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?

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

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]

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

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

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].

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

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.

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

Answer:

yes

Explanation:

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.

(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

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

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.

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:

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

Answer:

naruto and hinata

Explanation:

Explanation:

m=200kg

a=2m/s2

F=ma

F=200kg×2m/s2

=400kg.m/s2 or 400N

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)))

Answer:

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

Explanation:

umm yeah no sorry I tried

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.

Answer:

8.33 m/s

Explanation:

v=d/s, velocity = displacement/ time

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!

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.

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

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

Answer:

same

Explanation:

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

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

Answer:

18 m/s

Explanation:

1.8 meters / 0.1 seconds = 18 m/s

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.

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.

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