A ball of mass 0.600 kg is carefully balanced on a shelf that is 2.20 m above the ground. What is its gravitational potential energy?

Answers

Answer 1

Answer:

Explanation:

Gravitational potential energy = mass x height of object x g

g is gravitational acceleration .

Gravitational potential energy of the ball = .600 x 2.20 x 9.8

= 12.936 J .

12.9 J .


Related Questions

Are the refractive index and the speed of light in a vacuum direct propotional or inversley​

Answers

The refractive index of the medium is inversely proportional to the velocity of light in it. As the refractive index of a medium increases, the speed of light going through that medium decreases.

If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the internal energy of the gas?

Answers

Answer:

The change in the internal energy of the gas 1,595 J

Explanation:

The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)

This is summarized in the expression:

ΔU= Q - W

where the heat absorbed and the work done by the system on the environment are considered positive.

Taking these considerations into account, in this case:

Q= 500 cal= 2,092 J (being 1 cal=4.184 J) W=500 J

Replacing:

ΔU= 2,092 J - 500 J

ΔU= 1,592 J  whose closest answer is 1,595 J

The change in the internal energy of the gas 1,595 J

A ball is launched from ground level at 20 m/s at an angle of 40° above the
horizontal. A) How long the ball is in the air? B)What is the maximum
height the ball can reach?

Answers

(a) The ball's height y at time t is given by

y = (20 m/s) sin(40º) t - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :

0 = (20 m/s) sin(40º) t - 1/2 g t ²

0 = t ((20 m/s) sin(40º) - 1/2 g t )

t = 0   or   (20 m/s) sin(40º) - 1/2 g t = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 g t

t = (40 m/s) sin(40º) / g

t ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So

0² - ((20 m/s) sin(40º))² = 2 (-g) y

where y in this equation refers to the maximum height of the ball. Solve for y :

y = ((20 m/s) sin(40º))² / (2g)

y ≈ 8.4 m

What is the distance and the displacement of the race car drivers in the Indy 500?

Answers

Answer here
The displacement of the cars is somewhere near 0 miles since they virtually finish where they started. Yet the successful cars have covered a distance of 500 miles

The cars essentially finish where they started, their displacement is close to zero kilometers. However, the winning vehicles have traveled 500 miles.

What is displacement?

The separation between two places of an item in motion is known as displacement. Therefore, it relies on both the starting position and the ending position. Displacement is also the shortest distance between the initial and ultimate places.

The distinction between two locations of an object is known as displacement. Because it has both a direction and magnitude, it qualifies as a vector quantity. The symbol for it is an arrow pointing from the first position to the final position. The cars essentially finish where they started, their displacement is close to zero kilometers. However, the winning vehicles have traveled 500 miles.

Therefore,  their displacement is close to zero kilometers.

To learn more about displacement, here:

https://brainly.com/question/10919017

#SPJ2

show all work and round to nearest 100th thank you and you will get braineist ​

Answers

Answer:  16.78 miles

Explanation:

distance = rate * time

we're given the rate ( 5 m/s) and we're given the time it takes to get home (1.5 hrs). But notice how the units of hours don't match the seconds of the rate so we need to convert the hours into seconds.

1 hr = 3600 seconds so 1.5 hours = 5400 seconds

now we can plug it in and solve for the distance

distance = (5 m/s) * 5400 seconds

distance = 27000 m

now we have to convert meters to miles, so we divide our answer by 1609 and get 16.78

A block with a mass M = 4.85 kg is resting on a slide that has a curved surface. There is no friction. The speed of the block after it has slid along the slide sufficiently far for its vertical drop to be 19.6 m is:__________a. 19.6 m/s b. 384 m/s c. 93 m/s d. 43.2 m/s e. The problem cannot be solved because the shape of the curved slide is not given.

Answers

Answer:

The correct option is a

Explanation:

From the question we are told that

   The mass of the block is  [tex]m = 4.84 \ kg[/tex]

    The height of the vertical  drop is [tex]h = 19.6 \ m[/tex]

Generally from the law of energy conservation , the potential energy at the top  of the slide is equal to the kinetic energy at the point after sliding this can be mathematically represented as

        [tex]PE = KE[/tex]

i.e     [tex]m * g * h = \frac{1}{2} * m * v^2[/tex]

=>    [tex]gh = 0.5 v^2[/tex]

=>   [tex]v = \sqrt{\frac{9.8 * 19.6}{0.5 } }[/tex]

=>    [tex]v = 19.6 \ m/s[/tex]


Show the relation among MA, VR and n.

Answers

Answer:

good luck!!! sorry I just needed the points xoxo

Explanation:

umm yeah no sorry I tried

You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient
of friction of the floor is 0.53, what is the minimum angle, theta-min at which the ladder does nofip? What
do you think the maximum angle theta-max could be? Sketch and label your free body diagram.
(5 marks)​

Answers

Answer:

angle minimum   θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

    N - W = 0

    N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

     fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

     

the friction force formula is

     fr = μ N

     fr = μ W

we substitute

      μ m g l sin θ - m g l cos θ + mg l /2   cos θ = 0

      μ sin θ - cos θ + ½ cos θ= 0

         

       μ sin θ - ½ cos θ = 0

       sin θ / cos θ = 1/2 μ

       tan θ = 1/2 μ

       θ = tan⁻¹ (1 / 2μ)

       θ = tan⁻¹ (1 (2 0.57))

      θ = 41.3º


Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)

Answers

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then   m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

Just some Naruto couples having a Boxing Match.

Who do you think will win?! Naruto and Hinata or Pain and Konan?!

Answers

naruto and hinata for the win

Answer:

naruto and hinata

Explanation:

there’s a tornado warning where i live rn

Answers

Answer:

same

Explanation:

aww good luck, i hope u n ur family turn out okkk

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it around her head in a horizontal circle at 100 rpm. Assume the room temperature is 20 degrees Celsius. What are the highest and lowest frequencies heard by a student in the classroom?

Answers

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

Net force causes motion

Answers

Answer:

yes

Explanation:

If an object has a net force acting on it, it will accelerate. The object will speed up, slow down or change direction. An unbalanced force (net force) acting on an object changes its speed and/or direction of motion.

A track star runs a 100m race in 12s what is the velocity of the runner?

Answers

Answer:

8.33 m/s

Explanation:

v=d/s, velocity = displacement/ time

You use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0173 nm. What is the kinetic energy of an electron in the beam, expressed in electron volts?

Answers

Answer:

The kinetic energy of an electron in the beam is 5.04 keV.

 

Explanation:

We need to find the velocity of the electron by using the De Broglie wavelength:

[tex] \lambda = \frac{h}{mv} [/tex]

Where:

λ: is the wavelength = 0.0173 nm

v: is the velocity

m: is the electron's mass = 9.1x10⁻³¹ kg

h: is the Planck constant = 6.62x10⁻³⁴ J.s

[tex] v = \frac{h}{m\lambda} = \frac{6.62 \cdot 10^{-34} J.s}{9.1 \cdot 10^{-31} kg*0.0173 \cdot 10^{-9} m} = 4.21 \cdot 10^{7} m/s [/tex]

Now, we can find the kinetic energy:

[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*(4.21 \cdot 10^{7} m/s)^{2} = 8.06 \cdot 10^{-16} J*\frac{1 eV}{1.6 \cdot 10^{-19} J} = 5038 eV = 5.04 keV [/tex]

Therefore, the kinetic energy of an electron in the beam is 5.04 keV.

I hope it helps you!

What is the acceleration of gravity, in m/s2, on the surface (or outer limit) of Venus? The mass of Venus is 4.87 1024 kg and its radius is 6.05 106 m.

Answers

Answer:

8.9 m/[tex]s^{2}[/tex]

Explanation:

From Newton's law of universal gravitation,

F = [tex]\frac{GMm}{R^{2} }[/tex] .............. 1

and from Newton's second law of motion,

F = mg ........... 2

Equating the two expression,

mg = [tex]\frac{GMm}{R^{2} }[/tex]

g = [tex]\frac{GM}{R^{2} }[/tex]

Given that: mass of Venus = 4.87 x [tex]10^{24}[/tex] Kg, radius = 6.05 x [tex]10^{6}[/tex] and G = 6.67 x [tex]10^{-11}[/tex] N[tex]m^{2} Kg^{-2}[/tex]

Thus;

g = [tex]\frac{6.67*10^{-11}*4.87*10^{24} }{(6.05*10^{6} )^{2} }[/tex]

  = [tex]\frac{3.24829*10^{14} }{3.66025*10^{13} }[/tex]

 = 8.87450

g = 8.9 m/[tex]s^{2}[/tex]

the acceleration of gravity on the surface of Venus is 8.9 m/[tex]s^{2}[/tex].

Determine the force of gravitational attraction between a 92 kg student and a 550 g slice of pizza that are 25 cm apart

Answers

Answer:

F = 5.4 x 10⁻⁸ N

Explanation:

The gravitational force of attraction between two objects is given by Newton's Gravitational Law as follows:

F = Gm₁m₂/r²

where,

F = Gravitational Force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of student = 92 kg

m₂ = mass of pizza slice = 550 g = 0.55 kg

r = distance between student and pizza slice = 25 cm = 0.25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(92 kg)(0.55 kg)/(0.25 m)²

F = 5.4 x 10⁻⁸ N

You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat to the atmosphere at 27°C?

Answers

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

If a person Travels 100 metre due east and then returns to the same place his total displacement is 200. (needed ASAP)

A. True
B. False ​

Answers

I think it’s true because they traveled 100 meters east and 100 meters back so that adds up to 200 .

Distance is the total path covered by the object

Here, 200 m is the distance covered by the person and NOT the displacement

Displacement of an object is nothing more than the shortest path between the initial and the final point

If the person travelled 100m and came back, his initial and final point will remain the same which means that he will have a displacement of 0 m

Which possible component of initial energy is caused by molecular motion within a material?

Answer: thermal energy

Answers

Answer:

Thermal energy

Explanation:

The internal energy of a system is widely known as thermal energy. Now, thermal energy is also called heat energy and it is an internal energy of a component which is produced when an increase in temperature causes atoms and molecules within the component to move faster and start colliding with one other.

Therefore, the more heat the is applied to the component, the hotter the substance and the more its particles move which in turn leads to a higher thermal energy.

explain the relationship among visible light, the electromagnetic spectrum, and sight.

Answers

Explanation:

The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.

Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.

A cyclist is riding along at a speed of 20.7 when she decides to apply the brakes which gave a deceleration applied was a rate of -3.4 m/s2 over the span of 7.8 s. What distance does she travel over that period of time.

Answers

Answer:

The distance is 58.03 m

Explanation:

Constant Acceleration Motion

It occurs when the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

[tex]v_f=v_o+at[/tex]

The distance traveled by the object is given by:

[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]

The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.

The final speed is:

[tex]v_f=20.7+(-3.4)\cdot 7.8[/tex]

[tex]v_f=20.7-26.52[/tex]

[tex]v_f=-5.82\ m/s[/tex]

Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:

[tex]\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}[/tex]

[tex]\displaystyle x=161.46-103.43[/tex]

x=58.03 m

If you are pushing 200 kg of textbooks with acceleration of 2m/s2, how much net force are you exerting on the books? (Fnet=ma)
( There is more than one answer)

200N

100N

400Kg

400N

400 kg.m/s2


Answers

Explanation:

m=200kg

a=2m/s2

F=ma

F=200kg×2m/s2

=400kg.m/s2 or 400N

When you sweat, what is the external stimuli? I need help asap.

Answers

Answer:

An External Stimulus is a stimulus that comes from outside an organism. Examples: You feel cold so you put on a jacket. When you sweat, the external stimulus is either you're anxious or hot.

Explanation:

hope it helps! <3

The batter swings his bat 1.8 meters in 0.1 seconds. How fast is his bat speed in meters per second?

Answers

Answer:

18 m/s

Explanation:

1.8 meters / 0.1 seconds = 18 m/s

Find analytically the velocity of the object at the end point of the inclined plane for a certain angle Ө

Answers

I don't know if there is other given information that's missing here, so I'll try to fill in the gaps as best I can.

Let m be the mass of the object and v₀ its initial velocity at some distance x up the plane. Then the velocity v of the object at the bottom of the plane can be determined via the equation

v² - v₀² = 2 a x

where a is the acceleration.

At any point during its motion down the plane, the net force acting on the object points in the same direction. If friction is negligible, the only forces acting on the object are due to its weight (magnitude w) and the normal force (mag. n); if there is friction, let f denote its magnitude and let µ denote the coefficient of kinetic friction.

Recall Newton's second law,

F = m a

where the symbols in boldface are vectors.

Split up the forces into their horizontal and vertical components. Then by Newton's second law,

• net horizontal force:

F = n cos(θ + 90º) = m a cos(θ + 180º)

→  - n sin(θ) = - m a cos(θ)

→  n sin(θ) = m a cos(θ) ……… [1]

• net vertical force:

F = n sin(θ + 90º) - w = m a sin(θ + 180º)

→   n cos(θ) - m g = - m a sin(θ)

→   n cos(θ) = m (g - a sin(θ)) ……… [2]

where in both equations, a is the magnitude of acceleration, g = 9.80 m/s², and friction is ignored.

Then by multiplying [1] by cos(θ) and [2] by sin(θ), we have

n sin(θ) cos(θ) = m a cos²(θ)

n cos(θ) sin(θ) = m (g sin(θ) - a sin²(θ))

m a cos²(θ) = m (g sin(θ) - a sin²(θ))

a cos²(θ) + a sin²(θ) = g sin(θ)

a = g sin(θ)

and so the object attains a velocity of

v = √(v₀² + 2 g x sin(θ))

If there is friction to consider, then f = µ n, and Newton's second law instead gives

• net horizontal force:

F = n cos(θ + 90º) + f cos(θ) = m a cos(θ + 180º)

→   - n sin(θ) + µ n cos(θ) = - m a cos(θ)

→   n sin(θ) - µ n cos(θ) = m a cos(θ) ……… [3]

• net vertical force:

F = n sin(θ + 90º) + f sin(θ) - w = m a sin(θ + 180º)

→   n cos(θ) + µ n sin(θ) - m g = - m a sin(θ)

→   n cos(θ) + µ n sin(θ) = m g - m a sin(θ) ……… [4]

Then multiply [3] by cos(θ) and [4] by sin(θ) to get

- n sin(θ) cos(θ) + µ n cos²(θ) = - m a cos²(θ)

n cos(θ) sin(θ) + µ n sin²(θ) = m g sin(θ) - m a sin²(θ)

and adding these together gives

µ n (cos²(θ) + sin²(θ)) = m g sin(θ) - m a (cos²(θ) + sin²(θ))

µ n = m g sin(θ) - m a

m a = m g sin(θ) - µ n

m a = m g sin(θ) - µ m g cos (θ)

a = g (sin(θ) - µ cos (θ))

and so the object would instead attain a velocity of

v = √(v₀² + 2 g x (sin(θ) - µ cos (θ)))

your sister has gained admission into your former school write a letter giving her information about the school and advise her to behave well while in school​

Answers

Answer:

Find the letter below.

Explanation:

                                                                                   5183 Richmond Avenue,

                                                                                   Houston City,

                                                                                   Texas, U.S.A.

                                                                                   16th November 2020.

Dear Jane,

I received with joy the news of your admission into Texas City High School, my alma mater. This is a big feat and evidence of your hard work and commitment to your academics. To excel in this school, there are certain points which I would like you to note.

First, this is a new phase of your life and it would be important for you to clearly keep your purpose for going to school in mind. Your foremost purpose is to study and gain the needed knowledge that would help you in life.

Secondly, the school is a very sociable environment with lots of interesting activities such as sports, dancing, literary groups, and so many others. Find a group that appeals to you and participate actively in it. Extracurricular activities would help you socialize with your mates and teachers. It would also help you meet interesting people who might later serve some useful purpose in your life after graduation.

The teachers at Texas City High School are very kind and willing to help you excel in your academics and other chosen fields. Be sure to respect them and meet them when you have difficulties in any area of concern. There are also counselors that guide you and help you to remain on the path to excellence.

My dear Jane, life in Texas City High School would not always be rosy as there are bullies in school who would want to intimidate you or get you into fights. Avoid them by all means. Be cordial to all. Report any cases of harassment to the relevant authorities for they are seasoned in handling such matters maturely.

The cafeteria is also an interesting feature of the school. There you can find different kinds of snacks that I know you would love given how much you love snacks. I would only recommend that you make healthier choices so that you do not become overweight or fall ill due to overindulgence in unhealthy meals.

My dear baby girl, be assured of my love and kind wishes as you move on to this next stage of your life. Always know that I am here to give you my advice when you need it. Be a good girl my dear sister. Do have a lovely time at school.

Yours lovingly,

Queen Walters.

10points asap

A force of 30 N acts upon a 7 kg block. Calculate its acceleration.

Answers

Hello! Assuming that the only force acting on the mass is 30N...

Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
a = 30N / m
a = 30N / 7kg
a = 4.2857 m/s^2
a = 4 m/s^2

I hope this helps!

A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is launched to a maximum height 50.2 cm. How much should the spring be compressed to send the ball twice as high?

Answers

We know, by conservation of energy :

[tex]\dfrac{kx^2}{2}=mgh[/tex]

Therefore,

[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}[/tex]

Putting given values, we get :

[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm[/tex]

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 0.70 m/s. Determine the amplitude A of the motion.

Answers

Answer:

The amplitude of the motion is 0.0286 m.

Explanation:

Given;

mass of the object, m = 0.2 kg

spring constant, k = 120 N/m

maximum speed of the simple harmonic motion, [tex]V_m[/tex] = 0.70 m/s

The amplitude A of the motion is given by;

[tex]V_m = \omega A\\\\[/tex]

where;

ω is the angular velocity given as;

[tex]\omega = \sqrt{\frac{k}{m} }\\\\\omega = \sqrt{\frac{120}{0.2} }\\\\\omega =24.5 \ rad/s[/tex]

Now, substitute the value of angular velocity and solve the amplitude;

[tex]V_m = \omega A\\\\A = \frac{V_m}{\omega}\\\\A = \frac{0.7}{24.5}\\\\A = 0.0286 \ m[/tex]

Therefore, the amplitude of the motion is 0.0286 m.

Other Questions
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