What process makes this pattern on the ocean floor?

Answer:

It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.

Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.

This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.

The potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.

Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.

It is proven that if the potential at infinity is set to zero, the potential energy of an item on the planet's surface is [tex]\rm \frac{-GM}{R}[/tex].

The escape speed of such an item would necessitate kinetic energy will be [tex]\rm \frac{GM}{R}[/tex]. In all of these circumstances, the total energy is zero.

Consider the speed of an item falling from infinity towards the planet: if the total energy was zero before the thing began to descend, the total energy will stay zero.

Hence the potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.

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Answer:

I'm a Filipino hahahah I'm not

Answer:

52 / 4 = 13 m/s

Explanation:

nonmental

that should be the answer

Answer:

c

Explanation:

Answer:

Explanation:

Testing answer please do not delete

the diffrence in potential betwen two points that represents the work involved or the energy relesed in the transfe of a unit quantity of electicity from one point to the other

Answer: 39.2m/s

Explanation:

Due to the free falling body, the acceleration of the ball is 9.8 m/s^2, and its initial velocity is 0 m/s.

V(t) = g*t

The change of velocity from t=1 to t=5:

Delta V = 9.8*(5-1) = 39.2 m/s

Answer:

Straight Path

Answer:

0.4 g/cm

Explanation:

density (g cm ³) = mass (g)

÷

volume (cm³)

Answer:

C. 4 g/cm

Explanation:

Use the formula:

density = mass ÷ volume

Sub in the values:

density = 36 ÷ 9 = 4 g/cm

Answer = 4 g/cm

The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:

The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.

Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.

They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.

      Em₀ = U = m g h

Where the height is measured from the lowest point of the movement.

      h = L - L cos tea1 = L (1 cos tea1)

The second point of interest occurs for θ₂ = 30º.

At this point part of the energy is indica and part gravitational potential.  

     [tex]Em_f[/tex]  = K + U₂

      [tex]Em_f[/tex] = ½ m v² + m g h ’

There is no friction in the system, therefore mechanical energy is conserved.

       Em₀ = Em₀_f

       mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)

       v² = 2g L (cos θ₂ - cos θ₁)

Let's calculate.

       v² = 2 9.8 2.1 (cos 30 - cos 60)

       v² = 41.16 0.366

       v = 3.88 m / s

In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:

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Answer:

3.0 J

Explanation:

Just took the test

Answer:

D) 19.8 lbs

Explanation:

1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

1kg is also equal to 2.205 lbs. 9*2.205=19.8416

9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system

a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.

This leaves us with 19.8 lbs

Sorry, I don't know but I think the correct answer is the first option.

Explanation:

s=(v^2-u^2)/2g

=(0-3^2)/2*10

= 0.45m

I just need the points

Explanation:

But just pick any and say something like "it stans out to me most/more" or "it sounds/looks more intristing to me"

Answer:

31

Explanation:

No need

Answer:

Explanation:

My perdiction was correct. I predicted in Part A that the Kinetic Energy would increase. In Part B I predicted that the potential energy would decrease due to the kinetic energy increasing.

Answer:

Ac = v²/r

where, Ac is centripetal acceleration

v is velocity

r is radius

5 = (20)²/r

r = 400/5

r = 80m

from,

w = 2πf

where, w = angular velocity and f frequency

w = v/r

f = n/T

then,

v = 2πrn ,n = 1 to complete 1 cycle

T

T = 2πr/v

T = 2π*80/20

T = 25.14seconds.

The car would take 25.14s to complete 1 cycle of the roundabout.

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

The final velocity of the wood after the bullet hits is calculated as follows;

[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s[/tex]

The acceleration of the wood is calculated as follows;

[tex]\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2[/tex]

The distance traveled by the wood after the bullet emerges is calculated as follows;

[tex]v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m[/tex]

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

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The projectiles launch allows to find the results for the questions of the launch of the bomb are:

     a) The fall time is: t = 10.1 s

     b) The distance traveled is: x = 2.24 10³ m

Projectile launching is an application of kinematics where the acceleration on the x-axis is zero and the acceleration on the y-axis is gravity acceleration.

a) Let's find the time until we reach the ground.

The initial vertical velocity is zero, the initial height is I = 500 m, when reaching the ground its height y = 0.

        [tex]y = y_o + v_o_y - \frac{1}{2} g t^2[/tex]  

        0 = y₀ + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2y_o}{g}[/tex]  

        t = [tex]\sqrt{\frac{2 \ 500}{9.8} }[/tex]  

        t = 10.1 s

b) We look for the horizontal distance traveled.

Let's reduce to the international measurement system (SI).

       v = 800 km / h ([tex]\frac{1000m}{1km}[/tex]) ([tex]\frac{1h}{3600s}[/tex]) = 222.22 m / s

       x = vₓ t

       x = 22.22 10.1

       x = 2.24 10³ m

In conclusion, using the projectile launch relationships, we can find the results for the questions that are:

     a) The fall time is: t = 10.1 s

     b) The distance traveled is: x = 2.24 10³ m

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The amount of work done by the gas in the given expansion is 1800 J.

The given parameters;

The amount of work done by the gas in the given expansion is calculated as follows;

W = PΔV

where;

Substitute the given parameters and solve for the work done;

W = 1000(2.4  -  0.6

W = 1800 J

Thus, the amount of work done by the gas in the given expansion is 1800 J.

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Explanation:

this is actually not as simple as it sounds here.

quite some energy is lost in the deformation of the bodies of car and truck, and it also needs more energy to get a standing object going than to accelerate an already moving object.

but assuming the simple described circumstances, then the energy and impulse of the moving truck of 5000 kg is transferred to a new combined system of car and truck of now 5000 + 3000 = 8000 kg.

so, the 20m/s inertia energy of the truck is now distributed to the truck/car combination.

since the same energy has to move now more mass, it is clear that the combined speed will be lower.

20×5000 = x×8000

20×5 = x×8

x = 100/8 = 12.5 m/s

that is the resulting speed of the combined truck/car object.

Answer:

yes

Explanation:

Answer:

2.5 m/s

Explanation:

(60*1.5+60*3.5)/120=2.5

or

(1.5+3.5)/2=2.5

Explanation:

Rules

A match consists of the best of three games of 21 points.

The player/pair winning a rally adds a point to its score.

At 20-all, the player/pair which first gains a 2-point lead wins that game.

At 29-all, the side scoring the 30th point wins that game.

The player/pair winning a game serves first in the next game.

A badminton match can be played by two opposing players (singles) or four opposing players (doubles).

A competitive match must be played indoors utilising the official court dimensions.

A point is scored when the shuttlecock lands inside the opponent's court or if a returned shuttlecock hits the net or lands outside of the court the player will lose the point.

At the start of the rally, the server and receiver stand in diagonally opposite service courts.

A legal serve must be hit diagonally over the net and across the court.

A badminton serve must be hit underarm and below the server's waist height with the racquet shaft pointing downwards, the shuttlecock is not allowed to bounce. After a point is won, the players will move to the opposite serving stations for the next point.

The rules do not allow second serves.

During a point a player can return the shuttlecock from inside and outside of the court.

A player is not able to touch the net with any part of their body or racket.

A player must not deliberately distract their opponent.

A player is not able to hit the shuttlecock twice.

A 'let' may be called by the referee if an unforeseen or accidental issue arises.

A game must include two rest periods. These are a 90-second rest after the first game and a 5-minute rest after the second game.