Which statement or phrase best describes the currently accepted model of the atom?

An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?

OLEGan [10]

Answer:

500km per hour

Explanation:

if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.

5 0

3 years ago

Explain why water buckets are narrow at the base and wider at the top

polet [3.4K]

Answer:

This is to optimize storage and transport.

Explanation:

4 0

2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago

The minimum number of vectors of unequal magnitude required to

EleoNora [17]

Answer:

d

Explanation:

4 0

3 years ago

Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has

Anastasy [175]

-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is

   (3.95 x 10⁵m) + (4.2 x 10⁶m)

   = (3.95 x 10⁵m) + (42 x 10⁵m)

   = 45.95 x 10⁵ m

The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

   (91.9 π x 10⁵ m) / 2 hr

= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)

=    0.04 x 10⁵ m/sec

      =    4 x 10³    m/sec

   (4 kilometers per second)

6 0

3 years ago