Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless.
Identify all the action/reaction pairs of forces between the three skaters. Draw a free-body diagram for skater 2, in the middle.

Answer:

Kindly check explanation

Explanation:

The trampling and violation of human rights individuals, groups and corporate organizations is really alarming ad as such, the government who are charged to protect the right and interest of its citizen. In other to curtail the trending issues of human right violation, it is imperative if sensitization programmes could be organiz d in other to keep people informed of the various ways in which people's right may be trampled upon. With these education, the ignorance can be expunged leaving only those who genuinely decides to relate and threaten the right of his fellow country person.

The laws on human right violation should be reviewed and capital punishment metted on violators in other to send a strong warning to those who still nurture the intention.

Answer: its b bro

Explanation:

ajafa'jfbA'FJ

Answer:

A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...

Explanation:

One example of a physical entity that is quantized is:

The amount of money in your pocket.

The amount can't have any fraction of 1 cent.  

Its value must be an integer-multiple of cents, or 0.01 dollar.    

When it increases or decreases, it jumps from one integer number of cents to the next integer number.  It doesn't "slide" from one to the next.  It can never have a value between two integer numbers of cents.

Answer:

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]

Making K the subject;

[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]

[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Answer:

10.66 m/s

Explanation:

Applying,

The law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').............. Equation 1

Where m = mass of the first puck, u = initial velocity of the first puck, m' = mass of the second puck, u' = initial velocity of the second puck, V = Velocity of the center of the two pucks

make V the subject of the equation

V = (mu+m'u')/(m+m').............. Equation 2

Given: m = 0.18 kg, u = 16 m/s, m' = 0.14 kg, u' = 3.8 m/s

Substitute these values into equation 2

V = [(0.18×16)+(0.14×3.8)]/(0.18+0.14)

V = (2.88+0.532)/(0.32)

V = 3.412/0.32

V = 10.66 m/s

Answer:

[tex]t=2413s[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=1.5m/s[/tex]

Time [tex]t=30min=>30*60=>1800[/tex]

Distance [tex]d=24.3km[/tex]

Generally the Newton's equation for Speed going down the stream is mathematically given by

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]1.5+v=frac{24300}{1800}[/tex]

 [tex]v=12m/s[/tex]

Therefore

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]t=\frac{24300}{12-1.5}[/tex]

 [tex]t=2413s[/tex]

Answer:

Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.

Potential energy is when an object has some sort of potential eg. for motion such as in this example.

Answer:

A. Potential energy

Explanation:

Potential energy can be thought of as stored energy, which has the potential of becoming kinetic energy once it has been released.

Answer:

Magnitude = 15.86 units

direction = 69 degree below negative X axis

Explanation:

A = 20 units at 60.0° counterclockwise from the negative x - axis

B = 40 units at 30.0° counterclockwise from the positive x - axis

C = 35 units at 60.0° clockwise from the negative y - axis

Write the vectors in the vector form

[tex]\overrightarrow{A} =20 (- cos 60 \widehat{i} - sin 60 \widehat{j})=- 10\widehat{i} - 17.3 \widehat{j}\\\\\overrightarrow{B} =40 (cos 30 \widehat{i} + sin 30 \widehat{j})= 34.6\widehat{i} +20 \widehat{j}\\\\\overrightarrow{C} =35 (- sin 60 \widehat{i} - cos 60 \widehat{j})=- 30.3\widehat{i} - 17.5 \widehat{j}\\\\Now\\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = (- 10 + 34.6 - 30.3) \widehat{i} + (-17.3 + 20-17.5)\widehat{j}\\\\[/tex]

[tex]\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = - 5.7\widehat{i} -14.8\widehat{j}[/tex]

The magnitude is given by

[tex]= \sqrt{5.7^2 + 14.8^2} = 15.86 units[/tex]

The direction is given by

[tex]tan\theta = \frac{- 14.8}{- 5.7}\\\\\theta= 69^o[/tex]

below negative X axis.  

Answer:

3.83×10¯⁴ N

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +2.4x10¯⁸ C

Charge 2 (q₂) = +1.8x10¯⁶ C

Distance apart (r) = 1.008 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

The magnitude of the electrical force acting between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²

F = 0.0003888 / 1.016064

F = 3.83×10¯⁴ N

Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N

Answer:

a)  v_f = 5,06 m/s, b)  GAIN in kinetic energy.

Explanation:

a) For this exercise we will use the relationship between work and kinetic energy

           W = ΔK

Work is defined by

           W = F. d

bold indicates vectors

the displacement is

            d = x_f - x₀

            d = 4 -1

            d = 3i m

we calculate

          W = 20 10⁻² 3 i.i

let's remember that

         i.i = j.j = 1

         i.j = 0

          W = 6.0 10⁻¹ J

we substitute in the first equation

          W = K_f - K₀

          W = ½ m (v_f ² -v₀²)

          v_f ² = [tex]\frac{2W}{m} + v_o^2[/tex]

let's calculate

          v_f ² = 2 6.0 10⁻¹ /2 + 5²

          v_f = √25.6

          v_f = 5.06 m / s

b) we can see that the speed at the end of the movement is greater than the initial speed, therefore there is a GAIN in kinetic energy.

Answer:

[tex]a_2=3.88m/s^2[/tex]

Explanation:

From the Question we are told that:

Initial Mass [tex]m_1=1300kg[/tex]

Final mass [tex]m_2=1300+400=>1700kg[/tex]

[tex]a_1=3.0m/s^2[/tex]

Generally the equation for  Force  is mathematically given by

 [tex]F=m_1a_1[/tex]

 [tex]F=1300*5[/tex]

 [tex]F=6500N[/tex]

Generally the equation for Final acceleration  is mathematically given by

 [tex]F'=m_2*a_2[/tex]

 [tex]a_2=\frac{6500}{1700}[/tex]

 [tex]a_2=3.88m/s^2[/tex]

Answer:

The required wavelength is 1.19 μm

Explanation:

In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.

[tex]y_m = \dfrac{m \lambda D}{d}[/tex]

where;

m = fringe order

d = slit separation

λ = wavelength

D = distance between screen to the source

For the first bright fringe, m = 1, and we make (d) the subject, we have:

[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]

[tex]d = \dfrac{ \lambda D}{y_1}[/tex]

replacing the value from the given question, we get:

[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]

In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:

[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

The value of m in the dark fringe first order = 0

∴

[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

making λ the subject of the formula, we have:

[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]

[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]

Answer:

22.5 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

Answer:

In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid. In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid. 

Explanation:

this is the correct

Answer: 1 cal is 4.186 J, 1 kcal = 4186 J   A : 1014 m , B  200 m

Explanation:   A) Work done by climber is change in potential energy.

W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.

Solve h  =  160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m

B  Energy is only 20 %  :   Then  h  =  0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.

Actually, muscles also produce heat from most of the energy provided by food.

I am sure it help you with that much ☺️

Explanation:

pleasae give me some thanks please good morning sister

Answer:

The advantages of Hovercraft:

They can travel over almost any non-porous surface.

They can operate to and from any unprepared beach or slipway.

They take fast, direct routes compared to a conventional marine vessel.

Answer:

D because of POE

Explanation:

the reaction force is the ball exerting the same newtons of force back to your leg opposite of the ball, but the reaction force and the action force is never stronger than each-other. and action is only being done on the soccer ball, so process of elimination, the answer is D

Answer:

B

Explanation:

it can only have a different motion if it is acted upon a different ball

Answer:

OD. -9.8 m/s2

Explanation:

The only force vertical force that is acting on the skier is gravity and since its pulling him back it's a negative force down the y axis.

Answer:

The average power per day is 1008 kW.

Explanation:

Solar constant = 1.4 kW/m2

efficiency = 20 %

area, a = 1 square meter

time = 6 hours

Energy falling on the panel in 6 hours = 1.4 x 6 x 3600 kJ

The output is

= 20 % of 1.4 x 6 x 3600

= 0.2 x 1.4 x 6 x 3600

= 6048 kJ

Average power per day is

= 6048/6 = 1008 kW

Answer:

the charge of the bees must be of the same sign

Explanation:

The electric field is given by the relation

            E = k q / r²

This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.

The force generated by this field on a test charge is

            F = q E

Since the charge is a scalar, the direction of the force is the same as the electric field.

In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.

The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,

Consequently the charge of the bees must be of the same sign

Answer:

According to the data, factors influencing mining investment in South Africa's favour are the availability of labour and skills, the quality of the country's infrastructure, the quality of its geological database, and the State's environmental regulations.

Answer:

T = 56.11 N

Explanation:

Given that,

The equation of a wave is :

y = 0.08 sin(469t – 28.0x),

where x and y are in meters and t is in seconds

The linear mass density of the wave = 0.2 kg/m

The speed of wave is given by :

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Also,

[tex]v=\dfrac{\omega}{k}[/tex]

We have,

[tex]k=469\ and\ \omega=28[/tex]

Put all the values,

[tex]\dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}\\\\(\dfrac{\omega}{k})^2=\dfrac{T}{\mu}\\\\T=(\dfrac{\omega}{k})^2\times \mu[/tex]

Put all the values,

[tex]T=(\dfrac{469}{28})^2\times 0.2\\\\T=56.11\ N[/tex]

So, the tension in the string is 56.11 N.

Answer:

171.5 m

Explanation:

To find the distance, speed x time

342 x 0.5

171.5 m

Hope this helped!

Answer:

B. The Spring equinox

Explanation:

The vernal equinox marks the moment the sun crosses the celestial equator.  The vernal equinox happens on March 19, 20, or 21 every year in the Northern Hemisphere. In the Southern Hemisphere, this same event marks the beginning of fall.  (Source: What Exactly Is The Spring Equinox? - Dictionary.com)

Hopefully this helps.

Answer:

a.49 n

b. 63 n

c. 112 n

Explanation:

a.10 times 9.8 from gravity/2 = 49 n

b. 49n times 4.5/8-4.5 = 63 n

c 49n + 63 n = 112 n

Answer:

[tex]\mu_w=86\%[/tex]

Explanation:

From the question we are told that:

Velocity on Dry road [tex]V_d=10m/s[/tex]

Radius Dry [tex]r_d=4.8[/tex]

Radius wet [tex]r_w=11.8[/tex]

Generally the equation for coefficient of static friction on the dry day is mathematically given by

[tex]\mu mg=\frac{mv^2}{r_d}[/tex]

[tex]\mu g=\frac{v^2}{r_d}[/tex]

[tex]\mu_d 9.8=\frac{10^2}{4.8}[/tex]

[tex]\mu_d=2.125[/tex]

Generally the equation for the relationship between Radius & coefficient of static friction is mathematically given by

[tex]\frac{\mu_d}{\mu_w}=\frac{r_d}{r_w}[/tex]

[tex]\frac{\mu_w}{2.125}=\frac{4.8}{11.8}[/tex]

[tex]\mu_w=0.86[/tex]

Therefore

[tex]\mu_w=86\%[/tex]

Answer:

option C

Explanation:

as water vapor transfer heat, colored drops are seen inside the wrap.

Answer:

The area directly above the light bulb is warmest because of convection.

Explanation:

if all the sides of the bulb are equally close to the light source inside the bulb, all area of the bulb would be equally heated by conduction. however, convection heating mainly heats up the surface above the light source. in convection heating, the air above the surface of the light source get heated by the light source and expands, casuing it to be less dense and rise to the top of the bulb. colder denser air at the top of the bulb sink to the light source adn gain heat and expands, becoming less dense. this process repeats and the surface above the light source becomes the warmest due to convection heating