A weightlifter holds a 1,500 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m

Answers

Answer 1

Answer:

1800 J

Explanation:

We are given that

Weightlifter hold weight, w=1500N

Height of weightlifter from ground,h=1m

Length of chain=2m

Total weight of chain, w'=400 N

We have to find the work done required to lift the barbell to height of 2m.

Total work done=Work done to lift  a barbell and half of the chain+ work done  to lift the upper half of the chain

Total work done=[tex]1500+\frac{1}{2}(400)+\frac{1}{2}(400)(\frac{1}{2}(1))[/tex]

Total work done=[tex]1700+100[/tex]

Total work done=1800 J

Hence, work done is required to lift the barbell to a height of 2 m=1800 J


Related Questions

you describe a friend’s position by including distance, direction, and what other term?
Answer choices:
A. Acceleration
B.displacement
C.Average speed
D. Reference point
PLEASE HELP I NEED THIS IN AN HOUR

Answers

Answer:

Acceleration

Explanation:

Answer: Acceleration

Explanation:

Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 15.6 revolutions/s. Find the linear speed v of the car, assuming that the tires are not slipping against the ground.

Answers

Answer:

v = 32.345 m/s

Explanation:

given data

radius r = 0.330 m

angular speed = 15.6 revolutions/s

solution

we will get here linear speed v, that is express as

v = r × w        .....................1

here w = 17.5 × 2 × π    radians

so now put value in eq1

v = 0.330 × 15.6× 2 × π

v = 32.345 m/s

Problem 2: Estimate the electric power requirement, in kW, of a 1,400 ft2floor area (three bedroomshome) with three occupants. Using your home power estimate, predict the power requirementsfor a city of 300,000 people. Use these results to estimate the area (inkm2)of silicon solar cells requiredto satisfy the community power requirements. Assume, thepower requirements for an average single family home of 3 are 108.4 x 106BTU per year and solar panels insolation

Answers

Answer:

a) 3170 kw

b) 377 km^2

Explanation:

Estimate of electric power

a) Given :

Average power consumption for a family of 3 = 108.4 * 106 BTU per year  =  0.0317 kw = 31.7 watts

The power requirement for a city of 300000 people

= 31.7 watts * 100000 = 3170000 watts = 3170 kw

b) Given :

Average solar panel insulation = 8.4 W /m^2

Estimate the area of silicon solar cells required to satisfy community power requirement

= (1 * 3170) * (1000/8.4 )

= 377.380 * 10^3 m2 = 377 km^2

Which cell line is pointing to the body?

Answers

Answer:

The answer is B .........number 2

Explanation:

3. If the gravitational force between 2 objects is 50 N, what is the gravitational
force when the distance between the two is increased to four times the distance?

Answers

The gravitational  force : 3.125 N

Further explanation

Given

F₁ = 50 N

Required

F₂

Solution

Newton's Gravity Law:  

[tex]\rm F=G.\dfrac{m_1.m_2}{r^2}[/tex]

with F = gravitational force,  

G = gravitational constant,  

m1, m2 = mass of object,  

r = distance between two objects.  

The value of m and G are the same , so :

F₁ ≈ 1/r₁²

The distance between the two is increased to four times:

r₂ = 4r₁

F₂ = 1/(4r₁)²

F₂ = 1/16r₁²

F₂ = F₁ x 1/16

F₂ = 50 : 16

F₂ = 3.125 N

The study of heat is ____?

Answers

Explanation:

thermodynamics is the study of heat.

Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie

Explanation:

A basketball with a mass of 20 kg is accelerated with a force of 10 N. If resisting forces are ignored, what is the acceleration of the basketball?

Answers

I’m pretty sure it would be 10/20= 0.5m/s2

A wall clock has a minute hand with a length of 0.46 m and an hour hand with a length of 0.24 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x axis pointing to 3 o'clock and the positive y axis pointing to 12 o'clock. Write the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M. (Let vector D represents the displacement of the fly.)

Answers

Answer:

the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M is ; { 0.24 m(i) - 0.46 m(j) }

Explanation:

Given the data in the question;

as illustrated in the image below,

3:00 pm means

the hour hand is on 3 i.e along x-axis

while the minute hand is on 12 i.e along y-axis

so Displacement will be;

D = ( 0.24 + 0i) - ( 0 + 0.46j )

D = { 0.24 m(i) - 0.46 m(j) }

Therefore, the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M is ; { 0.24 m(i) - 0.46 m(j) }

What is the frequency of a wave of a light is with a wavelength of 4 x 10-7 m?

Answers

Answer:

7.5 × 10^14 Hz

Velocity of light = 3×10^8m/s

Frequency = (3×10^8)/(4 x 10^-7)

= 7.5 × 10^14 Hz

hmu if u brave shawtys

Answers

Answer:

BET, & done ✌

Answer:

boop

Explanation:

4. Sally applies a horizontal force of 462 N with a rope to drag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 56.00 . c.What work is done by the floor through force of friction between the floor and the crate

Answers

Answer:

-6,329.5Joules

Explanation:

Complete question:

Sally applies a horizontal force of 462N with a rope to drag a wooden crate across a floor with a constant speed the rope tied to the crate is pulled at an angle of 56.0degree and sally moves the crate 24.5m. What work is done by the floor through the force of friction between the floor and crate.

Work done = Fd cos theta

F is the horizontal force

d is the distance covered

theta angle of inclination

Substituting into the formula

Workdone  = 462(24.5)cos 56

Workdone = 11,319(0.5592)

Workdone = 6,329.5Joules

Hence the workdone by sally is 6,329.5Joules

The work done by friction will be opposite the work done by sally, hence work done by the floor through force of friction between the floor and the crate is -6,329.5Joules

A spacecraft is flying away from the moon toward earth.

What will be true of the moon’s gravitational pull on the spacecraft?

It will decrease.

It will increase.
It will repel the spacecraft.
It will remain the same.

Answers

Answer:

it will  decrease

Explanation:

According to the law of universal gravitation, the gravitational force exerted by the moon on the spacecraft is equal to the product of their masses and inversely proportional to the square of the distance that separates them. Therefore, as the spacecraft moves away, its distance increases and the force of attraction exerted by the moon decreases.

Answer:

A. It will increase

Explanation:

I took the quiz on K12 and this was the correct option.

Hope I helped

A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 rad/s . A monkey drops a 8.97 kg bunch of bananas vertically onto the platform. They hit the platform at 45 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 22.1 kg , drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.73 m .

Answers

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana [tex]m_{b}[/tex] = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey [tex]m_{m}[/tex] = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + [tex]m_{b}[/tex] ([tex]\frac{4}{5}[/tex] R)² + [tex]m_{m}[/tex]R² ]w

m/2 × ω₀ = [ m/2 + [tex]m_{b}[/tex] ([tex]\frac{4}{5}[/tex] )² + [tex]m_{m}[/tex] ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

a. As you coast down a hill on your bicycle, you accelerate at 0.5 m/s2. If the total mass of your body and the bicycle is 80 kilograms what is the net force pulling you down the hill (gravity - friction)?

Answers

According to Newton’s second law of motion:
F=ma
=80*0.5
=40N

The net force pulling you down the hill will be = 40 N

What is Newton's Second Law of motion?

The Second Law of motion states that the acceleration of an object depend upon the object and the mass of the object.

F = Mass * acceleration

Given

Total mass  : 80 kg

Acceleration : 0.5 m / s^2

Net Force = mass * acceleration      (Second Law of motion )

Net Force = 80 * 0.5  = 40 N

 The net force pulling you down the hill will be = 40 N

Learn more about  Newton's Second Law of motion:

https://brainly.com/question/11553356?referrer=searchResults

#SPJ2

Two ropes are connected to a 200 kg dinghy. Two cousins each take one rope and pull. When the cousins pull in the same direction, the dinghy accelerates at a rate of 1.31 m/s2 to the east. If they pull in opposite directions, the dinghy has an acceleration of 0.526 m/s2 to the west. Assume the ropes are horizontal, and ignore any other horizontal forces acting on the dinghy. What is the magnitude of the force each cousin exerts on the dinghy

Answers

Answer:

The magnitude of the force each cousin exerts on the dinghy is 183.6 N and 78.4 N.

Explanation:

When the cousins pull in the same direction we have:

[tex] F_{1} + F_{2} = ma_{e} [/tex]   (1)        

Where:

F₁ and F₂ are the forces exerted by the two boys.

m: is the mass of the dinghy = 200 kg

[tex]a_{e}[/tex]: is the acceleration in the east direction

When the cousins pull in opposite directions we have:

[tex] F_{1} - F_{2} = ma_{w} [/tex]   (2)            

By adding equation (1) and (2):                                              

[tex] 2F_{1} = m(a_{e} + a_{w}) [/tex]          

[tex] F_{1} = \frac{200 kg(1.31 m/s^{2} + 0.526 m/s^{2})}{2} = 183.6 N [/tex]    

Now, by entering F₁ into equation (1) we can find F₂:    

[tex] F_{2} = ma_{e} - F_{1} = 200kg*1.31 m/s^{2} - 183.6 N = 78.4 N [/tex]

Therefore, the magnitude of the force each cousin exerts on the dinghy is 183.6 N and 78.4 N.    

I hope it helps you!  

A DC motor connected to a switch-mode dc-dc converter goes into regenerative braking mode. The average current being supplied by the dc motor is 20 A. In the equivalent circuit of the dc motor, Ea=103 V, Ra=0.19 Ohms, and La=4 mH. Calculate the average power flow into the converter. Round answer to the nearest whole number.

Answers

Answer:

The correct answer is "2524 W".

Explanation:

The given values are:

Ia = 20 A

Ea = 103 V

Ra = 0.19 Ω

La = 4 mH

According to KVL,

⇒ [tex]V_0=E_a-I_aR_a[/tex]

On substituting the given values in the above equation, we get

⇒      [tex]=103-(20)(0.19)[/tex]

⇒      [tex]=103-3.8[/tex]

⇒      [tex]=126.2 \ V[/tex]

Now,[tex]=126.2\times 20[/tex]

The average power flow into the converter will be:

⇒ [tex]P=V_0 I_a[/tex]

On substituting the given values, we get

⇒     [tex]=(126.2\times 20)[/tex]

⇒     [tex]=2524 \ W[/tex]

What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/C

Answers

Answer:

[tex]q=-2.26\times 10^{-5}\ C[/tex]

Explanation:

Given that,

The mass of a particle, m = 1.48 g = 0.00148 kg

The electric field, E = 640 N/C

We need to find the charge of the particle when placed in a downward-directed electric field.

The force of gravity is balanced by the electric force such that,

mg = qE

Where

q is the charge of the particle

[tex]q=\dfrac{mg}{E}\\\\q=\dfrac{0.00148\times 9.8}{640}\\\\q=2.26\times 10^{-5}\ C[/tex]

q must be negative, the force must be upward (opposite direction of the electric field).

A student claims an object in motion must experience a force to stay in motion. Do you agree or disagree?

Answers

Answer:

agree because there is always a force that causes motion..

A box of bananas weighing 51.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.46 and the coefficient of kinetic friction is 0.23. Part A If no horizontal force is applied to the box and the box is at rest, how large is the frictional force exerted on the box by the surface

Answers

static friction force = The coefficient of static friction * normal force

static friction force = 0.46 * 51 = 23.46 N

kinetic friction force = The coefficient of kinetic friction * normal force

kinetic friction force = 0.23 * 51 = 11.73 N

the applied force acting on the object must be more than 23.5 N if the object was stationary to move it and must be more than 11.7 N during the movement to keep the object moving

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.70 m/s2. At the same instant a truck, traveling with a constant speed of 9.50 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck

Answers

Answer:

66.85 m

Explanation:

We are given that

Acceleration ,a=[tex]2.7m/s^2[/tex]

Speed of truck, v=9.5 m/s

We have to find the distance beyond which the traffic signal will the automobile overtake the truck.

Initial speed of automobile, u=0

We know that

[tex]s=ut+\frac{1}{2}at^2[/tex]

Using the formula

[tex]s=0+\frac{1}{2}(27)t^2=\frac{27}{2}t^2[/tex]

For constant speed

Acceleration, a=0

Again

[tex]s=vt+0=9.5t[/tex]

[tex]9.5t=\frac{27}{2}t^2[/tex]

[tex]t=\frac{9.5\times 2}{2.7}=7.037s[/tex]

Substitute the value of t

[tex]x=9.5(7.037)=66.85m[/tex]

Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m

An electric vehicle starts from rest and accelerates at a rate of 2.3 m/s2 in a straight line until it reaches a speed of 29 m/s. The vehicle then slows at a constant rate of 1.5 m/s2 until it stops. (a) How much time elapses from start to stop

Answers

Answer:

t = 12.6 seconds

Explanation:

Given that,

Initial velocity, u = 0

Acceleration of an electric vehicle, a = 2.3 m/s²

Final velocity, v = 29 m/s

We need to find the time elapses from start to stop. The acceleration of an object is given by the relation as follows :

[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{29-0}{2.3}\\\\t=12.6\ s[/tex]

So, 12.6 seconds is elapsed from start to stop.

You are concerned about the moon roof on your new sports car. It seems to flex when driving at high speeds. Calculate how much net force the moon roof must withstand and in what direction. Assume the moon roof is flat with an area of 0.5 m^2 and the pressure and velocity over the moon roof is constant. Your driving speed is 20 m/s, and the velocity over the moon roof is 30 m/s. The pressure inside the car is 90,500 N/m^2 and the freestream pressure and density in front of the car are 90,000 N/m^2 and 1.1 kg/m^3.

Answers

Answer:

Explanation:

We shall apply Bernoulli's formula to solve the problem . It is as follows .

P + ρ gh + 1/2 ρ v² = constant .

P₁ + ρ gh + 1/2 ρ v₁²  = P₂ + ρ gh + 1/2 ρ v₂²

P₁ +  1/2 ρ v₁²  = P₂  + 1/2 ρ v₂²

P₁ - P₂ = 1/2 ρ (v₂² - v₁² )

= .5 x 1,1 ( 30² - 20² )

= 275 N / m²

velocity over moon roof is high , pressure will be lower there by 275 N / m²

Given pressure difference already existing = 90500 - 90000 = 500 N / m²

Additional pressure difference due to velocity difference = 275 N / m²

Total pressure difference = 275 + 500 = 775 N / m²

Area of roof = .5 m²

Total force acting upwards on the roof

= .5 x 775 N

= 387.5 N .

15 points!!:-) Need help ASAP!

When does a compass NOT point towards magnetic north?

A.during a solar eclipse, which changes

Earth's magnetic field.

B. When there is another magnet close by.

C.When there is an unused battery close by.

D. When there is a coil of copper wire close by.

Answers

Answer:

b i think so because it makes senes

Answer: When there is another magnet close by.

Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.

A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did it take the car to reach this final velocity?

Answers

Answer:

t = 5 s

Explanation:

Data:

Initial Velocity (Vo) = 7 m/sAcceleration (a) = 3 m/s²Final Velocity (Vf) = 22 m/sTime (t) = ?

Use formula:

[tex]\boxed{t=\frac{Vf - Vo}{a}}[/tex]

Replace:

[tex]\boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}[/tex]

Solve the subtraction of the numerator:

[tex]\boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}[/tex]

It divides:

[tex]\boxed{t=5\ s}[/tex]

How much time did it take the car to reach this final velocity?

It took a time of 5 seconds.

A 70 kg Throckmorton was riding his skateboard when a small, but juicy 0.001 kg beetle hits the front of his helmet, splattering, obstructing his view. Who experienced the greatest change in momentum?

Answers

Answer:

The beetle

Explanation:

The beetle is correct because of the the force of the Throckmortan was a lot heavier and thus making the beetle feel the force.

Momentum before = momentum after

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(70.0 kg) (3.0 m/s) + (2.0 kg) (3.0 m/s) = (70.0 kg) (3.1 m/s) + (2.0 kg) v

210 kg m/s + 6 kg m/s = 217 kg m/s + (2.0 kg) v

-1 kg m/s = (2.0 kg) v

v = -0.5 m/s

The skateboard's velocity is 0.5 m/s west. Hope I helped and have a good day!

________________________________________________________

前の勢い=後の勢い

m₁u₁+m₂u₂=m₁v₁+m₂v₂

(70.0 kg)(3.0 m / s)+(2.0 kg)(3.0 m / s)=(70.0 kg)(3.1 m / s)+(2.0 kg)v

210 kg m / s + 6 kg m / s = 217 kg m / s +(2.0 kg)v

-1 kg m / s =(2.0 kg)v

v = -0.5 m / s

スケートボードの速度は西に0.5m / sです。私が助けて、良い一日を過ごせることを願っています!

what kind of lens curve inward toward its center​

Answers

Answer:

A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.

Answer: A concave lens

A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.

Hopes this helps :)

Why do birds not get shock when they
sit on high power live wire but we do?

Answers

Answer:

Their bodies don't conduct electricity like we do.

Explanation:

Answer:

birds dont get shocked because they sit on their talons and their talons are a different type of skin then the rest of their body

Explanation:

Which elements have one valence electron?
A. Sodium
B. Carbon
C. Fluorine
D. Magnesium


The answer is A

Answers

Answer:

Well, you said the answer is A, so it’s A!

Haha answer is a you said it

two students sit on a see-saw. archie is a hulking football player with a mass of 120 kg. clementine is a dainty cheerleader with a mass of 40 kg. the see-saw is 3.5 m in total length with the fulcrum at the center. if clementine sits at the end on one side, where must archie sit relative to the center to keep the see-saw balanced

Answers

Answer:

Archie must sit 0.58 m relative to the center to keep the see-saw balanced

Explanation:

Given the data in the question;

Mass of Archie [tex]m_{a}[/tex] = 120 kg

Mass of clementine [tex]m_{c}[/tex] = 40 kg

total length of see-saw L = 3.5 m

as illustrated on the image below, Fulcrum is at the center,

suppose Archie sits at a distance x  from center then for balancing, we will have;

[tex]m_{a}[/tex] × x = [tex]m_{c}[/tex] × ( one end = 3.5/2 = 1.75)

so we substitute

120kg × x = 40kg × 1.75m

x12okg = 70 kg.m

x = 70 kg.m / 120 kg

x = 0.58 m

Therefore, Archie must sit 0.58 m relative to the center to keep the see-saw balanced

Please Help! Will mark brainliest.

Answers

Answer:W = m*g*h

19*9.8*32.4 = 6,032.9 rounded

honestly, I do not know if this is correct so please don't come back at me

hopefully this helps

Explanation: [do the following, if you think I am wrong]

just pick a formula,

plug in the number to the mass, gravity, and height

than multiply

get your answer, but don't forget to round to the nearest tenth