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stiv31 [10]
3 years ago
13

A wave travels down the length of a 25-meter rope in 5.0 seconds. The speed of the wave is _____.

Physics
2 answers:
Luden [163]3 years ago
7 0

Answer:

The speed of the wave is 50 m/s

Explanation:

The wave traveled through the length in 5 seconds,  the speed is the distance  that the wave travel at one second.

For know the speed we will use the following equation:

V = X/T

Where V is speed, X is distance and T is time).

From the information in the statement:, we know that 25 meters is our distance  and 5 seconds is our time .

Replacing this values in the initial equation, we get:

V = X/T

V= 25 meters/5 seconds

V= 5 meters/s econd or 5m/s

Finally, the speed of the wave is 50 m/s.

Alborosie3 years ago
4 0
Speed
= (distance covered) / (time to cover the distance)

= (25 m) / (5.0 sec) = 5.0 m/s .
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3 years ago
When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

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A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont
jeka94

Answer:

1)a. It is constant the whole time the ball is in free-fall.

2)b. = 14 m/s

3) e. = 19.6 m/s

Explanation:

1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.

2) speed = distance/time

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-u = u - 4(9.8)

-2u = -4(9.8)

u = -4(9.8)/-2

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