A 2.0kg object is dropped from a height of 30m.
After it drops for 2.0 seconds, what is its kinetic
energy and what is its potential energy?
(Assume no air resistance.)

Answers

Answer 1

Answer:

1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules

2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J

Explanation:

1) The given mass of the object, m = 2.0 kg

The height from which the object is dropped, h = 30 m

The kinetic energy of the object after it drops for 2.0 seconds = Required

Kinetic energy, K.E. = (1/2)·m·v²

Where;

v = The velocity of the object

The kinematic equation for finding the velocity of the object is presented as follows;

v = u + g·t

Where;

u = The initial velocity of the object = 0

g = The acceleration due to gravity of the object ≈ 9.81 m/s²

t = The time of motion of the object = 2.0 seconds

∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s

The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;

[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J

2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;

The total mechanical energy, M.E. = P.E. + K.E.

M.E. = m·g·h + (1/2)·m·u²

∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J

M.E. = 588.6 J

Given that the total mechanical energy, M.E., is constant, we have;

At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.

∴ P.E. = 588.9 J - 384.9 J ≈ 204 J

The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J


Related Questions

A force of 10 N is making an angle of 30° with the horizontal. Its horizontal component will be:
A. 4N
B. 5N
C. 7N
D. 8.7 N

Answers

Answer:The answer is A

Explanation:just did it on a test

Answer: D

Explanation

To find the horizontal and vertical components of a force, we use cos the angle x the force and sin the angle x the force respectively.

So in this case we use cos the angle x the force to find the horizontal component.

1) cos30 x 10
2) 0.866 x 10
3) 8.66
4) ≈ 8.7 N

Hope this helps:)

Which of the following can be correct units for acceleration?
A. miles/hr/m
B. Km/s/hr
C. m/s/m
D. km/m/s

Answers

Answer:

B. Km/s/hr

Explanation:

An electromagnetic wave has a frequency of 6.0 x 10^18 Hz. What is the
wavelength of the wave? Use the equation 2 = and the speed of light as 3.0
x 108 m/s.

Answers

Answer:

Wavelength = 5 * 10^{-11} meters

Explanation:

Given the following data;

Frequency = 6.0 x 10^18 Hz

Speed = 3 * 10⁸ m/s

To find the wavelength of the wave;

Mathematically, the wavelength of a wave is given by the formula;

[tex] Wavelength = \frac {speed}{frequency} [/tex]

Substituting into the formula, we have;

[tex] Wavelength = \frac {3 * 10^{8}}{6.0 x 10^{18}} [/tex]

Wavelength = 5 * 10^{-11} meters

what will be the effect on the acceleration due to gravity of the earth if it is compressed to a the size the moon?​

Answers

Answer:

it gravitional pull on earth will increased becauste it is compress to a form of moon which is comperatively smaller so the gravitonal pull on per cm of earth will incrased so we can say that there will be change in acceleration due to gravity

What do you mean by unit?​

Answers

Unit is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature.

Answer:

The standard known quantity which is used to measure a physical quantities is known as unit.

A monatomic gas is measured to have an average speed of 1477 m/s. If the
total amount of the gas is 2 mol (which equates to a mass of 0.008 kg), what
is the approximate temperature of the gas? (Recall that the equation for
kinetic energy due to translation in a gas is KEtranslational = 1 mv2 = 3 nRT,
2
and R = 8.31 J/(mol-K).)
2

Answers

Answer:

The temperature of the gas is 350.02 K.

Explanation:

The average speed is related to the temperature as follows:

[tex] \bar v = \sqrt{\frac{3RT}{M}} [/tex]   (1)

Where:

[tex] \bar v [/tex]: is the average speed = 1477 m/s

R: is the gas constant = 8.31 J/(K*mol)                                                  

T. is the temperature =?

M: is the molar mass

First, let's find the molar mass:

[tex] M = \frac{m}{n} [/tex]

Where:

m: is the mass of the gas = 0.008 kg

n: is the number of moles = 2 mol

[tex] M = \frac{m}{n} = \frac{0.008 kg}{2 mol} = 0.004 kg/mol [/tex]

Hence, by solving equation (1) fot T we have:

[tex] T = \frac{\bar v^{2}*M}{3R} = \frac{(1477 m/s)^{2}*0.004 kg/mol}{3*8.31 J/(K*mol)} = 350.02 K [/tex]

Therefore, the temperature of the gas is 350.02 K.

I hope it helps you!            

(This is for other people with this Question i hope you find this when you need help) Need answer for this Help!!!
Question 7 of 20 You plan to use a slingshot to launch a ball that has a mass of 0.025 kg. You want the ball to accelerate straight toward your target at 19 m/s2. How much force do you need to apply to the ball? O A. 19.03 N OB. 0.48 N O C. 4.51 N D. 760.00 N​

Answers

Answer:

0.48N

Explanation:

according to the second law of motion

force=mass×acceleration

the mass in this question is 0.025,the acceleration 19

therefore f=0.025×19

=0.48N

I hope this helps

why do black holes have a large gravitational pull that even light cannot escape from

Answers

Answer:

Because matter has been squeezed into a tiny space.

Explanation:

According to NASA, this can happen when a star is dying.

A black hole has no more gravity than the same amount of matter in any other form.

But remember that the gravitational forces are stronger as you get closer to the center of the body. The mass of a black hole is packed into such a small size (theoretically zero !) that you can get very close to its center. THAT'S where its gravity is hugely strong.

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​

Answers

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, [tex]a_c[/tex], is given as follows;

[tex]a_c = \dfrac{v^2}{r}[/tex]

Therefore, the centripetal acceleration of the stone found as follows;

[tex]a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2[/tex]

The centripetal acceleration of the stone, [tex]a_c[/tex] = 5 m/s².

"Measurement is essential in our life. Justify the statement.​

Answers

Explanation:

Measurement plays an important role in our daily lives because they are useful to do basic tasks, such as taking the temperature of a child with the help of a thermometer,making time estimations,measure out medicines and find out weights, areas and volumes of different substances

Ignore friction. A 20. Lb. Object was lifted to a height of 10 feet by a force of 30. Lb. (a) How much work (ft.Lb) is done by the force? (b) Find the change in potential energy (ft.Lb) of the object

Answers

a)
W = Fd
W = (30Lb)(10 ft)
W = 300ft.Lb

b)
work-energy theorem: W = ∆E
∆E = 300ft.Lb

(a) The work done by force in lb-ft is 300 lb-ft.

(b) The change in potential energy of the object is 200 lb-ft.

Work done by force

The work done by force in lb-ft is calculated as follows;

W = Fd

W = 30 x 10

W = 300 lb-ft

Change in potential energy

The change in potential energy of the object is calculated as follows;

ΔP.E = P.Ef - P.Ei

ΔP.E = 20 x 10 - 0 = 200 lb-ft

Learn more about work done here: https://brainly.com/question/8119756

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Hai điện tích đặt cách nhau một khoảng R trong không khí thì lực tương tác
giữa chúng là 2.10−3N. Nếu khoảng cách đó mà đặt trong môi trường điện môi thì
lực tương tác giữa chúng là 10−3N. Để lực tương tác giữa hai điện tích đó khi đặt
trong môi trường điện môi bằng lực tương tác giữa hai điện tích đó khi đặt trong
không khí thì khoảng cách giữa 2 điện tích là bao nhiêu?

Answers

what language are you typing

f an object has a mass of 200 kg and a weight of 1000 N, what is g?

Answers

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

g = 5 m/s square

20.In case the conductor is a heating appliance, then this energy(w) is converted into heat(H) i.e

A. w=H C. w=I2Rt

B. w=VIt D. all of above​

Answers

Answer:

not sure just need points

Explanation:

a+b+c

A horse is running on a circular path with constant speed but its direction is changing at every point. It is

Answers

Answer:

accelerating.

Explanation:

A horse is running on a circular path with constant speed but its direction is changing at every point. Thus, it is accelerating.

.........도움......
Is this correct? (the option is marked)​

Answers

Answer:

sorry I don't understand this language

Please please helpjejejnebwbww​

Answers

Answer:

C. 540 N

Explanation:

Let suppose that system formed by the athlete and the load are in equilibrium. By Newton's Laws of Motion, we use the moment equation with respect to the feet of the athlete to determine the upward force exerted by his two arms:

[tex]\Sigma M = -F\cdot r_{1} + m_{L}\cdot g \cdot r_{2} + m_{A}\cdot g \cdot r_{3} = 0[/tex]

[tex]F = \frac{(m_{L}\cdot r_{2} + m_{A}\cdot r_{3})\cdot g}{r_{1}}[/tex] (1)

Where:

[tex]m_{L}[/tex] - Mass of the load, in kilograms.

[tex]m_{A}[/tex] - Mass of the athlete, in kilograms.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]r_{1}[/tex], [tex]r_{2}[/tex], [tex]r_{3}[/tex] - Distances with respect to the feet, in meters.

[tex]F[/tex] - Upward force exerted by his two arms, in newtons.

If [tex]m_{L} = 6\,kg[/tex], [tex]r_{2} = 1.20\,m[/tex], [tex]m_{A} = 70\,kg[/tex], [tex]r_{3} = 0.90\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r_{1} = 1.30\,m[/tex], then the upward force is:

[tex]F = \frac{[(6\,kg)\cdot (1.20\,m)+(70\,kg)\cdot (0.90\,m)]\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{1.30\,m}[/tex]

[tex]F = 529.578\,N[/tex]

The upward force exerted by his two arms is 529.578 newtons. (Right answer: C)

3 An un calibrated mercury in glass thermometer immersed in melting ice. The length of the mercury thread is 25 mm when the thermometer immersed in steam from pure water boiling under a pressure of 1 atmosphere the length of the thread is 200 mm what is the temperature in degree centigrade when the length of the thread is 95mm.​

Answers

Answer:

25 mm = 0 deg C

200 mm = 100 deg C

200 - 25 = 175 = change in thread per 100 deg C

95 - 25 = 70 mm - change in thread from 0 deg C

70 / 175 * 100 = 40 deg C    final temperature at 95 mm

A car increase its speed steadily from 30km/hr to 60km/hr in 1 min A what is the average speed during this time

Answers

Explanation:

initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s

final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s

time taken(t) = 1 minutes

= 60 seconds

Now,

Average velocity = u+v/2

= 8.33 m/s + 16.67m/s÷2

=12.5 m/s

A. Tick (1) the best alternatives. 1. What is the acceleration due to gravity on the surface of moon ? (a) 9.8m/s (b)1.6m/s2 (c) 6.67x10-1m/s (d) 9.8m/s? ​

Answers

Answer:

[tex] \green{ \sf \: \: 1.6 \: m {s}^{ - 2} \: \: \: is \: the \: correct \: answer}[/tex]

Explanation:

[tex] \sf \: \huge{g } _{ \small{moon}} = \frac{ {\huge{g}}_{earth}}{6} \\ \\ \sf \implies \: \sf \: \huge{g } _{ \small{moon}} = \frac{ 9.8}{6} = 1.6 \: \: m {s}^{ - 2} [/tex]

which species are near extinctions and where do they live now?

Answers

Answer:

Amur Leopard, Southeastern Russia, and northern china

Explanation:

Answer:

Amur Leopard    China/Russia

Hawks bill Turtle   live in Oceans

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.

Answers

[tex] \\ \tt \longmapsto \: h = - 4.8 {t}^{2} + 8.2t + 1.8 \\ \\ \tt \longmapsto \: h = - 4.8(1.7) {}^{2} + 8.2 \times 1.7 + 1.8 \\ \\ \tt \longmapsto \: - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 17.06[/tex]

If the final ​velocity is 0. third equation of motion will be

Answers

Answer:

vf²=vi²+2a∆x

Explanation:

The third equation of motion gives the final velocity of an object under uniform acceleration given the distance traveled and an initial velocity: v 2 = v 0 2 + 2 a d . v^2=v_0^2+2ad. v2=v02+2ad. The graph of the motion of the object.

A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an inclined plane. The surface of the incline is 2.75 m long, and is tilted at an angle of 22.0∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed v2 at the bottom of the ramp.

Answers

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

[tex]K_i[/tex] + [tex]U_i[/tex] = [tex]K_f[/tex] + [tex]U_f[/tex]

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K =  (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×[tex]v_f[/tex]² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75×[tex]v_f[/tex]²

[tex]v_f[/tex]² = 77.93/(7/10 × 4.75)

[tex]v_f[/tex] ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, [tex]v_f[/tex] ≈ 4.84 m/s.

Question 3 of 10
Which image shows an example of the strong nuclear force in
action?
A.
B.
C.
D.

Answers

Answer: The answer is B

Explanation:

There are 4 fundamental forces that hold matter together.

- Gravitational Force

- Electromagnetic Force

- Strong Nuclear Force

- Weak Nuclear Force

We have barely just scratched the info about nuclear forces but the reason why B is the answer to the question is that Strong nuclear force actually holds the protons and neutrons together in the nucleus of an atom, much like the picture in B.

Answer: B

Explanation:

You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to the right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.

Please help me with this

Answers

Answer:

Explanation:

From your other post, the complete question is:

"You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to the right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.

Find the sum of the force on the sled.

Determine the acceleration of the sled .

If the sled has an initial velocity 2m/s to the right, how fast will it be traveling after 5 seconds?

"

Given the rope exerting a force of 50 N to the right and the snow exerts a friction force of 30 N to the left, the sum of forces

= 50 - 30

= 20N to the right

The mass of sled is 50 kg and force = mass * acceleration.

Acceleration = Force / mass

= 20 / 50 = 0.4 m/s^2 to the right

If the sled has an initial velocity 2m/s to the right, after 5 seconds it will be traveling at initial velocity + acceleration * time

= 2 + 0.4*5

= 4 m/s to the right

Answer:

Explanation:

net force = applied force - friction force

= 50-30

= 20N to right

acceleration = net force/mass

= 20/50

= 2/5m/s tp right

final velocity = initial velocity + acceleration*time

= 2+2/5*5

= 4m/s to right

Tính công của dòng điện

Answers

Answer:

CG gh sure er go b vh pxuh FPI OO c AM h kh

How can a simulation study be used to learn about a planet we have never visited?
In a simulation study you imitate some conditions and characteristics of the real planet. The simulation can be used as a substitute for real data collected on the planet.
Simulation studies can be used for planets we have already visited.
In a simulation study you imitate as many conditions and characteristics as possible of the real thing, in this case the planet you have not visited. By simulating as many conditions as possible you can visualize what the planet might be like and learn things about it without actually visiting it.
Simulation studies cannot be used to learn about a planet we have never visited because we do not know anything about the planet.

Answers

Hi! I think the answer is C. as we have never visited the planet we can’t be 100% sure but we can have an idea of what the planet might me like. With experimenting with the simulation it can help you get an understanding or an idea of what the planet might be like exactly like option C. I hope this helped Goodluck :)

A simulation study that is used to learn about a planet by simulating as many conditions as possible, you can visualize what the planet might be like and learn things about it without actually visiting it. The correct option is c.

What is simulation?

A simulation is a method of simulating a process or change in the actual world to forecast future events or to explain past events and their causes. These days, simulations are frequently carried out using computers. Scientists use simulations to find answers and to test complicated systems.

A fire drill is used in this instance to get everyone ready for an impending event. In fire drills, the fire alarm is sounded even when there isn't actually a fire.

Therefore, the correct option is c.

To learn more about simulation, refer to the link:

https://brainly.com/question/16359096

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a sky driver jumps from an aircraft. the mass of sky driver sky driver is 70kg. state the equation linking weight,mass,and gravitional field strenght.

Answers

Answer:

Weight = Mass × Gravitational field strength

Explanation:

The mass of the skydiver = 70 kg

The weight of the skydiver, W, is given as follows;

W = m × g

Where-

g = The gravitational field strength ≈ 9.81 m/s²

Therefore, the equation linking weight mass and gravitational field strength is presented as follows;

Weight = Mass × Gravitational field strength


What the velocity from the graph given above?

Answers

Answer:

i think its 4 or 35

Explanation:

its in the middle of 40 and 30

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