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arlik [135]
2 years ago
8

If one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter on

e. (ii) What happens to the frequency? Choose from the same possibilities. (a) It increases.(b) It decreases.(c) It is constant.(d) It changes unpredictably.
Physics
1 answer:
Kamila [148]2 years ago
3 0

We can actually deduce here that if one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter one. What happens to the frequency that it increases.

Thus, option A. It increases is the answer.

<h3>What is frequency?</h3>

Frequency is actually known to be the amount or number of waves that passes a fixed point per unit time. In physics, frequency is also known as the number of cycles or vibrations per unit time.

When the wave move from the heavy rope into the lighter one, the wave speed will definitely increase. This is because when a wave is created from the end of the heavy rope, the frequency will start small and then increases as it gets to the end of the lighter rope.

Learn more about frequency on brainly.com/question/254161

#SPJ4

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3 years ago
Read 2 more answers
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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3 years ago
With time running out in a game, Rachel runs towards the basket at a speed of 2.5 meters per second and from half-court, launche
finlep [7]

Answer:

Rachel(2.5,0)

ball(6.5,4.7)

b.R=10.15m/s, 27.57deg

Explanation:

The reference angle of Rachel is 00^{0} while the ball is at 36^{0}

resolving rachel's speed to the horizontal, we have

Ux=2.5cos0

Ux=2.5m/s

resolving rachel's speed to the vertical we have,

Uy=2.5sin0

Uy=0

for the ball

resolving the speed to its horizontal component

Ux=8cos36

Ux=6.5m/s

Uy=8sin36

Uy=4.7m/s

Rachel(2.5,0)

ball(6.5,4.7)

To get the resultant of their speed

Add the horizontal speed of rachel to that of the ball to get the total horizontal speed

Add the vertical speed of rachel and the ball to get the total vertical speed component

TUx=2.5+6.5=9M/S

TUy=0+4.7=4.7m/s

R=\sqrt{(TUx^2+TUy^2}

R=\sqrt{(9^2+4.7^2}

R=\sqrt{(103)}

R=10.15m/s

the direction

tan\alpha=TUy/TUx

tan\alpha=4.7/9

\alpha=tan^-1(0.522)

\alpha=27.57deg

4 0
3 years ago
Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l
OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\

<u>F = 85696.5 N = 85.69 KN</u>

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3 years ago
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