Why do planets speed up as they get closer to the sun?:
A. air resistance
B. friction
C. gravity
D. tension

The new speed of car is 10.9 m/s

According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

[tex]m_1u_1 + m_2u_2 = (m_1 + m_2) v[/tex]

[tex]v =\frac{m_1u_1}{m_1+m_2}[/tex]

[tex]v = \frac{7950*15}{7950+2950}[/tex]

[tex]v= 10.9 m/s[/tex]

Therefore, the new speed of care is 10.9 m/s

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Answer:

Impulse is defined as change in momentum of an object divided by time interval.

at t= 0s

initial velocity = u , initial momentum = mu

at some time t .

final velocity = V, final momentum= mv.

now, change in momentum= ( final - initial) = ( mv-mu)

time interval = (t-0) = t

impulse force = (mv-mu)/ ( t)

Ft = (mv-mu) proved .

this law is known as Newton's second law.

The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

V = (zm/ρN)

where;

V = (2 x 95.96) / (10.28 x 6.02 x 10²³)

V = 3.10 x 10⁻²³ cm³

a³ = V

a = (V)^¹/₃

a = ( 3.10 x 10⁻²³)^¹/₃

a = 3.142 x 10⁻⁸ cm

a = 3.142 x 10⁻¹⁰ m

a = 314.2 x 10⁻¹² m

a = 314.2 pm

Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

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The workout of Peter that involves jumping rope, doing push-ups, kicking a ball through a pattern of cones, doing sit-ups, and then repeats the cycle is an example of circuit training.

A workout is any activity that requires much physical or mental effort, or produces strain.

There are several approaches that can be employed to achieve workout procedures and they are as follows:

Circuit training is a combination of six or more exercises performed with short recovery periods between them for either a set number of repetitions or a prescribed amount of time.

On the other hand, interval training is similar to circuit training but differs in the sense that a period of rest is allowed in between the exercises.

Therefore, the workout of Peter that involves jumping rope, doing push-ups, kicking a ball through a pattern of cones, doing sit-ups, and then repeats the cycle is an example of circuit training.

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Answer:

potential energy = (mass × height × acceleration due to gravity.)

750 = mass × 3 × 10

hence, mass = 750/30 = 25 kg.

Answer:

by using a magnetic field

Use a magnet to attract the iron fillings from the sand!

Hope this helps!

The distance from the wall you would stand and hear no sound at all is 0.83 m.

The speed of a wave is the rate of change of distance traveled by a wave with time.

The distance at which the wave will have zero amplitude, there will be no sound at all since amplitude of a sound is proportional to intensity of the sound.

The point of zero amplitude, L = λ/₂

Where;

λ is wavelength of the wave

The wavelength of the wave is calculated as follows;

λ = V/f

where;

λ = 332/200

λ = 1.66 m

L = 1.66/2

L = 0.83 m

Thus, the distance from the wall you would stand and hear no sound at all is 0.83 m.

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Answer:

Nat gas

Explanation:

Approx:

Natural gas 38%       coal 21%       oil  19%

Answer:

e^-4t e^-5t

Explanation:

solving s²+9s+20 quadratically we have (s+4)(s+5)

x(s) can be written as x(s) =(1/s+4)(1/s+5)

if we take the laplace inversve

L-¹ (s)=L-1(1/s+4) L-1(1/s+5)

we have e^-4t e^-5t

This is a series of analysis of Angular velocity as relates to an asteroid's orbit. See the explanation below.

Asteroids are stony bodies that circle the Sun.

Although asteroids circle the Sun in the same way as planets do, they are far smaller.

Hence, from the information given:

A) The square of the period is proportional to the cube of the semimajor axis, according to Kepler's third law. As a result, the period of Y equals (E) the period of Z.

B) Angular momentum is preserved here, hence it is equivalent (E).

C) As eccentricity increases, so does the angular momentum. In this case, Y and Z have the same period, and both satellites cover the same proportion of the territory in the same length of time.

This indicates that a satellite on Z must cover a lesser area in a given period of time than a satellite on Y. The area swept is approximately 1/2 the radius times the tangential displacement.

Because both satellites have the same "radius" at point y, the satellite on Z must have a lower tangential velocity than the one on Y. As a result, Y has more angular momentum than (G) Z.

D) Using Kepler's third law, X's period is bigger than (G) Z's.

E) In a circular orbit, the angular velocity is constant. As a result, the angular velocity of Y at y equals (E) that at s.

F) Z's angular velocity at c is smaller than (L) at i.

G) Y's angular velocity at y is larger than (G) Z's at y.

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Answer:

The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation |→τ|=r⊥F | τ → | = r ⊥ F , where r⊥ is the perpendicular distance from the axis to the line upon which the force vector lies.

Explanation:

:)

The orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

The law of gravitation states that the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between the masses. Mathematically;

F = GMm/r²

where

M and m are the mass of ice cube and

Recall that;

s = Gm1/r^2

Also;

F = sm²

Substitute to have;

s = m²/F

For the centripetal acceleration

a = v²/r

Such that;

v²/r = Gm/r²

v² = Gm/r

v = √Gm/r

Substitute the given parameters into the formula to have:

V =  √6.67×10^-11 *  5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s = 3.56 * 10^6 m/s

Therefore the orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

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Answer:

(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?

No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum

(b)Is it possible for only one to be at rest after the collision?

Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.

Free Fall under gravity and time of meeting is used to find out the required answer.

∴ Let us assume, the upward direction to be positive and and downward direction to be negative.

Given, Velocity of P, = 40m/s

Distance travelled by P,

∴Using First Equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s  ; which is the time taken by P to rise up

Now, Maximum Height(s) reached by particle P,

∴Using second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) Particle P is falling when Q is shot up after 4s from the initial time

∴ Using third equation of motion using Free Fall under Gravity,

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Hence, Particle P and Q meet at a distance (H₂) from ground,

∴ Height, H₂ = s - H

⇒ H₂ = 80 - 11.25 = 68.75m

Hence, Particle P and Q meet at a distance of 68.75m from ground.

∴ Using First equation of motion for Particle P using the Time of Meeting,

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s  ; which is equal to the fall time of P and Rise time of Q

b) For particle Q

∴ Using second equation of motion,

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Hence, the Velocity with which Particle Q was shot is 15m/s in the upward direction.

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The two evidence are the car's engine gets hot as it burns gasoline are the car remains full of gasoline if its engine stays off.

The car's engine gets hot as it burns gasoline and the car remains full of gasoline if its engine stays off are the two pieces of evidence support the conclusion that the total amount of energy in a close system remains the same because when the gasoline burns, the engine gets hotter and when the car remains full of gasoline if its engine stays off which means no energy leaves the system.

So we can conclude that The two evidence are the car's engine gets hot as it burns gasoline are the car remains full of gasoline if its engine stays off.

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Answer:

Electromagnetic waves

Answer:

100 368 J

Explanation:

Heat lost by the water = heat gained by the air ( which the 'frige works to expel and keep at 4 degrees)

specific heat for water ( in joules) :  4182 J / kg-C

    3 kg *  4182 J / kg-C  * (12-4 C) = 100 368 J

The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

a)

The center of mass of a three-particle system is expressed as

[tex]r_{cm}=[/tex] [tex]$$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$$[/tex]

When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )

[tex]$$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$$[/tex]

b)

When the two particles are shifted

[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]

Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

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Answer:

Astrobiologists have put in tremendous efforts till now in the search for extra-terrestrial life. They are searching not only for tremendously developed life forms like humans, but also for microbes that are found to inhabit even the harshest of environments on Earth. Due to the tremendous development in technology, the odds of finding life beyond Earth are increasing day by day. Nasa's Kepler mission was launched to hunt for Earth like planets beyond our Solar System, and till now this mission has established the presence of 2,337 exoplanets, along with 1,284 new planets. This hugely increases the chances of finding another Earth like planet outside the limits of our Solar system. NASA's Astrobiology researchers are presently studying the most severe environments on Earth for improved understanding of the kind of conditions that would be able to support life in every part of the universe. Moreover, programs for instance SETI are keeping an eye on the galaxy for electromagnetic communications occurring or situated between stars from the societies on new worlds.

There is still a tremendous possibility to find life in our solar system. By the discovery of life forms in the extreme environments here on Earth, scientists have realized that the presence of solar energy is not an absolute necessity for the subsistence of life. Energy from chemical reactions can also trigger the organic elements to interact and start the formation of life. In the Solar system, Jupiter's moons Io and Europa and Saturn's moons Enceladus and Titan have been realized to the very strong candidates for the presence of microscopic life forms.

Explanation:

I hold a very optimistic view on the presence of life beyond Earth. It is highly probable that life on Earth is not an exceptional case but it should be a general one in the Universe. While we are always excited to imagine that we might encounter species as much developed or more highly developed than us, there is very good chance that the Universe is teeming with microscopic life forms. Also, to discover another intelligent life form would also be highly welcome.

(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

(d) The potential difference between A and B is 56.7 V.

The electric force on the sphere is calculated as follows;

F = Eq

where;

F = 300 x (44 x 10⁻⁹)

F = 1.32 x 10⁻⁵ N

The direction of the force is towards the right.

W = Fd

W = 1.32 x 10⁻⁵ N  x 0.189 m

W = 2.5 x 10⁻⁶ J

The change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.

VB − VA =  Ed

VB − VA =  300 N/C  x   0.189 m

VB − VA =  56.7 V

Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.

The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.

The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.

The potential difference between A and B is 56.7 V.

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Answer:

Warm weather may be found in the tropical zone since the sun's heat is concentrated in regions close to the equator. At 23.5 degrees north of the equator and 23.5 degrees south of the equator, this area is nearly in the middle of the world. It has a monthly average temperature of 18 degrees Celsius or higher.

Explanation:

Answer:

because he's fast and speedy

Explanation:

(a) The speeds of the tips of both rotors; main rotor 181.02 m/s and  tail rotor 223.8 m/s.

(b) The speed of the main rotor is 52.8 % speed of sound, and the speed of the tail rotor is 65.2 % speed of sound.

v = ωr

where;

v(main rotor) = (449 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.7 m)

v(main rotor) = 181.02 m/s

v(tail rotor) = (4,150 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.03 m)

v(tail rotor) = 223.8 m/s

% speed (main motor) = 181.02/343 = 0.528 = 52.8 %

% speed (tail motor) = 223.8/343 = 0.652 = 65.2 %

Thus, the speed of the main rotor is 52.8 % speed of sound, and the speed of the tail rotor is 65.2 % speed of sound.

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The minimum necessary height of the IV bag above the position of the needle is 0.37 m.

The minimum necessary height of the IV bag above the position of the needle is calculated as follows;

P = ρgh

where;

h = P/ρg

h = (4759.609) / (9.8 x 1308)

h = 0.37 m

Thus, the minimum necessary height of the IV bag above the position of the needle is 0.37 m.

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Answer: Both total and annular eclipses depend on the distances between Earth, the Moon, and the Sun. A total solar eclipse occurs when the Moon is closer to Earth, thus taking up a larger portion of the sky. The Moon is then able to cover the entire Sun.

Explanation: Edmentum answer

Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position

Displacement is the distance travelled in a specific direction. It is a vector quantity.

Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.

a) Let P be the Aster's final position from her initial position?

We can use bearing by using Cosine formula to solve this question.

P² = 70² + 82² - 2 × 70 × 82 cos 73

P² = 4900 + 6724 - 11480 cos 73

P² = 11624 - 3356.43

P² = 8267.57

P = √8267.57

P = 90.9 m

P = 90.9 - 28

P = 62.9 m

We can get the angle by using Sine rule

82/ sin Ф = 90.9 / sin 73

sin Ф = 0.8627

Ф = [tex]Sin^{-1}[/tex] (0.8627)

Ф = 59.6°

Ф = 60°

b)  She will head north west direction to return to her initial position

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(a) The angular acceleration will be 26.035 rev/[tex]min^{2}[/tex].

(b) The final angular velocity is expected to be 33.846 rev/min.

Given.

t=1.3 min, Θ=22 rev, [tex]ω_{i}[/tex]=0

We know, Θ= [tex]ω_{i}[/tex]t+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]t^{2}[/tex]

22=0+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]1.3^{2}[/tex]

[tex]\alpha[/tex]=26.035 rev/[tex]min^{2}[/tex]

[tex]ω_{f} =ω_{i}+\alpha t[/tex]=0+26.035*1.3=33.846 rev/min

An object's rate of change in angular position or orientation over time is depicted by its angular velocity, rotational velocity, or both ( or ), also known as the angular frequency vector (i.e. how quickly an object rotates or revolves relative to a point or axis). The direction of the pseudovector is normal to the instantaneous plane of rotation or angular displacement, and its magnitude denotes the angular speed, or the rate at which the item rotates or revolves. It is customary to use the right-hand rule to specify the direction of angular motion. A general definition of angular velocity is "angle per unit time" (angle replacing distance from linear velocity with time in common). Radians per second is how angles are measured in the SI.

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(a) The magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.

(b) The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.

(c) The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.

Apply the principle of conservation of linear momentum as follows;

m₁u₁  +  m₂u₂ = m₁v₁x  +  m₂v₂x

where;

750(0) + 1000(13) = 750(4 cos 30)   +   1000v₂x

13000 = 2,598.1  +   1000v₂x

10,401.9 = 1000v₂x

v₂x  =  10.4 m/s

m₁u₁  +  m₂u₂ = m₁v₁y  +  m₂v₂y

750(0) + 1000(0) = 750(4 sin 30)   +   1000v₂y

0 = 1500 +  1000v₂y

v₂y  = -1500/1000

v₂y  = -1.5 m/s

v = √(v₂ₓ² + v₂y²)

v = √[(10.4)² + (-1.5)²]

v = 10.5 m/s

tanθ = v₂y/v₂ₓ

tanθ = -1.5/10.4

tanθ =  -0.144

θ = arc tan(-0.144)

θ = 8.2 ⁰ north west

K.Ei = 0.5m₁u₁²  +  0.5m₂u₂²

K.Ei = 0.5(750)(0)²  +  0.5(1000)(13)²

K.Ei = 84,500 J

K.Ef = 0.5(750)(4)²  +  0.5(1000)(10.5)²

K.Ef = 61,125 J

K.Ef/K.Ei = 61,125/84,500

K.Ef/K.Ei = 0.72

Thus, the magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.

The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.

The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.

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