Please help!
You release a pendulum of mass 1 kg from a height of 0.75 m.
A. If there is no air resistance, how fast is the pendulum going when it reaches the bottom?
B. If the pendulum loses 18% of its initial energy by the time it reaches the bottom, how fast is it going when it reaches the bottom?
C. If the pendulum loses another 7% of its remaining energy by the time it reaches the other side, how high does it go?
D. After a few minutes, the pendulum is no longer swinging at all explain why this happens, in terms of energy.

Answers

Answer 1

(A) The speed of the pendulum when it reaches the bottom is 3.83 m/s.

(B) The speed of the  pendulum when it reaches the bottom after losing 18% of its energy is 3.47 m/s.

(C) The height reached by the pendulum after losing another 7% of its energy is 0.57 m.

(D) When the pendulum stops swinging, it has used all its energy to overcome frictional force of air.

Speed of the pendulum when it reaches the bottom

Apply the principle of conservation of energy;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 0.75)

v = 3.83 m/s

Speed pendulum after losing 18% of the its initial energy

K.E = (100 -  18)P.E

¹/₂mv² = 0.82mgh

V = √(0.82 x 2gh)

v = √(0.82 x 2 x 9.8 x 0.75)

v = 3.47 m/s

Height reached when its looses another 7%

K.E = 0.5(1)(3.47)² = 6.02 J

When it losses 7% = 6.02 - (0.07 x 6.02) = 5.598 J

Height reached:

mgh = 5.598

h = 5.598/mg

h = 5.598/(1 x 9.8)

h = 0.57 m

Final energy of the pendulum

When the pendulum stops swinging, it has used all its energy to overcome frictional force of air.

Thus, the speed of the pendulum when it reaches the bottom is 3.83 m/s.

The speed of the  pendulum when it reaches the bottom after losing 18% of its energy is 3.47 m/s.

The height reached by the pendulum after losing another 7% of its energy is 0.57 m.

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Related Questions

Why might the measurements of ""true distances"" be different for different groups even though they are measuring the same distance?

Answers

While measuring , there are a lot of errors which can happen due to which ""true distances"" be different for different groups even though they are measuring the same distance.

While measuring , there are a lot of errors which can happen , to find the most correct value in order to make the result precise and accurate number of measurements needed to be done in order to reduce the error

As nothing can be measured with full precision and accuracy, different measurements for different groups even though they are measuring the same distance.

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A bungee jumper jumps from a bridge and starts accelerating towards a lake below. What energy transfer is he experiencing?

Group of answer choices

a. Elastic potential to kinetic

b. Kinetic to elastic potential

c. Kinetic to gravitational potential

d. Gravitational potential to kinetic

Answers

D. The energy transfer he is experiencing is Gravitational potential to kinetic.

Energy transferred experienced by the bungee jumper

The bungee jumper possesses gravitational potential energy due to his position above the ground level (on a bridge).

As he starts accelerating towards a lake below, his gravitational potential energy will be converted into kinetic energy.

Thus, the energy transfer he is experiencing is Gravitational potential to kinetic.

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What happens within a protostar to create a star?

Answers

Clouds of gas in space


A protostar resembles a star, but its core is still too cool for fusion to occur. The protostar's heating as it contracts is the only source of brightness. The light that protostars release is typically blocked by dust, making them challenging to study in the visible spectrum.

By the time a protostar is produced, the cloud has flattened and a protostellar disk is spinning around it. The cloud starts spinning as it collapses. These disks occasionally form planetary systems and are thought to slow the protostar's rotation. The protostar produces a powerful magnetic field as it revolves.

Additionally, a strong protostellar wind—a movement of particles into space—is produced by the magnetic field. A lot of protostars also release gas into space in the form of fast-moving streams or jets.

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How does the distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n?

Answers

The distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n because:

Option C: The energy difference between adjacent orbit radii decreases with increasing values of the principal quantum number.

What determines the distance of the electrons from the nucleus?

In an atom, an electron is known to be attracted to a given nucleus by the use of "electromagnetic force".

Note that  similar to a baseball, the faster the electron is said to go, the farther away from the nucleus it is known to be seen. Therefore, the electrons in an atom are known to be in a state where they are moving a lot and very fast, so they are said to be far away from their nucleus.

Note also that energy difference between what we call consecutive levels tends to often decreases as well as increases.

Therefore, The distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n because:

Option C: The energy difference between adjacent orbit radii decreases with increasing values of the principal quantum number.

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See full question below

Consider the hydrogen atom. How does the distance between adjacent orbits in a hydrogen atom vary with increasing values of the orbital number n?

The distance difference between adjacent orbit radii varies with increasing values of the principal quantum number.

The distance difference between adjacent orbit radii increases with increasing values of the principal quantum number.

The distance difference between adjacent orbit radii remains constant with increasing values of the principal quantum number.

The distance difference between adjacent orbit radii decreases with increasing values of the principal quantum number.

an interplantetary speedcarft moving at 20000m/s.how far will it travell in one day?(give your answer in km)

Answers

Explanation:

Step I: 1 day means 24 hours * 60 minutes * 60 seconds.

So 1 day have 86400 seconds.

Step I:

Now,

To calculate the travel distance 20,000m/s* 84600 is 1728000000m

Step Ill:

Now convert the meter in kilometer

Because 1 km = 1000 m

So, = 1728000000/1000 = 172800OKm

2. Find rectangular coordinates for the point (6, 240°).

Answers

Answer:

(-3, -3√3)

Explanation:

To convert from polar coordinate to rectangular coordinate, first you have to know these two equations:

[tex]\displaystyle{x=r\cos \theta}\\\displaystyle{y=r\sin \theta}[/tex]

We know that (x,y) is in rectangular coordinate form while (r,θ) is in polar coordinate form.

Therefore, substitute r = 6 and θ = 240° in both equations:

[tex]\displaystyle{x=6\cos 240^{\circ}}\\\displaystyle{y=6\sin 240^{\circ}}[/tex]

After evaluating, you'll get:

[tex]\displaystyle{x=-3}\\\displaystye{y=-3\sqrt{3}}[/tex]

Therefore, substitute x and y in rectangular coordinate form - hence, the answer is (-3, -3√3)

A 150 g piece of metal has a specific heat capacity of 0.845 J/g°C. If it takes 3.30x 103 J to heat the metal to 120°C, the initial temperature of the metal was

Answers

The initial temperature of the metal was 94.24⁰C.

Initial temperature of the metal

The initial temperature of the metal is calculated as follows;

Q = mcΔθ

where;

Δθ is change in temperature

Δθ = Q/mc

Δθ = (3,300) / (0.854 x 150)

Δθ = 25.76⁰C

Δθ = T₂ - T₁

T₁ = T₂ - Δθ

T₁ = 120 - 25.76

T₁ = 94.24⁰C

Thus, the initial temperature of the metal was 94.24⁰C.

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a bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away

Answers

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 ×  t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between  when the bat emits the sound and when it hears the echo is 0.05 s.

How does a bat know how far away something is?

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

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Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

If the ocular lens magnifies a specimen 10x, and the objective lens used magnifies the specimen 35x. What is the total magnification being used to observe the specimen?

Answers

Answer:

350x^2

Explanation:

I'm assuming you're saying you're using them together, like stacking them on top of each other, so if that's the case then this is a simple multiplication problem, which can be written as 10x(35x). Solve it and you get 350x^2.

For vibrational motion, what term denotes the maximum displacement from the equilibrium position?.

Answers

For vibrational motion, the amplitude denotes the maximum displacement from the equilibrium position.

What is vibrational motion?

The motion in which there are some vibrations about the fixed position called mean position is termed vibrational motion. For example, the motion of a wave on the string has perpendicular vibration.

What is amplitude?

The amplitude is defined as the maximum displacement of the vibration from its mean position. For vibrational motion, the mean position is the equilibrium position. So amplitude is the term that denotes the maximum displacement from the equilibrium position. For example, the displacement of the string wave from its equilibrium position is the amplitude of the string wave.

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Does your experimental density of the unknown liquid agree with the true value? if no, give reasons for the discrepancy. Do not include operator errors

Answers

The reasons  for the discrepancy in the experimental density of the unknown liquid agree with the true value can be instrument not having same levels , Temperature Effects , Kind of scales used and  Shape of objects used for measure .

According to the question

Discrepancy:

The state or quality of being discrepant or in disagreement, as by displaying an unexpected or unacceptable difference; inconsistency.

Density:

Density is mass per unit volume, so when measuring density, you find the mass of the object and divide it by its measured volume.

All measurements include some uncertainty, however, and certain kinds of errors can increase the uncertainty in your calculation.

The reason of discrepancy in the experimental density of the unknown liquid agree with the true value can be:

1. Use of instrument not having same levels

2. Temperature Effects as Density varies with temperature .

3. Kind of scales used for measures

4. Shape of objects used for measure .

Hence, the reasons  for the discrepancy is instrument not having same levels , Temperature Effects , Kind of scales used and  Shape of objects used for measure .

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A construction worker pushes a 25 kg load in a wheelbarrow for a distance of 5.0 m, using a horizontal force of 50.0 N. How much work is done by the worker on the wheelbarrow?
Group of answer choices

a. 55 J

b. 250 J

c. 1250 J

d. 10 J

Answers

B. The amount of work done by the worker on the wheelbarrow is 250 J.

Work done by the worker

The amount of work done by the worker on the wheelbarrow is calculated as follows;

W = Fd

where;

F is applied forced is displacement

W = 50 x 5

W = 250 J

Thus, the amount of work done by the worker on the wheelbarrow is 250 J.

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A vehicle that weights 400 n on the surface of the earth is traveling in outer space at a speed of 400 m/s. It can be stopped by applying a constant force of 20 n for:_____.

Answers

A vehicle that weights 400 n on the surface of the earth is traveling in outer space at a speed of 400 m/s. It can be stopped by applying a constant force of 20 n for 800 secs.

Newtons 3 law whilst one object exerts pressure on any other object, the second object exerts a force on the first item this is identical in significance, but opposite in course; for each movement, there is an identical, but opposite response; referred to as the law of action-reaction.

You can use the equal formulation d = rt because of this distance equals the price instance's time. To solve for speed or rate use the components for pace, s = d/t this means that pace equals distance divided by way of time. To clear up for time use the components for time, t = d/s this means that time equals distance divided by way of velocity.

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5. a new type of hoverboard has been designed to levitate over a copper surface. the
developers had a skateboard icon test out the hoverboard on a copper halfpipe.
what type of field is responsible for levitating the hoverboard and rider as it
traverses the half pipe?
a-magnetic field
b-gravitational field
c-electric field
d-tesla field

Answers

The magnetic field is responsible for levitating the hoverboard and rider as it traverses the half pipe.

What is a magnetic field?The magnetic field is defined as the field the magnetic materials generate or when an electric charge moves in a field region that generates the magnetic field.The surface is made up of from the copper and copper is conductor of electricity.When an electrical charge moves, it generates a magnetic field. As with electrical current running via a wire, the spinning and orbiting of an atom's nucleus produces a magnetic field.Thus, the magnetic field is responsible for levitating the hoverboard and rider as it traverses the half pipe.What is the characteristics of Magnetic fields?Magnetic field lines never cross one another.They begin (from the north pole) and end (to the south pole) perpendicular to the surface of the magnet. Magnetic field lines also run through the magnet.The magnetic field is denser at the poles and less dense as we move away from them.A single magnetic pole does not exist. Magnetic poles are always found in pairs of opposing poles.

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A 5 cm radius conducting sphere has a charge density of 2.0x10-6 C/m2 on its surface. Find the electric potential of the sphere.

Answers

noob anyways I can't get u vape because first reason I lost ur money and second I felt bad for losing ur and I will not on yourself as I am thinking of a number my number is a multiple of 6.What other numbers must my number is a multiple of 6.What other numbers

If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the ______ ______ is firmly in place. Select one: Mouthpiece plug Alternate-air-source retainer Dust cap None of the above

Answers

If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.

What is Dust cap?

A dust cap is a gently curved dome mounted either in concave or convex orientation over the central hole of most loudspeaker diaphragms.

Thus, if you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.

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A substance has life of 40 years. In how many years will the substance disintegrate to 1/8 of its initial size

Answers

Answer:

5 years

Explanation:

[tex]\frac{1}{8}(40)\\\\= 5[/tex]

An astronaut in training is seated at the end of a horizontal arm 8 meters long. How many revolutions per second must the arm make for the astronaut to experience a horizontal acceleration of 4.0g

Answers

Answer:

a = v^2 / R        where a is the centripetal acceleration and R the radius

v = 2 π R N / t        where (N / t) is the revolutions / sec giving distance / time

a = 4 g         horizontal acceleration

4 g = (2 π R N / t)^2 / R = 4 π^2 * R * (N / t)^2

(N / t)^2 = g / (π^2 * R)

N / t = 1 /  π * (g / R)^1/2 = .318 * (9.80 m / s^2 / 8 m)^1/2  = .352 / sec

If a 100 ω resistor is placed across a 0. 10 μf charged capacitor which is initially charged to 3 v. How long does it take it to discharge to 2v to 1v?

Answers

1. The time taken to discharge to 2 V is 2×10⁻⁹ s

2. The time taken to discharge to 1 V is 5×10⁻¹⁰ s

Energy stored in a capacitor

The energy stored in a capacitor is given by

E = ½CV²

But

E = Pt

Thus,

Pt = ½CV²

Where

E is the energy C is the capacitorV is the voltageP is the power t is the time

With the formula (Pt = ½CV²), we can determine the time in each case. Detail below:

1. How to determine the time required to discharge to 2 V

Data obtained from the question include:

Power (P) = 100 wCapacitor (C) = 0.10 μF = 1×10⁻⁷ FVoltage (V) = 2 VTime (t) = ?

Pt = ½CV²

100 × t = ½ × 1×10⁻⁷ × 2²

Divide both sides by 100

t = (½ × 1×10⁻⁷ × 2²) / 100

t = 2×10 s

Thus, the time required to discharge to 2 V is 2×10⁻⁹ s

2. How to determine the time required to discharge to 1 V

Data obtained from the question include:

Power (P) = 100 wCapacitor (C) = 0.10 μF = 1×10⁻⁷ FVoltage (V) = 1 VTime (t) = ?

Pt = ½CV²

100 × t = ½ × 1×10⁻⁷ × 1²

Divide both sides by 100

t = (½ × 1×10⁻⁷ × 1²) / 100

t = 5×10⁻¹⁰ s

Thus, the time required to discharge to 1 V is 5×10⁻¹⁰ s

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A military surveillance satellite is in circular orbit around the Earth at an altitude of 1,000 km above the surface. If the Earth's mass is 5.97 x 1024 kg and its radius is 6,370 km, what is the satellite's orbital speed in m/s

Answers

The orbital speed of the satellite is 7.35*10^3 m/s.

What is orbital speed?

The speed of the satellite in its orbit is termed the orbital speed.

The orbital speed is given by the formula,

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

where G is the universal gravitational constant, M is the mass of the planet and r is the distance of the satellite from the center of the planet.

Here the distance of the satellite from the center of the planet is the sum of the planet's radius and the height attained by the satellite above the ground. So

r=6370 + 1000

r=7370 km

Given the mass of the planet is 5.97*10^24 kg and the value of the gravitational constant is 6.67*10^(-11) N m^2 kg^(-2), substitute these values in the formula of the orbital speed.

Note: 1 km=1000 m

[tex]v=\sqrt{\frac{6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\times5.97\times10^{24} \text{ kg}}{7370 \text{ km}}} \\ v=\sqrt{\frac{6.67\times10^{-11}\text{ N m}^2\text{kg}^{-2}\times5.97\times10^{24} \text{ kg}}{7370\times 1000 \text{ m}}} \\ v= 7.35\times 10^3 \text{ m/s}[/tex]

Hence the orbital velocity of the satellite is 7.35*10^3 m/s.

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a 1200 kg car rolling on a horizontal surfact has a speed of 25 m/s when it strikes a horizontal coiled spring and is brought ot rest in a distance of 2.5m. what is the spring constant of this spring

Answers

The required spring constant:

The spring constant of the spring is [tex]12\times 10^4 \text{ N/m}[/tex].

Calculation:

The mass of the car is m=1200 kg, the speed of the car is v=25 m/s, and after colliding the spring is brought to rest at a distance of x=2.5m. Let the spring constant of the spring is, k.

From the conservation of energy,

Total initial kinetic energy= Total final potential energy of the spring

Therefore,

[tex]\frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]

Now, substituting the values of the mass of the car, speed of the car, and displacement, we get:

[tex]$$\begin{aligned}1200\times(25)^2&=k\times (2.5)^2\\\Rightarrow k&=\frac{1200\times(25)^2}{(2.5)^2}\\&=12\times 10^4 \text{ N/m}\end{aligned}$$[/tex]

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Which would require the greater energy; slowing down of the orbital speed of the Earth so it crashes into the sun, or speeding up the orbital speed of the Earth so it escapes the sun

Answers

Answer: Speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Explanation: To find the answer, we need to know more about the Orbital and escape velocities.

What is Orbital and Escape velocity?Orbital velocity can be defined as the minimum velocity required to put the satellite in its orbit around the earth.The expression for orbital velocity near to the surface of earth will be,

                   [tex]V_o=\sqrt{gR}[/tex]

Escape velocity can be defined as the minimum velocity with which a body must be projected from the surface of earth, so that it escapes from the gravitational field of earth.The expression for orbital velocity will be,

                    [tex]V_e=\sqrt{2gR}[/tex]

If we want to get into the sun, we want to slow down almost completely, so that your speed relative to the sun became almost zero.We need about twice the raw speed to go to the sun than to leave the sun.

Thus, we can conclude that, the speeding up the orbital speed of earth so it escapes the sun require the greater energy.

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What does the word apparent mean in apparent visual magnitude

Answers

The word 'apparent' mean in apparent visual magnitude is, how bright the star appears when viewed from Earth.

What is visual magnitude?

The brightness of a celestial body is determined by eye estimation with or without optical aid or by other instrumentation equivalent to the eye in spectral sensitivity.

Apparent magnitude:

Apparent magnitude is a measure of how bright the star appears when viewed from Earth.It is a measure of the brightness of a star or other astronomical object observed from Earth. An object's apparent magnitude depends on its intrinsic luminosity and its distance from Earth.

Hence,

The word 'apparent' mean in apparent visual magnitude is, how bright the star appears when viewed from Earth.

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A copper penny has a mass of 3.0 g. A total of 4.0 × 1012 electrons are transferred from one neutral penny to another. If the electrostatic force of attraction between the pennies is equal to the weight of a penny, what is the separation between them?

Answers

Answer:

F = K Q1 Q2 / R^2 = m g

Q1 = - Q2 = (4E12 * 1.60E-19) = 6.4E-7   coulombs

Q^2 = 4.10E-13

R^2 = 9.0E9 * 4.1E-13 / (.003 * 9.80) = 36.9E-4 / 2.94E-2

R^2 = .126

R = .35 m

F = K Q1 Q2 / R^2 = m g, Q1 = - Q2 = (4E12 * 1.60E-19) = 6.4E-7   coulombs Q^2 = 4.10E-13, R^2 = 9.0E9 * 4.1E-13 / (.003 * 9.80) = 36.9E-4 / 2.94E-2 R^2 = .126 and R = .35 m.

What is Electrostatic force?

Due to their electric charges, particles can be attracted to or repelled by electrostatic forces. This force is also known as the Coulomb force or Coulomb contact, and it was first described in 1785 by French physicist Charles-Augustin de Coulomb.

Over a distance of 10-16 meters, or roughly one-tenth the diameter of an atomic nucleus, the electrostatic force is present. While opposite charges attract one another, like charges repel one another.

For instance, two positively charged protons, two negatively charged electrons, or two anions all repel one another. Cation and anions, as well as protons and electrons, are all attracted to one another.

Therefore, F = K Q1 Q2 / R^2 = m g, Q1 = - Q2 = (4E12 * 1.60E-19) = 6.4E-7   coulombs Q^2 = 4.10E-13, R^2 = 9.0E9 * 4.1E-13 / (.003 * 9.80) = 36.9E-4 / 2.94E-2 R^2 = .126 and R = .35 m.

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The radial velocity method preferentially detects: Choose one: A. all of the above described planets equally well. B. small planets far from the central star. C. small planets close to the central star. D. large planets close to the central star. E. large planets far from the central star.

Answers

The radial velocity method preferentially detects large planets close to the central star

what is the Radial velocity:

The radial velocity technique is able to detect planets around low-mass stars, such as M-type (red dwarf) stars.

This is due to the fact that low mass stars are more affected by the gravitational tug of planets.

When a planet orbits around a star, the star wobbles a little.

From this, we can determine the mass of the planet and its distance from the star.

hence we can say that,

option D is correct.

The radial velocity method preferentially detects large planets close to the central star

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A car goes around a curve at a constant speed. what is the direction of the net force on the car?

Answers

Answer:

The direction of the net force is at a tangent to the circle.

What is accurate about the planet’s climate system?

Answers

The accurate about the planet’s climate system is the wind

because heating near the equator blows the wind to drive the convection cells in the atmosphere, and the friction created by the rotation of the spherical planet in the atmosphere causes the wind to appear to bend left or right across the surface of the planet. ..

The climate system is a highly complex global system consisting of five major components: the atmosphere, the ocean, the cryosphere (cryosphere), the land surface, the biosphere, and the interactions between them.

Solar energy drives the climate by heating the surface of the earth unevenly. Ice also reflects incoming sunlight, further cooling the poles. Temperature differences move the ocean and atmosphere as they work together to disperse heat throughout the globe.

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If a car increases its velocity from +6 m/s to +30 m/s in 6 seconds, its acceleration in m/s2 is__________.

Answers

Answer:

4m/s^2

Explanation:

v = 30m/s, u = 6m/s, t = 6s

Change in velocity = v(final velocity) - u (initial velocity)

v-u = 30-6 = 24m/s

acceleration = (v-u)/t

(24m/s)/6s = 4m/s^2

A pulley system is used to lift a 4500 n weight using a 900 n force. if the mass moves 1.0m while the pulley is pulled 6.0m, what is the efficiency of the machine?

Answers

The efficiency of the machine is determined as 83.3 %.

Mechanical advantage of the pulley

M.A = load/Effort

M.A = 4500/900 = 5

Velocity ratio of the pulley

V.R = distance moved by effort/distance moved by load

V.R = 6/1 = 6

Efficiency of the pulley

E = M.A/V.R x 100%

E = (5/6) x 100%

E = 83.3 %

Thus, the efficiency of the machine is determined as 83.3 %.

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Arrange these objects according to size, largest to smallest. Group of answer choices traditional dwarf elliptical, ultra-compact dwarfs, globular clusters ultra-compact dwarfs, traditional dwarf galaxies, globular clusters globular clusters, traditional dwarf elliptical, ultra-compact dwarfs globular clusters, ultra-compact dwarfs, traditional dwarf elliptical

Answers

Among the given objects, the largest one is dwarf elliptical, then ultra-compact dwarf and then globular clusters. Thus, option a is correct.

To find the correct answer, we need to know about the galaxies.

Are galaxies bigger than globular clusters?Globular clusters are spherical groups of stars that are primarily located in the spiral galaxy' stellar halo. There are 150 or so globular clusters in our Milky Way galaxy, some of which house our galaxy's oldest stars.Globular clusters are much smaller than galaxies.a system of gas and dust, as well as millions or billions of stars, kept together by gravitational attraction is called galaxy.Ultracompact dwarfs are more extended and have higher surface brightness than typical dwarf nuclei.Dwarf elliptical galaxies are elliptical galaxies that are smaller than ordinary elliptical galaxies. They are quite common in galaxy groups and clusters and are usually companions to other galaxies.

Thus, we can conclude that, option a is correct.

Learn more about the galaxies here:

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