The columns in the periodic table are called groups. What do the elements in group 18 have in common
A. They are found only in compounds in nature

B. They tend to form ions of a -1 Electric charge

C. They tend to form ions with a +1 electric charge

D. They rarely react with any other elements

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

   Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

  vo = 15.5 m/s

The speed of car driving is 15.5 m/s as the car is parked and drive away.

Speed is defined as a measurement of the length of time it takes for an object to travel a certain distance. You can determine an object's speed if you know how far it moves in a given amount of time. Time does not move, hence there is no concept of a speed of time. Time refers to how we move through the temporal realm. Speed is a unit of measurement for how quickly something is moving. A change in velocity results in a change in speed.

To calculate the speed we use the formula

f₀ = f (v + vo) / (v + vs)

Δf = f approach - f depart

69.5 = (769(343 + vo) / (343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

vo = 15.5 m/s

Thus, the speed of car driving is 15.5 m/s as the car is parked and drive away.

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Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree

During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.

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The ratio of the primary turns to the secondary turns is 1/3

The correct answer to the question is Option A. 1/3

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

Ratio = Nₚ/Nᵣ

Nₚ/Nᵣ = 225 / 675

Nₚ/Nᵣ = 1/3

Therefore, the ratio of the primary turns to the secondary turns is 1/3

Complete question:

See attached photo

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At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is  4.51 sec.

The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.

Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,

force = ma = μmg, putting the value,

a = - 1.175 m/sec²

now using equation of motion

v = u + at

t = 4.51 sec

So, time taken by the puck to come to rest is 4.51 sec.

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Answer:

11.573

Explanation:

f = m*a

where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration

so, we solve for a

a = f/m

a = 22/1.901

a = 11.573

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

Answer:

When buying a margarine, people should look for trans fat and saturated fat content. If it is over 2%, then it is not healthy. If it is lower that means it is healthy and fresh!

Explanation:

Answer:

Which extended definition would be most helpful to add to this body paragraph?

Laughter Yoga, a new form of yoga, is becoming increasingly popular.

Laughter Yoga, which is becoming more and more popular, was created in the 1990s.

Laughter Yoga is modeled after traditional yoga but includes laughter exercises.

Laughter Yoga is a form of yoga that was created by Dr. Madan Kataria in the 1990s.

Explanation:

c

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

The wavelengths of the constituent travelling waves is calculated as follows;

[tex]L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}[/tex]

for first mode: n = 1

[tex]\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm[/tex]

for second mode: n = 2

[tex]\lambda = \frac{2L}{2} = L = 100 \ cm[/tex]

For the third mode: n = 3

[tex]\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm[/tex]

For fourth mode: n = 4

[tex]\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm[/tex]

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

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Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

The locations where the riders feel less than their normal weight are Location A, Location C and Location D.

The given parameters;

The normal weight of the riders is calculated by applying Newton's second law of motion as follows;

W = mg

W = 9.8m

The apparent weight of the riders for the upward acceleration is calculated  as follows;

[tex]R = m(g + a)[/tex]

The apparent weight of the riders for the downward acceleration is calculated  as follows;

[tex]R = m(g - a)[/tex]

The apparent weight of the riders at location A is calculated as follows;

[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]

The apparent weight of the riders at location B is calculated as follows;

[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]

The apparent weight of the riders at location C is calculated as follows;

[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]

The apparent weight of the riders at location D is calculated as follows;

[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]

The apparent weight of the riders at location E is calculated as follows;

[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]

Thus, the locations where the riders feel less than their normal weight are;

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This question involves the concepts of the thin lens formula, focal length, and image distance.

The image of the house is "16 ft" away from the lens.

According to the thin lens formula:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

where,

f = focal length = 8 ft

p = object distance = 16 ft

q = image distance = ?

Therefore,

[tex]\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\[/tex]

q = 16 ft

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homework

or just play some 2012 math game

Answer:

I think calculating distance (if thats an option)

Explanation:

Really hope this helped!

Answer:

do it got a picture

Explanation:

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

The wavelength will remain unchanged.

Explanation:

The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is

[tex]v = \lambda\nu[/tex] (1)

so if we double both the velocity and the frequency, the equation above becomes

[tex]2v = \lambda(2\nu)[/tex] (2)

Solving for the wavelength from Eqn(2), we get

[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]

We would have gotten the same result had we used Eqn(1) instead.

Answer:

the wavelength increases

Explanation:

Answer:

Explanation:

I will us θ as the angle above horizontal as I don't know how to make your chosen symbol for angle.

           KEy = PE

½m(vsinθ)² = mgh

                h = (vsinθ)²/2g

Answer:

Explanation:

v² = u² + 2as

0² = u² + 2(-4.7)(235)

u² = 2209

u = 47 m/s

Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.

Explanation:

We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:

Vt = ωr with ω being the angular velocity.

We now calculate ω:

ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s

Now that we have ω we can calculate the centripetal aceleration:

a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2

Answer:

60w

Explanation:

The power required is 30 Watt.

Let us recall that power is defined as the rate of doing work. Hence, we can write as follows;

Power = Work done/ time taken

Now;

work done =  Force × distance

Force = 30.0 N

Distance = 10.0 m

work done = 30.0 N × 10.0 m = 300 J

The power expended = 300 J/10.00 s = 30 Watt

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The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1.

     a) The stopping distance is: x = 270 m

     b) The initial velocity is: v₀ = 17.3 m / s

     c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

Given parameters

To find

Question 1.

    a) Minimum braking distance.

    b) If the distance is x = 40.0 m, what speed should the vehicles have?

    c)  Conclusive importance of physics in daily life.

Question 2.

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

         v² = v₀² - a2 x

Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.

Newton's second law states that the net force is proportional to the mass and the acceleration of the body.

          F = ma

Where F is force, m is mass and acceleration.

In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body

        fr = m a

        a =[tex]\frac{fr}{m}[/tex]  

        a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]  

        a = 3.75 m / s²

This acceleration is in the opposite direction to the speed.

Let's find the distance needed to stop, the final speed is zero.

          0 = v₀² - 2 ax

           x = [tex]\frac{v_o^2 }{ 2a}[/tex]  

Let's calculate.

          x = [tex]\frac{45^2 }{2 3.75}[/tex]  

          x = 270 m

This is the minimum distance that the two vehicles must separate to avoid a collision.

b) We look for speed.

        v₀ = [tex]\sqrt{2ax}[/tex]  

        v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]  

        v₀ = 17.3 m / s

c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.

           0 = v₀² - 2ax

Let's substitute.

             a = [tex]\frac{v^2}{2d}[/tex]  

They ask the distance traveled if this car traveled from an initial speed 2v.

             0 = v² - 2 a x

             x = [tex]\frac{v^2}{2a}[/tex]  

We substitute

            x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]  

            x = 4 d

In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1

       a) The stopping distance is: x = 270 m

       b) The initial velocity is: v₀ = 17.3 m / s

        c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

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Answer:

B

Explanation:

The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.

Answer:

b it is b part answer i think so

Answer:

Explanation:

For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.

For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.

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Answer:

more likely it is to have gravity

Answer:

true answer this question