if is a gamma random variable with parameters (n,1), approximately how large

The efficiency of the machine is determined as 83.3 %.

M.A = load/Effort

M.A = 4500/900 = 5

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

V.R = 6/1 = 6

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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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

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.

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

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The measure of the angle EFD = 34°. That is option A.

The radii of the given circle are CE and CD.

The measure of the given angle of the arc is = 68°

But the central angle of an intercepted arc measure the double of the inscribed angle by the same arc.

Therefore, the measure of angle of an intercepted arc is 68/2 = 34°

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Answer:

34

Explanation:

edge 2023

Answer:

5 years

Explanation:

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

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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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.

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)

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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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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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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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.

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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Answer:

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

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, 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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i. The work done by the load is load x distance moved by load.

ii. The work done by effort is effort applied x distance moved by effort.

iii. The mechanical advantage of the simple machine is Load/effort.

iv. The velocity ratio of the simple machine is 2.

v. The efficiency of the machine is M.A/V.R x 100%.

The work done by the load is the product of the load and the distance through which the load is moved. The magnitude is calculated as follows;

Work done by the load = load x distance moved by load

The work done by the effort is the product of the effort and the distance through which the effort is applied. The magnitude is calculated as follows;

Work done by effort = effort applied x distance moved by effort

M.A = Load/Effort

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

V.R  = 30 cm/15 cm

V.R = 2

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

Thus, the work done by the load is load x distance moved by load.

The work done by effort is effort applied x distance moved by effort.

The mechanical advantage of the simple machine is Load/effort.

The velocity ratio of the simple machine is 2.

The efficiency of the machine is M.A/V.R x 100%.

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

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

W = Fd

where;

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 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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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)

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

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Answer:

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

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

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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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

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.

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 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:

Hence,

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

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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

The energy stored in a capacitor is given by

E = ½CV²

But

E = Pt

Thus,

Pt = ½CV²

Where

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

Data obtained from the question include:

Pt = ½CV²

100 × t = ½ × 1×10⁻⁷ × 2²

Divide both sides by 100

t = (½ × 1×10⁻⁷ × 2²) / 100

t = 2×10⁻⁹ s

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

Data obtained from the question include:

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

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:

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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.

The initial temperature of the metal was 94.24⁰C.

The initial temperature of the metal is calculated as follows;

Q = mcΔθ

where;

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

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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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

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.

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

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

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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