07:11 AM

Thanks for sharing. The density of the sample

is 5 g/cm³. Do you have a question to ask ?

ANSWERED AT 06/12/2019 - 07:03 AM

QUESTION POSTED AT 06/12/2019 - 07:03 AM

On Earth, the acceleration of gravity is 9.8 m/s² downward.

This means that any falling object gains 9.8 m/s of speed downward

for every second it falls.

Since the penny fell for 1.5 seconds, its downward speed at the

end of that time is (1.5 x 9.8) = 14.7 m/s faster than at the beginning

of that time.

Since it was dropped gently from somebody's hand ... and not THROWN

down ... its speed at the beginning is zero. So its speed after 1.5 seconds

is 14.7 m/s, and its velocity after 1.5 seconds is

ANSWERED AT 06/12/2019 - 07:03 AM

QUESTION POSTED AT 06/12/2019 - 07:03 AM

The answer would be B. Zero

ANSWERED AT 06/12/2019 - 05:42 AM

QUESTION POSTED AT 06/12/2019 - 05:42 AM

The weight of the book is (mass) x (gravity) = (1kg) x (9.8m/s²) = 9.8 newtons

(about 2.2 pounds)

The upward force on the book is the sum of the

upward force from each cable (each person).

The forces from the two people must be equal, otherwise the book

would move sideways one way or the other.

Now for the big cookie: If the cables are within 1 degree of

being horizontal, then how much of each person's force winds

up being vertical to lift the book ?

If a force 'F' is directed 1 degree above horizontal, then

its strength in the horizontal direction is F x cos(1), and

its strength in the vertical direction is F x sin(1).

Each person is pulling with a force F.

His vertical component is F x sin(1).

Two of these hold the 1-kg book up.

2F sin(1) = 9.8 newtons

Divide each side by 2 : F sin(1) = 4.9 newtons

Divide each side by sin(1) F = 4.9 / sin(1) =

(about 63 pounds)

This is the tension in each cable.

____________________________________________

That's why, technically, it's impossible to make the cable or string

perfectly horizontal, even when there's nothing hanging from it.

When the force is perfectly horizontal, its vertical component is

zero, so it would take an infinite force to hold anything horizontal,

even the weight of the string itself.

From another viewpoint ... If the string or cable is almost horizontal,

then the tension in it is already so great that you can break it with

just a light push in the center with your finger.

ANSWERED AT 06/12/2019 - 05:31 AM

QUESTION POSTED AT 06/12/2019 - 05:31 AM

(1,500 meters) x (1 sec / 330 meters) = (4 and 18/33) seconds

(4.55 sec, rounded)

ANSWERED AT 06/12/2019 - 05:27 AM

QUESTION POSTED AT 06/12/2019 - 05:27 AM

ANSWERED AT 06/12/2019 - 04:42 AM

QUESTION POSTED AT 06/12/2019 - 04:42 AM

A force vector F1 points due
east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The
resultant of the two vectors has a magnitude of 400 newtons and points along
the due east/west line. Find the magnitude and direction of F2. Note that there
are two answers.
The given values are

F1 = 200 N F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N

F2 = 400 - 200

**F2 = 200
N**

F1 = 200 N F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N

F2 = 400 - 200

ANSWERED AT 06/12/2019 - 04:07 AM

QUESTION POSTED AT 06/12/2019 - 04:07 AM

No. They are different. The bird's distance for that period of time is double the height of the nest above the ground. His displacement for the same time is zero ... the length between the start point and end point.

ANSWERED AT 06/12/2019 - 03:59 AM

QUESTION POSTED AT 06/12/2019 - 03:59 AM

1500m

-----------

330ms-1

= 4.5455 s

-----------

330ms-1

= 4.5455 s

ANSWERED AT 06/12/2019 - 03:55 AM

QUESTION POSTED AT 06/12/2019 - 03:55 AM

4 km/hr. There are four quarters in an hour, so just divide 16 by 4 and you get 4. It is directly proportional.

ANSWERED AT 06/12/2019 - 03:38 AM

QUESTION POSTED AT 06/12/2019 - 03:38 AM

Given:

25.0 m/s -speed of his car

2.0 m/s^2 for 5s - acceleration

10s- constant velocity

Asked: total distance traveled during the entire 15 seconds.

( see attached file for the solution)

25.0 m/s -speed of his car

2.0 m/s^2 for 5s - acceleration

10s- constant velocity

Asked: total distance traveled during the entire 15 seconds.

( see attached file for the solution)

ANSWERED AT 06/12/2019 - 03:11 AM

QUESTION POSTED AT 06/12/2019 - 03:11 AM

38 years older miss =) im possitive

ANSWERED AT 06/12/2019 - 02:55 AM

QUESTION POSTED AT 06/12/2019 - 02:55 AM