What are the two thing change when an object is in motion

The volcano olympus mons is an unusual example of the cone type called

enot [183]

A shield volcano i believe. hope this helped

5 0

3 years ago

A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surfac

andrew11 [14]

Answer:

The bird appears farther

Explanation:

This is because as the light from the bird travels into the water which has a higher refractive index than air, light rays from the kingfisher bird bend towards the normal at the water surface and thus enter the eye of the scuba diver. Now, if we project the light rays from the eyes of the scuba diver into the air, we see that they appear to come from a point farther than that of the actual kingfisher bird perched on the branch.

<u>So, the bird appears to the diver to be farther from the surface than the actual bird</u>

3 0

3 years ago

The speed of a rocket just after being launched is 12 m / s.

drek231 [11]

This question involves the concepts of the equations of motion, kinetic energy, and potential energy.

a. The kinetic energy of the rocket at launch is "3.6 J".

b. maximum gravitational potential energy of the rocket is "3.6 J".

<h3>a. KINETIC ENERGY AT LAUNCH</h3>

The kinetic energy of the rocket at launch is given by the following formula:

$K.E=\frac{1}{2} m v_i^2$

where,

K.E = initial kinetic energy = ?
m = mass of rocket = 0.05 kg
$v_i$ = initial speed = 12 m/s

Therefore,

$K.E=\frac{1}{2}(0.05\ kg)(12\ m/s)^2$

K.E = 3.6 J

<h3>b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY</h3>

First, we will use the third equation of motion to find the maximum height reached by rocket:

$2gh=v_f^2-v_i^2$

where,

g = -9.81 m/s²
h = maximum height = ?
vf = final speed = 0 m/s

Therefore,

2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²

h = 7.34 m

Hence, the maximum gravitational potential energy will be:

P.E = mgh

P.E = (0.05 kg)(9.81 m/s²)(7.34 m)

P.E = 3.6 J

Learn more about the equations of motion here:

brainly.com/question/5955789

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

6) The temperature of 10 kg of a substance rises by 55°C when heated. Calculate the

Anna [14]

Answer:

121

Explanation:

8 0

3 years ago