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2 years ago

How far does light travel in the time it takes sound to travel 1 cm in air at 20°c?

1 answer:

8 0

The speed of sound at $20^{\circ}C$ is approximately v=343 m/s. The distance covered by the sound wave is
$s=1 cm=0.01 m$
And the time it takes is
$t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s$

Now we want to find how far the light travels during this time. Light travels at speed $c=3 \cdot 10^8 m/s$ , therefore the distance it covers during this time is
$S=ct = (3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8700 m$

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A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2. 00 m making 7. 00 revolut

mr Goodwill [35]

Answer:

F = M ω^2 R is the centripetal force on the ball

f = 7 / 10 = .70 / sec revolutions / sec

ω = 2 pi f = 4.40 / sec

ah = F / m = 4.4^2 * 2 = 38.7 m/s^2

horizontal acceleration = 38.7 m/s^2

vertical acceleration = 9.80 m/s^2

a = (38.7^2 + 9.8^2)^1/2 = 39.9 m/s^2 total acceleration

8 0

1 year ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

8 0

3 years ago

A device that converts mechanical energy into electricity is a

IgorLugansk [536]

Answer:

An electric generator is a device that converts a form of energy into electricity.

Explanation:

7 0

2 years ago

a car traveling at a speed of 72km/h is uniformly returded and comes to rest after 8 seconds if the car with the occupants has a

cricket20 [7]

Answer:

the breaking force is 12

Explanation:

7 0

2 years ago

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

1 year ago