¿Cuáles son los fenómenos que no se pueden explicar con la teoría corpuscular de la luz?

A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

3 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

blondinia [14]

Answer:

1.2 m/s

0.31 m

0.15 m

Explanation:

Time period is

$T=2.5\times 2\\\Rightarrow T=5\ s$

Frequency is

$f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{5}\\\Rightarrow f=0.2\ Hz$

Velocity is given by

$v=f\lambda\\\Rightarrow v=0.2\times 6\\\Rightarrow v=1.2\ m/s$

The waves are traveling at 1.2 m/s

Amplitude is given by

$A=\dfrac{d}{2}\\\Rightarrow A=\dfrac{0.62}{2}\\\Rightarrow A=0.31\ m$

Amplitude is 0.31 m

If d = 0.3 m

$A=\dfrac{0.3}{2}=0.15\ m$

The amplitude would be 0.15 m. The speed would remain the same.

8 0

3 years ago

2. Adelia holds a shiny steel spoon with its back (convex surface) facing her eyes at a distance

My name is Ann [436]

Answer:

(a) The convex mirror image, is always upright at all positions, while images formed by concave mirrors are always inverted when the object distance from the mirror is more than the mirrors focal length.

(b) An upright image is not seen for object at a distance from a concave mirror further than the focal length of the mirror, which is the spoon in the question

Therefore, the location of her eyes of approximately, 30 cm, from the mirror is more than the mirror's focal length

Explanation:

3 0

3 years ago

A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v

Tpy6a [65]

Current will be $I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz$
now just pluf in the values and Voila..

7 0

3 years ago

A power supply is connected to a 59 Ohm resistor and a 53 Ohm resistor in series. The total current is found to be 0.15 A. What

kirill [66]

the answer is 47265.dug

Explanation:

im bot sure

4 0

2 years ago