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ElenaW [278]
3 years ago
5

During the motion of the slinky in a transverse wave, what do the particles of the slinky coil do?

Physics
1 answer:
Mazyrski [523]3 years ago
5 0

Answer:

C.) The slinky particles move up and down

Explanation:

<u>Transverse Wave</u>-

<em>A wave that has a disturbance perpendicular to the wave motion</em>

<em></em>

<em>Hello! This is the correct answer! Have a blessed day! :)</em>

<em>If you are in K12, please review the lesson! :)  It will give you some very helpful definitions!  I hope this helped!</em>

<u />

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A small metal sphere has a mass of 0.11 g and a charge of -21.0 nC . It is 10 cm directly above an identical sphere with the sam
GrogVix [38]

Answer:

A. the magnitude of the force between the spheres is 3.97 x 10⁻⁴ N

B. the magnitude of its initial acceleration is 5.83 m/s²

Explanation:

given information:

metal sphere's mass, m = 0.1 g = 1 x 10⁻⁴ kg

charge, q = -21 nC = -2.1 x 10⁻⁸

r = 10 cm = 0.1 m

What is the magnitude of the force between the spheres?

F₁₂ = k q₁q₂/r²

     = ( 9 x 10⁹) (-2.1 x 10⁻⁸)²/(0.1)²

     = 3.97 x 10⁻⁴ N

If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?

mg - F₁₂ = ma

a = g - (F₁₂/m)

   = 9.8 - (3.97 x 10⁻⁴/1 x 10⁻⁴)

   = 5.83 m/s²

5 0
3 years ago
A train accelerates from 23m/s to 190m/s in 54 seconds. What was its acceleration?
Ierofanga [76]

Answer:

Use the method on the image and solve it.

3 0
4 years ago
A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a
worty [1.4K]

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

7 0
3 years ago
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the
Law Incorporation [45]

Answer:

y and length is directly relation

Explanation:

Given data

A single-slit diffraction pattern is formed on a distant scree

angles involved = small

to find out

what factor will the width of the central bright spot on the screen change

solution

we know that  for single slit screen formula is

mass ƛ /area = sin θ and y/L = sinθ

so we can say mass ƛ /area =  y/L

and y = mass length  ƛ / area       .................1

in equation 1 here we can see y and length is directly relation so we can say from equation 1 that  the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled

8 0
3 years ago
You walk 10 miles north, 20 miles south, and 5 miles north. Taking north to be the positive y direction, define a position vecto
gulaghasi [49]

Considering north as positive Y axis

       First  10 miles north means a positive 10 displacement along y direction

      Second 20 miles south means a negative 20 displacement along y direction

       Third 5 miles south means a positive 5 displacement along y direction

So total displacement = 10 - 20 + 5 = -5 miles displacement

So position vector of the final position = - 5 j

3 0
4 years ago
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