A person lifts a heavy load to a vertical height of 2.0 m in 3 seconds. If he/she had done this more slowly in 6 seconds, the

work on the load would have been:

Four times as great

half as great

the same

twice as great

Answers

Answer 1

Answer:

If the heavy load had been lifted more slowly, the work done on the load would have been the same.

Explanation:

Work done on an object is given as;

W = Fd

where;

F is the force applied on the object

d is the displacement of the object

for the given question, the applied force on the load = mg (mass of the load multiplied by acceleration due to gravity).

Also, the displacement of the object = vertical height the load was lifted.

W = mgh

The work done on the load is independent of time.

Thus, if the heavy load had been lifted more slowly, the work done on the load would have been the same.

Answer 2

A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.

A person lifts a heavy load to a vertical height of 2.0 m in 3 seconds.

We want to compare the work done with the one that he/she would have done if the process had taken 6 seconds.

What is work?

In physics, work (W) is the energy transferred to or from an object via the application of force (F) along a displacement (s).

W = F × s

Given the displacement is the same (2.0 m) and the force needed is also the same (weight of the object), the work is the same for both processes.

A person lifting a heavy load to a vertical height of 2.0 m in 3 seconds does the same work as if he/she lifts it in 6 s.

Learn more about work here: https://brainly.com/question/25064916


Related Questions

1. Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s,​

Answers

Answer:

104653.13J

Explanation:

Given parameters:

Mass of roller coaster  = 625kg

Speed  = 18.3m/s

Unknown:

Kinetic energy  = ?

Solution:

The kinetic energy is the energy due to the motion of a body.

     Kinetic energy  = [tex]\frac{1}{2}[/tex] x m x v²  

m is the mass

v is the speed

    Kinetic energy  =  [tex]\frac{1}{2}[/tex]  x 625  x  18.3²   = 104653.13J

A bullet has a mass of 0.06 kg. Starting from rest, after the gun's trigger is pulled, a constant force acts on the bullet for the next 0.025 seconds until the bullet leaves the barrel of the gun with a speed of 992 m/s.

What is the change in momentum of the bullet?

Answers

The change in momentum of the bullet : 59.52 kg m/s

Further explanation

Given

m=0.06 kg

Δt=0.025 s

vo=0(from rest)

vt= 992 m/s

Required

The change in momentum

Solution

The change in momentum  = ΔP

ΔP  =m(vt-vo)

ΔP =0.06(992-0)

ΔP =59.52 kg m/s

which is true about the way air flows

A. high pressure to low pressure

B. low pressure to high pressure

C. cold air to hot air

D. hot air to cold air

Answers

Answer:

A High-to-Low

Explanation:

its like water running down a hill.

A block of mass m is hung from the ceiling by the system of massless springs consisting of two layers. The upper layer consists of 3 strings in paralle, and the lower layer consists of 2 strings in parallel. The horizontal bar between the two layers has negligible mass. The force constants of all springs are k. Calculate the period of the vertical oscillations of the block.

Answers

Answer:

  T₀ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]           T = [tex]\sqrt{\frac{5}{6} }[/tex] T₀

Explanation:

When the block is oscillating it forms a simple harmonic motion, which in the case of a spring and a mass has an angular velocity

        w = [tex]\sqrt{k/m}[/tex]

To apply this formula to our case, let's look for the equivalent constant of the springs.

Let's start with the springs in parallels.

* the three springs in the upper part, when stretched, lengthen the same distance, therefore the total force is

       F_total = F₁ + F₂ + F₃

the springs fulfill Hooke's law and indicate that the spring constant is the same for all three,

       F_total = - k x - k x - kx = -3k x

         

therefore the equivalent constant for the combination of the springs at the top is

      k₁ = 3 k

* the two springs at the bottom

following the same reasoning the force at the bottom is

        F_total2 = - 2 k x

the equivalent constant at the bottom is

         k₂ = 2 k

now let's work the two springs are equivalent that are in series

the top spring is stretched by an amount x₁ and the bottom spring is stretched x₂

            x₂ = x -x₁

            x₂ + x₁ = x

if we consider that the springs have no masses we can use Hooke's law

            [tex]-\frac{F_{1} }{k_{1} } - \frac{F_{2}}{k_{2} } = \frac{F}{k_{eq} }[/tex]

therefore the equivalent constant is the series combination is

             [tex]\frac{1}{k_{eq} } = \frac{1}{k_{1} } + \frac{1}{k_{2} }[/tex]

we substitute the values

             \frac{1}{k_{eq} }  = \frac{1}{3k } + \frac{1}{2k }  

             \frac{1}{k_{eq} }  = \frac{5}{6k} }  

              k_eq = [tex]\frac{6k}{5}[/tex]  

therefore the angular velocity is

             w = [tex]\sqrt{\frac{6k}{5m} }[/tex]  

           

angular velocity, frequency, and period are related

           w = 2π f = 2π / T

           T = 2π / w

            T = 2π [tex]\sqrt{\frac{5m}{6k} }[/tex]

           T₀ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

           T = [tex]\sqrt{\frac{5}{6} }[/tex] T₀

a string attached to a 60.0 Hz vibr.ator creates a standing wave with 5 loops. What frequency would make 7 loops? (Unit = Hz)

Answers

Answer:

F=84.0 Hz

Explanation:

Using the equation f= n (v/2L),  frequency equals number of loops times velocity over 2 times the length, in order to get 60.0 Hz of frequency from 5 loops, v/2L would have to equal 12. (12*5=60) v/2L is constant, so to find the frequency of 7 loops you would times 7 by 12 to get 84.0.

Hope this helped! :)

Which of these is another name for Newton's
first law?
A. the law of action-reaction
B. the law of force and acceleration
C. the law of gravity
D. the law of inertia

Answers

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
D. The law of inertia

Determine the absolute pressure on the bottom of a swimming pool 30.0 mm by 8.4 mm whose uniform depth is 1.9 mm .

Answers

Answer:

=101343.62N/m^2

Explanation:

absolute pressure on the bottom of a swimming pool= atmospheric pressure +( 2 ×ρ ×g)

( 2 ×ρ ×g)= guage pressure

atmospheric pressure= 101325pa

h= height= 1.9 mm = 1.9×10^-3m

ρ = density of water

= 1000kg/m^3

g= acceleration due to gravity= 9.8m/s^2

Then substitute, we have

absolute pressure on the bottom of a swimming pool= 101325+ [0.0019 ×1000 × 9.8)]

=101343.62N/m^2

Hence, the absolute pressure on the bottom of a swimming pool is =101343.62N/m^2

please help thank you ​

Answers

YOUR QUERY :

which of the following statements BEST describes the difference between an atom and an ion ?

Answer:

well the correct answer is

d. An atom contains equal numbers of protons and electrons whereas an ion contains unequal numbers of protons and electrons .

Explanation:

A charged atom is known as an ion, well it can be negative as well as positive charge.

if atom has more protons than electrons then it get positively charged and known as cation

if the atom has more electrons that the number of protons then the atom get negatively charged and known as anion

Need help y’all ASAP please...physics

Answers

Answer:

t = 3/8 seconds

Explanation:

h=-16t^2 - 10t+6

h= 0 when it hits the ground

0=-16t^2 - 10t+6

factor out a -2

0= -2(8t^2 +5t -3)

divide by -2

0 = (8t^2 +5t -3)

factor

0=(8t-3) (t+1)

using the zero product property

8t-3 = 0    t+1 =0

8t = 3         t= -1

t = 3/8     t= -1

t cannot be negative  ( no negative time)

t = 3/8 seconds

5) Choose the best revision of the following statement: "All the isotopes of a particular element decay radioactively by
emitting electrons."
A. All the isotopes of a particular element are stable and do not decay.
B. Some isotopes are stable and others are unstable. Unstable isotopes decay by emitting various subatomic
particles and radiation
C. Some isotopes are stable and others are unstable. Unstable isotopes decay by emitting protons or
electrons.
D. The statement is correct as it is currently written.

Answers

Answer:

B. some isotopes are stable and others are unstable. unstable isotopes decay by emitting various subatomic particles and radiation.

Explanation:

test gave me the answer so yeah :/ XD

Please answer the questions... I will surely mark you as the brainliest according to me :)

Answers

Answer:

(a) You can tell that have the same strength because they have attracted the same amount of paper clips.

(b) Iron is used in electromagnets because steel retained magnetic properties after the power was turned off, but in the iron, the paper clips dropped off right away.

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