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3 years ago

How might a collision between Neptune and Halley's Comet affect Neptune's orbit?

1 answer:

4 0

The effect would be minuscule ... imperceptible and essentially unnoticeable,
a lot like a collision between Earth and a few tons of meteoroids.

-- Neptune's diameter is a little more than 2000 times the comet's diameter.

-- Neptune's mass is a little more than 465 billion times the comet's mass.

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An attacker at the base of a castle wall 3.65 m high throws a rock straight up with speed 7.4m/s from a height of 1.55m above th

Natali5045456 [20]

a) Yes, the rock will reach the top

b) The final speed is 3.7 m/s

c) The change in speed is 2.4 m/s

d) The change in speed in the two situations do not agree

e) Because the kinetic energy depends quadratically on the speed, $K\propto v^2$

Explanation:

The mechanical energy of the rock at the moment it is thrown from the ground is equal to the sum of its kinetic energy and its potential energy:

$E=KE_i + PE_i = \frac{1}{2}mu^2 + mgh_i$

where

m is the mass of the rock

u = 7.4 m/s is the inital speed

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

$h_i = 1.55 m$ is the initial height of the rock

Substituting, we find the initial mechanical energy of the rock

$E=\frac{1}{2}m(7.4)^2 + m(9.8)(1.55)=42.6m$ [J]

In order to reach the top of the castle, the rock should have a mechanical energy of at least

$E' = mgh'$

where

h' = 3.65 m is the heigth of the top

Substituting,

$E'=m(9.8)(3.65)=35.6m$ [J]

Since E > E', it means that the rock has enough mechanical energy to reach the top.

The final mechanical energy of the rock at the top is

$E=mgh'+ \frac{1}{2}mv^2$ (1)

where:

v is the final speed of the rock at the top

Since the mechanical energy is conserved, this should be equal to the initial mechanical energy:

$E=42.6 m$ [J] (2)

Therefore, equating (1) and (2), we can find the final speed of the rock:

$mgh' + \frac{1}{2}mv^2 = 42.6m\\v=\sqrt{2(42.6-gh')}=\sqrt{2(42.6-(9.8)(3.65))}=3.7 m/s$

Since the motion of the rock is a free fall motion (constant acceleration equal to the acceleration of gravity), we can use the following suvat equation:

$v^2 - u^2 = 2as$

where

v is the final speed, at the bottom

u = 7.4 m/s is the initial speed of the rock, at the top

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

s = 3.65 - 1.55 = 2.1 m is the vertical displacement of the rock

Solving for v, we find the final speed:

$v=\sqrt{u^2+2as}=\sqrt{7.4^2 + 2(9.8)(2.1)}=9.8 m/s$

Therefore, the change in speed is

$\Delta v = v-u = 9.8 - 7.4 =2.4 m/s$

In the first situation (rock thrown upward), we have:

u = 7.4 m/s (initial speed)

v = 3.7 m/s (final speed)

So the change in speed is

$\Delta v = v-u =3.7 - 7.4 = -3.7 m/s$

While the change in speed in the second situation (rock thrown downward) is

$\Delta v = 2.4 m/s$

Therefore, we see that their magnitudes do not agree.

In both situations, the change in kinetic energy of the rock is equal in magnitude to the change in gravitational potential energy, since the total mechanical energy is conserved.

The change in gravitational potential energy in the two situations is the same (because the change in height is the same), therefore the change in kinetic energy in the two situations is also the same.

However, the kinetic energy of the rock is not directly proportional to the speed, but to the square of the speed:

$K\propto v^2$

Since the initial speed is the same for both situation (7.4 m/s), but the change in kinetic energy has opposite sign in the two situations (negative when the rock is thrown upward, positive when thrown downward), the situation is not symmetrical, therefore in order to have the same magnitude of change in the kinetic energy, the change in speed must be larger when the kinetic energy involved is lower, so in the first situation.

Learn more about kinetic energy and about potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

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#LearnwithBrainly

6 0

4 years ago

List and describe the main functions of the lobes and the human brain

sladkih [1.3K]

Answer:

Each side of your brain contains four lobes. The frontal lobe is important for cognitive functions and control of voluntary movement or activity. The parietal lobe processes information about temperature, taste, touch and movement, while the occipital lobe is primarily responsible for vision.

6 0

3 years ago

Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck

erastovalidia [21]

Answer:

$3.3\cdot 10^3\:\mathrm{N}$

Explanation:

Impulse on an object is given by $\mathrm{[impulse]}=F\Delta t$ .

However, it's also given as change in momentum (impulse-momentum theorem).

Therefore, we can set the change in momentum equal to the former formula for impulse:

$\Delta p=F\Delta t$ .

Momentum is given by $p=mv$ . Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:

$p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}$ .

Now we plug in our values and solve:

$\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}$ (two significant figures).

6 0

3 years ago

A fire engine is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the fire engin

ValentinkaMS [17]

Answer:

The frequency increases, and the pitch increases

Explanation:

Doppler's law of sound is applicable in such case when the observer or the sound source or both are moving relative to each other.

In such a case due to space-time constraint the waveform of the sound adjust themselves so as to obey the law of conservation of energy.

The apparent frequency of the sound for the observer is given by:

$f_o=(\frac{s+v_o}{s+v_s} )f$ ....................................(1)

where:

$f_o=$ observed frequency

$f=$ original source frequency

$s=$ speed of sound

$v_s=$ speed of source relative to the observer (taken negative when approaching towards the observer and vice-versa)

$v_o=$ speed of observer relative to the source (taken negative when moving away from the source and vice-versa)

According to the given situation, eq. (1) becomes:

$f_o=(\frac{s}{s-v_s} )f$

Since, $\frac{s}{s-v_s} >1$

Therefore

$f_o>f$

Pitch is very closely related to the frequency, it means that how fast is the amplitude of sound varying with time.

7 0

3 years ago

The regions of the Moon that are of higher elevation are known as the __________. These regions have a high albedo and display a

REY [17]

Answer:

lunar highlands

Explanation:

Seeing the moon from the earth we can see that some parts are more illuminated than others, these different regions of the moon have been assigned a name , and have different properties.

The brightest parts are called lunar highlands, because they are at a higher elevation than the darkest parts, called lunar marias.

The lunar highlands are seen from the earth a whiter color than the rest of the moon due to its altitude.

4 0

4 years ago