If you were to come back to our solar system in 6 billion years, what might you expect to find?

A) a red giant star

B) a rapidly spinning pulsar

C) a white dwarf

D) a black hole

E) Everything will be essentially the same as it is now

Answers

Answer 1

Answer:

A)a red giant star


Related Questions

Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?

Answers

Answer:

Fluoroscopy

Explanation:

A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.

An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

pls give brainliest

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

                      [tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]

Substitute,  [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]

                   [tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]

Hence, The new length of aluminum rod is 17.435 cm.

Learn more:

https://brainly.com/question/19495810

The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA

Answers

Answer and Explanation:

Data provided as per the question is as follows

Speed at point A = 20 m/s

Acceleration at point C = [tex]5 m/s^2[/tex]

[tex]r_A = 25 m[/tex]

The calculation of the magnitude of the acceleration at A is shown below:-

Centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

now we will put the values into the above formula

= [tex]\frac{20^2}{25}[/tex]

After solving the above equation we will get

[tex]= 16 m/s^2[/tex]

Tangential acceleration is

[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]

In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force

Answers

Answer:

E = VdB

Explanation:

This is because canceling the electric and magnetic force means

q.vd. B= we

E= Vd. B

Calculate the density of the following material.

1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³

Answers

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer:

The answer is

0.179 kg/m³

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Density = 0.179 kg/m³

Hope this helps you

Suppose a certain laser can provide 82 TW of power in 1.1 ns pulses at a wavelength of 0.24 μm. How much energy is contained in a single pulse?

Answers

Answer:

The energy contained in a single pulse is 90,200 J.

Explanation:

Given;

power of the laser, P = 82 TW = 82 x 10¹² W

time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s

the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m

The energy contained in a single pulse is calculated as;

E = Pt

where;

P is the power of each laser

t is the time to generate the power

E = (82 x 10¹²)(1.1 x 10⁻⁹)

E = 90,200 J

Therefore, the energy contained in a single pulse is 90,200 J

Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have

Answers

Answer:

The planet that is farther from the star must have a time period greater.

Explanation:

We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex]     (1)

Where:

r: is the radius of the circular orbit of the planet and the star

T: is the period

G: is the gravitational constant

M: is the mass of the planet

From equation (1) we have:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex]   (2)          

Where k is a constant

From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.

I hope it helps you!

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz

Answers

Answer:

380 kHz

Explanation:

The speed of sound is taken as 1500 m/s

The length of the fetus is 1.6 cm long

The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.

For this 1.6 cm baby, the wavelength must not exceed

λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =

0.4 cm = 0.004 m   this is the wavelength of the required ultrasonic sound.

we know that

v = λf

where v is the speed of a wave

λ is the wavelength of the wave

f is the frequency of the wave

f = v/λ

substituting values, we have

f = 1500/0.004 = 375000 Hz

==> 375000/1000 = 375 kHz ≅ 380 kHz

Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?

Answers

Answer:

29.5°

Explanation:

To find the distance d

d = 1E10^-2/8500lines

= 1.17x 10-6m

But wavelength in first order maximum is 577nm

and M = 1

So

dsin theta= m. Wavelength

Theta= sin^-1 (m wavelength/d)

= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6

= 493*10^-3= sin^-1 0.493

Theta = 29.5°

If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?

Answers

Answer:

The wavelength is  [tex]\lambda = 589 nm[/tex]

Explanation:

From the question we are told that

    The  distance of the mirror shift  is  [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]

      The number of fringe shift is  n =  792

       

Generally the wavelength producing this fringes is mathematically represented as

               [tex]\lambda = \frac{ 2 * k }{ n }[/tex]

substituting values

              [tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]

             [tex]\lambda = 5.885 *10^{-7} \ m[/tex]

            [tex]\lambda = 589 nm[/tex]

An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?

a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V

Answers

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.

a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above

Answers

Answer:

None of the above

Explanation:

The formula of the magnetic field of a point next to a wire with current is:

B = 2×10^(-7) × ( I /d)

I is the intensity of the current.

d is the distance between the wire and the point.

● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T

A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of carbon (5.4 dis/min*gC). If living organisms have a decay rate of 15.3 dis/min*gC, how old is this skull

Answers

Answer:

9.43*10^3 year

Explanation:

For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation

To start with, we use the formula

t(half) = In 2/k,

if we make k the subject of formula, we have

k = in 2/t(half), now we substitute for the values

k = in 2 / 5730

k = 1.21*10^-4 yr^-1

In(A/A•) = -kt, on rearranging, we find out that

t = -1/k * In(A/A•)

The next step is to substitite the values for each into the equation, giving us

t = -1/1.21*10^-4 * In(5.4/15.3)

t = -1/1.21*10^-4 * -1.1041

t = 0.943*10^4 year

A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm

Answers

Answer:

a

   [tex]F = 67867.2 \ N[/tex]

b

  [tex]\Delta L = 2.6 \ mm[/tex]

Explanation:

From the question we are told that

      The Young modulus is  [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]

      The stress is  [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]

      The  diameter is  [tex]d = 2.40 \ cm = 0.024 \ m[/tex]

The radius is mathematically represented as

       [tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]

The cross-sectional area is  mathematically evaluated as

        [tex]A = \pi r^2[/tex]

         [tex]A = 3.142 * (0.012)^2[/tex]

        [tex]A = 0.000452\ m^2[/tex]

Generally the stress is mathematically represented as

        [tex]\sigma = \frac{F}{A}[/tex]

=>     [tex]F = \sigma * A[/tex]

=>    [tex]F = 1.50 *10^{8} * 0.000452[/tex]

=>    [tex]F = 67867.2 \ N[/tex]

Considering part b

      The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]

Generally Young modulus is mathematically represented as

           [tex]E = \frac{ \sigma}{ strain }[/tex]

Here strain is mathematically represented as

         [tex]strain = \frac{ \Delta L }{L}[/tex]

So    

       [tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]

        [tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]

=>     [tex]\Delta L = \frac{\sigma * L }{E}[/tex]

substituting values

       [tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]

       [tex]\Delta L = 0.0026[/tex]

Converting to mm

      [tex]\Delta L = 0.0026 *1000[/tex]

      [tex]\Delta L = 2.6 \ mm[/tex]

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?

Answers

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]

So, the force has a magnitude of 4212 N

Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m

Answers

Answer:

0.03/π m/min

Explanation:

See attached file pls

"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"

Answers

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

What is the direction of the net gravitational force on the mass at the origin due to the other two masses?

Answers

Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *

Answers

Answer:

 cost = $ 243.00

Explanation:

This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m

                 #_tables = 3 m (1 / 0.20 m)

                #_tables = 15 tables

Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?

              Cost = 15 (16.20 / 1)

              cost = $ 243.00

Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit

Answers

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)

d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)

  = 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047

  = 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away (in m) is the planet Venus if the echo time is 900 s? m (b) What is the echo time (in µs) for a car 80.0 m from a Highway Patrol radar unit? µs (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 11.5 m? ns

Answers

Answer:

a) 1.35 x 10^11 m

b) 0.53 µs

c) 8 ns

Explanation:

Radar involves the use of radio wave which has speed c = 3 x 10^8 m/s

a) for 900 s,

The distance for a round trip = v x t

==>  (3 x 10^8) x 900 =  2.7 x 10^11 m

The distance of Venus is half this round trip distance = (2.7 x 10^11)/2 = 1.35 x 10^11 m

b) for a 80.0 m distance of the car from the radar source, the radar will travel a total distance of

d = 2 x 80 = 160 m

the time taken = d/c = 160/(3 x 10^8) = 5.3 x 10^-7 s = 0.53 µs

c) accuracy in distance Δd = 11.5 m

Δt = accuracy in time = Δd/c = 11.5/(3 x 10^8) = 3.8 x 10^-8 = 38 ns

Please help!
Much appreciated!​

Answers

Answer:

F = 2.7×10¯⁶ N.

Explanation:

From the question given:

F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²

Thus we can obtain the value value of F by carrying the operation as follow:

F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²

F = 2.7648×10¯⁷ / 0.1024

F = 2.7×10¯⁶ N.

Therefore, the value of F is 2.7×10¯⁶ N.

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification

Answers

Answer:

The magnification is  [tex]m = 12[/tex]

Explanation:

From the question  we are told that

   The object distance is [tex]u = 36.2 \ cm[/tex]

     The focal length is  [tex]v = 39.5 \ cm[/tex]

From the lens equation we have that

         [tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]

=>     [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]

substituting values

       [tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]

       [tex]\frac{1}{v} = -0.0023[/tex]

=>   [tex]v = \frac{1}{0.0023}[/tex]

=>   [tex]v =-433.3 \ cm[/tex]

The magnification is mathematically represented as

         [tex]m =- \frac{v}{u}[/tex]

substituting values

        [tex]m =- \frac{-433.3}{36.2}[/tex]

         [tex]m = 12[/tex]

         

A charge of 15 is moving with velocity of 6.2 x17 which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 06.0T.
b. 08.0T.
c. 07.0T.
d. 05.0 T.

Answers

Complete question:

A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.

a. 0.06 T

b. 0.08 T

c. 0.07 T

d. 0.05 T

Answer:

The magnitude of the magnetic field is 0.07 T.

Explanation:

Given;

magnitude of the charge, q = 15C

velocity of the charge, v = 6.2 x 10³ m/s

angle between the charge and the magnetic field, θ = 48°

the force on the particle, F = 4838 N

The magnitude of the magnetic field can be calculated by applying Lorentz force formula;

F = qvBsinθ

where;

B is the magnitude of the magnetic field

B = F / vqsinθ

B = (4838) / (6.2 x 10³ x 15 x sin48)

B = 0.07 T

Therefore, the magnitude of the magnetic field is 0.07 T.

A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______

a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.

Answers

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf

Answers

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

|E| = 90 V

Therefore, the magnitude of the average induced emf is 90V

Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R

Answers

Answer:

The value of resistance of each resistor, R is 2.25 Ω

Explanation:

Given;

voltage across the three resistor, V = 1.5 V

power dissipated by the resistors, P = 3.00 W

the resistance of each resistor, = R

The effective resistance of the three resistors is given by;

R(effective) = R/3

Apply ohms law to determine the current delivered by the source;

V = IR

I = V/R

I = 3V/R

Also, power is calculated as;

P = IV

P = (3V/R) x V

P = 3V²/R

R = 3V² / P

R = (3 x 1.5²) / 3

R = 2.25 Ω

Therefore, the value of resistance of each resistor, R is 2.25 Ω

What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round window

Answers

Perilymph of scala vestibule; endolymph of cochlear duct; perilymph of scala tympani

help... Please help!!!!!!!!!!!

Answers

Answer:

a) 6.8--5.10 thats equal 11.9

b) m=ris/run +10 equal 0.06/8 =7.5*10^-3

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?

Answers

Answer:

The value is  [tex]\mu = 0.76[/tex]

Explanation:

From the question we are told that

    The  acceleration is [tex]a = 7.4 \ m /s^2[/tex]

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

          [tex]F = ma[/tex]

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

        [tex]F_f = F[/tex]

Here  [tex]F_f[/tex] is the frictional force which is mathematically represented as

         [tex]F_f = \mu * m * g[/tex]

substituting into above equation

         [tex]\mu * m * g = ma[/tex]

=>        [tex]\mu = \frac{a}{g}[/tex]

substituting values

           [tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]

           [tex]\mu = 0.76[/tex]

         

Other Questions
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