Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a mass of 0.5 kilograms, falls off the edge of the scaffolding. Calculate the bucket’s kinetic and potential energy when it is 4 meters above the ground. Also calculate its velocity at this point.

Answers

Answer 1

Answer:

.............................


Related Questions

If a person is pushing a cart with a force of 40 Newtons and it
accelerates at 0.5 m/s², what is the mass of the cart?

Answers

Answer:

80kg

Explanation:

A = Fm

0.5 = 40m

m = 40/0.5

m = 80kg

List examples of how the Bill of Rights protects you:
.
.
.
.
.

Answers

guaranteeing freedom of speech, press, assembly, and exercise of religion

A person who sits on the right-hand seat of the car that is making a left turn slides over to the right. What is the possible reason for this?

Answers

Answer:

zsjgbjidasngwhugbhwuabvhuvbwhuebewghvwev

Explanation:

11111111111111111111111111111111111111111111111111111

To minimize signal distortion, at each end of the J-1939 CAN-bus there is a(n)_____________resistor.

Answers

5-ohm
Extra
Variable
120-ohm
Variable
Pg. 614

please help me guys please​

Answers

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Solution is in attachment ~

I hope that you got what you were looking for, and if there's different data then go through the same procedure, using same formula with different values and you will get your answer ~

[tex]\mathrm{✌TeeNForeveR✌}[/tex]

Answer:1.9 × 10³⁰

Explanation:

100N force is applied to a ball with a mass of 0.25kg. Calculate the ball acceleration.

Answers

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

The Maximum and Minimum resultant of two forces are 35N and 51N. What is the possible Value of the two forces ?​

Answers

hplhevsdvakxjb1zonsvaka shdv

URGENT NEED GIVING POINTS question is how do i find R and the angle THANK YOU

Answers

Answer:

Theta = 26.93°

R = 0.38m

Explanation:

Theta = cos^-1(0.74m/0.83m)

Theta = 26.93°

R = sqrt(0.83²-0.74²)

R = 0.38m

Does the mass of an object always equal the sum of its parts?

yes

no

Answers

I think the correct answer is no

A car accelerates from 20m/s to 30m/s in 10 sec. Find the cars acceleration using v=u+at

Answers

Answer:

1

Explanation:

because when we use v=u+at we derived v-u÷t

Mjchddjnfndndnddnndnddn

Answers

Answer:b

Explanation:

Listed below are four activities. In which activity is friction useful?
a. climbing up a cliff
b. running on the ground
c. opening a bottle
d. catching a ball

Answers

Answer:

a. climbing up a cliff

Explanation:

Rory uses a force of 25 N to lift her grocery bag while doing 50 J of work. How far did she lift the
grocery bag?

Answers

Answer:

[tex]W=Fd \\ 50 = 25d \\ d = \frac{50}{25} \\ \color{yellow} \boxed{d = 2m}[/tex]

A 22 kg body is moving through space in the positive direction of an x axis with a speed of 190 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 13 kg, moves away from the point of explosion with a speed of 130 m/s in the positive y direction. A second part, with a mass of 2.2 kg, moves in the negative x direction with a speed of 460 m/s. What are the (a) x-component and (b) y-component of the velocity of the third part

Answers

Answer:   Our notation is as follows : the mass of the original body is M=20.0kg ; its initial velocity is  

ν

 

0

=(200m/s)  

i

^

 ; the mass of one fragment is m  

1

=10.0kg ; its velocity is  

ν

 

1

=(100m/s)  

j

^

 ; the mass of the second fragment is m  

2

=4.0kg ; its velocity is  

ν

 

2

=(−500m/s)  

i

^

 ; and , the mass of the third fragment is m  

3

=6.00kg . Conservation of linear momentum requires  

          M  

ν

 

0

=m  

1

 

ν

 

1

m  

2

 

ν

 

2

+m  

3

 

ν

 

3

 .  

The energy released in the explosion is equal to ΔK , the change in kinetic energy .  

(a) Using the above momentum -conservation equation leads to  

             

ν

 

3

=  

m  

3

 

M  

ν

 

0

−m  

1

 

ν

 

1

−m  

2

 

ν

 

2

 

 

             =  

6.00kg

(20.0kg)(200m/s)  

i

^

−(10.0kg)(100m/s)  

j

^

−(4.0kg)(−500m/s)  

i

^

 

 

          =(1.00×10  

3

m/s)  

i

^

−(0.167×10  

3

m/s)  

j

^

.  

The magnitude of  

ν

 

3

 is  

                        ν  

3

=  

(1000m/s)  

2

+(−167m/s)  

2

 

=1.01×10  

3

m/s  

It points at θ=tan  

−1

(−167/1000)=−9.48  

 (that is at 9.5  

 measured clockwise from the +x axis) .  

(b) The energy released is ΔK :  

             ΔK=K  

f

−K  

i

=(  

2

1

m  

1

ν  

1

2

+  

2

1

m  

2

ν  

2

2

+  

2

1

m  

3

ν  

3

2

)−  

2

1

Mν  

0

2

=3.23×10  

6

J

Explanation:

If F = 4.0 N and m = 2.0 kg, what is the magnitude a of the acceleration for the block shown below? The surface is frictionless. *

Answers

Since f=ma when f=4.0 and m=2.0 so the magnitude of acceleration for the block is 2.0m

The magnitude of the acceleration of the block is 3.5 [tex]\bold{m/s^2.}[/tex]

The correct option is (C) 3.5 [tex]\bold{m/s^2.}[/tex]

What do you mean by acceleration?

Acceleration is the rate of change of velocity with time.

By Newton’s Second law

This law establishes the relationship between the net applied force on an object with the resultant acceleration. For a constant magnitude of force, the more is the mass of the object, the lesser will be its acceleration.

Given,

The mass of the block is 2.0 kg.

The force in the horizontal direction on the block is [tex]\bold{4.0 N + 4.0 N \cos 40{}^\circ.}[/tex]

a is the acceleration in the horizontal direction.

The acceleration for the block can be determined using Newton’s Second law,

[tex]\begin{aligned} \Sigma F&=ma \\ F+F\cos 40{}^\circ &=ma \\ a&=\frac{F\left( 1+\cos 40{}^\circ \right)}{m} \\ &=\frac{4.0\text{ N}\left( 1.766 \right)}{2.0\text{ kg}} \\ &=3.5\text{ m/}{{\text{s}}^{2}} \end{aligned}[/tex]

Thus, the acceleration is 3.5 [tex]\bold{m/s^2.}[/tex]

Learn more about acceleration, here:

https://brainly.com/question/2437624

Imagine you have a ball tied to the end of a string. You hold the other end of the string and swing it around. Suppose the string breaks, what direction will the ball travel

Answers

It will move in a curving motion starting in the direction where it snapped from. In the end it goes down :D

A child is stationary on a swing.(a)The child is given a push by his brother to start him swinging.His brother applies a steady force of 84 N over a distance of 0.25 m.(i) Calculate the work done by this force.(2)..............................................................................................................................................(ii) State how much energy is transferred by this force.(1)..............................................................................................................................................(iii) After several more pushes, the child has a kinetic energy of 71 J.The mass of the child is 27 kg.Show that the velocity of the child at this point is about 2.3 m/s.(2)(iv) Which one of these quantities changes in both size and direction while he is swinging?Put a cross ( ) in the box next to your answer.(1)Ahis gravitational potential energyBhis momentumCthe force of gravity acting on himDhis kinetic energy

Answers

Answer:

Work done = Fs

Explanation:

(i) W = Fxs

= 84N x0.25m

= 21 Joules

(ii) Ek = 1/2mv^2

71 = 1/2×27×v^2

71×2/27 = v^2

v = 23m/s

The work done by this force is 21 Joule.

Energy transferred by the force is 21 Joule.

The boy's momentum changes periodically, that is,  in both size and direction.

What is force?

An external force is an agent that has the ability to change the resting or moving condition of a body. It has a direction and a magnitude. So, it is  a vector quantity. Newton is the SI unit of force (N).

(i) Given parameter:

Steady force applied by his brother, F = 84 N.

Displacement, d = 0.25 m.

The work done by this force, W = F.d = 84×0.25 = 21 Joule.

(ii) Amount of work done is transferred by this force as an energy. So, amount of energy transferred by the force is 21 Joule.

(iii) Given parameter:

Mass of the child, m= 27 kg.

Kinetic energy, E = 71 J.

Let, velocity of the child = v, then kinetic energy is,

E = 1/2 × mv²

⇒ v = √(2E/m)

= √(2×71/27)

= 2.3 m/s.

Hence, the velocity of the boy is 2.3 m/s. (proved).

(iv) During swing, his velocity changes periodically. That's why, his momentum changes periodically, that is,  in both size and direction.

Learn more about force here:

https://brainly.com/question/13191643

#SPJ2

Explain when acceleration remains constant.​

Answers

Explanation:

acceleration remains constant when velocity does not increase or decrease

which acts as a transverse wave with particle motion perpendicular to wave motion?

Answers

Answer:

transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.


2. Given what you know about the acceleration of Earth's gravity (g = 9.8 m/s2), is this number accurate?
accurate. If not explain why you think it is not accurate. Pleaseee help mee

Answers

Answer:

it is correct

Explanation:

Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate

Asap

A cyclist needs to spin 100 meters with a velocity of 5 m/s. How long will the cyclist take?

Answers

Answer:

[tex]v = \frac{d}{t} \rightarrow t= \frac{v}{d} \\ = 100m \div 5m. {s}^{ - 1} = \frac{100}{5} \\ \color{green} \boxed{ t = 20s}[/tex]

Find the gravitational potential energy of a body of mass 25kg,kept at a height of 4m

Answers

Answer:

Massm=2.5kg

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mgh

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15

Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15U=25/10×10×=375j

Jefferson refers to
_____ rights in the Declaration of Independence.
Explain what these rights are:

Answers

Answer: natural

Explanation:

Answer:

natural i just copy tha answer in there i just explain it(づ ̄ 3 ̄)づ

Explanation:

The meaning of the term “Pursuit of Happiness.” In the Declaration of Independence, Thomas Jefferson announced that every human being has “certain unalienable rights,” among which are those to “life, liberty, and the pursuit of happiness.” What did he mean by “the pursuit of happiness”?

a comet or asteroid impact on the earth is a frequently suggested cause for many of the mass extinctions. for which of the major mass extinctions does the most evidence exist that this was the main trigger for the event?

Answers

Answer:

Cretacious-Tertiary

Explanation:

I hope this helps!

Find the number of moles in a 28.0g sample of NH3.

PLS URGENTLY

Answers

Answer:

Molar mass of NH3=14+3=17g

[tex]mole = \frac{given \: mass}{molar \: mass} \\ mole = \frac{28}{17} \\ mole = 1.65[/tex]

BRAINLIST

What is the weight, on Earth, of a book with a mass of 1. 5 kg? 1. 5 N 6. 5 N 11. 3 N 14. 7 N.

Answers

Answer:

The weight, on Earth, of a book with a mass of 1.5kg is 14.7N.

Explanation:

Hi there!

Weight of an object = mass (kg) × acceleration due to gravity (N/kg)

Plug in the given values using the g constant g = 9.8 N/kg:

W = 1.5 × 9.8 = 14.7 N

How to find resistance

Answers

Answer:

R = V/I

Explanation:

Example:

V = IR

R = V/I

If we were to use V = 120 and I = 0.25

R = 120/0.25

R = 480 Ohms

Your resistence would equal Ohms

IF THE VECTOR COMPONENTS OF VECTOR A ARE MULTIPLIED BY 9 THEN THE MAGNITUDE OF VECTOR A IS INCREASES BY A FACTOR OF .........

Answers

Answer:

it increased by 29 sorry hope this helps

what is the average speed of the toy car during the two trials to the nearest tenth of m/s

wait nvm the answer is D

Answers

Answer:

what is the average speed of the toy car during the two trials to the nearest tenth of m/s

wait nvm the answer is B

Explanation:

CARRY ON LEARNING

1 point
What is the potential energy of a bird flying, if the bird has 1000 J of total
energy and 450 J of kinetic energy? (only put the number, no units or
commas)

Answers

Answer:

the bird has 550J of potential energy

Explanation:

PEtotal=PE+KE

1000J=PE+450J

subtracting the kinetic energy from the total we get:

1000J-450J=PE

550J=PE

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