phil leans over the edge of a cliff and throws a rock upward at 5 m/s. how far below the level from which it was thrown is the rock 2 seconds later?
A. 10 m B. 5 m C. 15 m D. 20 m

The new magnitude of the force of attraction will be 6 times the original force of attraction

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

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She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

What is focal length?

The focal length is a measure of how strongly the system converges or diverges light.

A positive focal length indicates that a system converges light, while a negative focal length indicates that the system diverges light.

For a standard rectilinear lens,

FOV = 2 arctan (x/2f)

FOV ∝ 1 / f

where x is the diagonal of the film.

Focal length (f) and field of view (FOV) of a lens are inversely proportional.

From the equation we can say that,

A shorter focal length gives you a wide angle of view which allows more view to fit in the frame.

Hence,

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

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The electric field due to the nucleus at the location of the electron is 4.101 × 10¹² N/C.

Given:

Distance between the electron and nucleus, r = 26.5 × 10⁻¹² m

Charge on an electron, Q = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the Helium atom (He²⁺) has an atomic number Z = 2  and an atomic mass number A = 2 ( 2 protons + 2 neutrons). Thus the charge of a helium atom can be calculated as:

Q' = +2e = 2(+1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C

We know that the electric field on charge Q due to the nucleus is given as:

E = kQ' / r²

where E is the electric field

           k is Coulomb's force constant

           Q' is a  charge ona  Helium atom

           r is the separation distance between the electron and nucleus

Applying values in the above equation we get:

E = (9×10⁹ Nm²/C²) (3.2 × 10⁻¹⁹ C) / (26.5 × 10⁻¹² m)²

  = 4.101 × 10¹² N/C

Thus, the electric field on electrons due to the Helium nucleus is of magnitude 4.101 × 10¹² N/C.

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Answer:

Approximately [tex]\$ 3.02[/tex].

Explanation:

Note that the electric rate in this question is in the unit dollar-per-[tex]{\rm kWh}[/tex], where [tex]1\; {\rm kWh}[/tex] is the energy to run an appliance of power [tex]1\; {\rm kW}[/tex] for an hour.

Number of minutes for which the air conditioner is running in that day: [tex]15 \times 24 = 360\; \text{minute}[/tex]. Apply unit conversion and ensure that this time is measured in hours (same as the unit of the electric rate.)

[tex]\begin{aligned} \text{time} &= 360\; \text{minute} \times \frac{1\; \text{hour}}{60\; \text{minute}} = 6\; \text{hour} \end{aligned}[/tex].

The power of this air conditioner is:

[tex]\begin{aligned} \text{power} &= \text{voltage} \times \text{current} \\ &= 240\; {\rm V} \times 21\; {\rm A} \\ &= 5040\; {\rm W} \\ &= 5.04\; {\rm kW} \end{aligned}[/tex].

Thus, the energy that this air conditioner would consume would be:

[tex]\begin{aligned}\text{energy} &= \text{power} \times \text{time} \\ &= 5.04\; {\rm kW} \times 6\; \text{hour} \\ &= 30.24\; {\rm kWh} \end{aligned}[/tex].

At a rate of [tex]0.1[/tex] dollar-per-[tex]{\rm kWh}[/tex], the cost of that much energy would be approximately [tex]3.02[/tex] dollars (rounded to the nearest cent.)

Answer:

The requried number of outer-orbital electrons in an atom of Na is 1, P is 5, Br is 7, I is 7, Te is 6, and Sr is 2.

An orbital is a region of space around the nucleus of an atom where there is a high probability of finding an electron. The electrons in an atom are arranged in specific energy levels or shells, and each shell contains one or more subshells, which in turn contain one or more orbitals.

Here,
To determine the number of outer-orbital electrons in an atom, we need to look at the position of the element in the periodic table. The number of outer-orbital electrons is equal to the group number of the element for groups 1, 2, and 13 to 18. For groups 3 to 12, we need to subtract the group number from 10.

a) Na is in group 1, so it has 1 outer-orbital electron.

b) P is in group 15, so it has 5 outer-orbital electrons.

c) Br is in group 17, so it has 7 outer-orbital electrons.

d) I is in group 17, so it has 7 outer-orbital electrons.

e) Te is in group 16, so it has 6 outer-orbital electrons.

f) Sr is in group 2, so it has 2 outer-orbital electrons.

Therefore, the number of outer-orbital electrons in an atom of Na is 1, P is 5, Br is 7, I is 7, Te is 6, and Sr is 2.

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V = 0.9375 m3 is minimum volume .

Given,

            1000 kg/m3 = density of water

             920 kg/m3 = density of ice

BF = Woman Weight + Slab Weight

Pvg = 75 + Pvg

V = 75 kg / (1000 - 920)

V = 75 / 80

V = 0.9375 m3

Therefore, V = 0.9375 m3 is minimum volume .

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It would take 531 minutes for a single electron in the electron sea to drift from one end to the other end of a wire 35 cm long.

What is electric current?

A rate of electric charge flowing past a place or region is called an electric current. A net passage of electric charge through a region creates an electric current.

Here,

ρ is the density of electrons = 8.4×10²⁸ kg/m³

i is the electric current =9.08×10¹⁸ electrons/sec

d is the diameter of the wire =3.6 mm = 0.0036 m

r is radius, r = d/2, r = 1.8 mm = 0.0018 m

l is the length of wire = 34 cm = 0.34 m

v is The volume of the wire

We know that Volume:

V = π r^2* l

V = (3.14* 0.0018^2) 0.34

V = 3.45* 10^-6 m³

Now,

no of electrons= ρ × V

n =  8.4×10²⁸ * 3.45* 10^-6

n = 2.898 * 10^ 23

The time required for the drifting electron is equal to the ratio of the number of electrons to the electric current flows.

T = n/i

T = 2.898 * 10^ 23 / 9.08×10¹⁸

T = 0.3191 * 10^5 s

T = 531 minutes

Hence,

It would take 531 minutes for a single electron in the electron sea to drift from one end to the other end of a wire 35 cm long.

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when they meet the 40-kg boy will have moved a distance of 6 m.

Friction is the force that resists motion when the surface of one object comes in contact with the surface of another.

It is the average position of all the parts of the system, weighted according to their masses.

here,

masses of the boys

M=40 kg,

m=60 kg

x1 = 10 m

The displacement of the 40 kg boy is calculated from the principle of center mass.

XM = ( mx1 + Mx2) / ( m + M )

X(40) = (60 x 10 +  40 x 0) / (40 kg + 60 kg)

X(40) = (600)/(100)

X(40) = 6 m

X(60) = (60 x 0 + 40 x 10 m) / (40 kg + 60 kg)

X(60) = (400)/(100)

X(60) = 4 m

Thus,

when they meet the 40-kg boy will have moved a distance of 6 m.

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A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300 then its inventories will be 12300

Inventory is the accounting of items, component parts and raw materials that a company either uses in production or sells

The quick and current ratios are liquidity ratios that help investors and analysts gauge a company's ability to meet its short-term obligations. The current ratio divides current assets by current liabilities. The quick ratio only considers highly-liquid assets or cash equivalents as part of current assets.

current ratio = current assets / current liabilities

current assets = current ratio  * current liabilities

                        = 1 * 12300 = 12300

since , inventory is a current asset for accounting purpose , hence inventories will be 12300

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0.07  is the magnitude of the magnetic field at a distance of 2d from the wire

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The tangential speed of a lug nut on a wheel of a car will be 4.68 m/s

Given:

Distance of lug nut = 0.114 m

wheel is rotating at speed = 6.53 rev/sec

To Find:

tangential speed of a lug nut

Solution: Tangential velocity is the linear speed of any object moving along a circular path. A point on the outside edge of a turntable moves a greater distance in one complete rotation than a point near to the center

Vr = r.ω

(0114 m)(41.03 rad/s) = Vr

Vr = 4.68 m/s

Hence, tangential speed of lug nut on wheel of car is 4.68 m/s

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The forces listed are considered fundamental forces (2) gravitational (4) strong nuclear (6) electroweak.

Fundamental force, also called fundamental interaction, in physics, is any of the four primary forces—gravitational, electromagnetic, sturdy, and weak—that govern how gadgets or particles interact and how certain particles decay. All of the regarded forces of nature may be traced to those essential forces.

Now, physicists say they have discovered viable symptoms of a fifth essential force of nature. The findings come from studies performed at a laboratory near Chicago. The four fundamental forces govern how all of the objects and particles inside the Universe engage with each other.

The strong nuclear pressure pulled definitely and negatively charged quarks collectively to form undoubtedly charged protons and neutrally charged neutrons. The strong nuclear force also binds protons and neutrons within the nucleus of atoms. The weak nuclear force enabled complicated atoms to form through nuclear fusion.

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Galaxy X contains more total mass within 25,000 light-years of its center than Galaxy Y.

A light-year, a significant unit of length used to express astronomical distances, is roughly 9.46 trillion kilometers, or 5.88 trillion miles. A light-year is the distance that light travels in a vacuum in one Julian year, according to the International Astronomical Union. The name "light-year" is occasionally misunderstood as a unit of time since it contains the time measurement word "year". When expressing distances to stars and other distances on a galactic scale, the light-year is most frequently employed, especially in non-specialized situations and popular science publications. The parsec, which comes from astrometry and is the distance at which one astronomical unit subtends an angle of one second of arc, is the unit that is most frequently used in professional astronomy.

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Answer:

initial velocity is the constant in horizontal projectile

Hydrogen, the element that is by far the most prevalent in the universe, exhibits radio emissions with a wavelength of 21 centimetres.

This makes it practical for using radio waves to explore space objects. An object in space emits radio waves that are more noticeable the more hydrogen it has.

The majority of a disk-shaped galaxy's mass is located on the galactic plane. The galactic poles are shown by directions that are perpendicular to the galactic plane.

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The rarest type of stars is those that are more than 50 solar masses.

How are stars categorized?

Stars are categorized on the basis of mass, spectral behavior, and temperature.

There are 7 spectral types of stars:

Thus, the rarest type of stars are the O-type stars that have more than 50 times the solar mass of the sun.

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True

In place of photographic plates or films, charge coupled devices, or CCDs, are sensitive photon detectors that can be employed in telescopes to create images. CCDs were created in the late 1960s and are now found in various devices, including digital cameras and photocopiers.

CCDs are more light-efficient, catch a greater proportion of incoming photons, can store and transfer digital data, and don't require development like film does. They respond linearly and capture up to 80% of the light that strikes them.

I understand the question you are looking for is this:

One of the primary advantages of CCDs over photographic plates is the former's high efficiency in detecting light. True or false.

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Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

                                          [tex]T=2\pi \sqrt{\frac{m}{k} }[/tex]

               [tex]T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s[/tex]

              [tex]T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\[/tex]

               [tex]m=\frac{5}{8} =0.625kg[/tex]

Thus, we can conclude that, the mass m will be 0.625kg.

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Option b) Stomach is most likely to have the most muscles.

There are five main muscles in stomach pyramidalis, rectus abdominus, external obliques, internal obliques, and transversus abdominis.

Muscles circulate frame parts by contracting after which relaxing. muscular tissues can pull bones, however, they cannot push them to return to their unique role. so they paintings in pairs of flexors and extensors. The flexor contracts to bend a limb at a joint.

The muscle groups that circulate our frame elements are called skeletal muscles, and they're a form of striated muscle. We will actively manipulate those with our brains. another sort of striated muscle is those that maintain our hearts pumping, which we are not able to actively control.

Muscles pull at the joints, allowing us to move. a couple of muscular tissues are used for any sort of motion of a bone. Although we are sitting flawlessly nevertheless, muscle tissues at some point in the frame are constantly shifting.

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The distance traveled by the sprinter in meters is determined as 1.88 m.

The acceleration of the sprinter is the rate of change of velocity of the sprinter with time.

The acceleration of the sprinter is calculated as follows;

Apply Newton's second law of motion as follows;

F = ma

a = F/m

where;

a = 693 N / 64 kg

a = 10.83 m/s²

The distance traveled by the sprinter is calculated as follows;

s = ut + ¹/₂at²

where;

s = ¹/₂at²

where;

s = (0.5)(10.83)(0.59²)

s = 1.88 m

Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.

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Sport-type motorcycles typically feature a higher than average power-to-weight ratio.

What is scrambler motorcycle?

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Answer: A small mailbag is released from a helicopter that is descending steadily at 1.53 m/s. Then,

a) after 4s, the speed of the mailbox will be, 40.73m/s downwards.

b)78.4m downwards.

c) v=37.67m/s downward. and d=78.4m downward.

Explanation: To find the answer, we need to know more about the equations of uniformly accelerated motion.

                                [tex]v=u+at\\S=ut+\frac{1}{2}at^2\\ v^2-u^2=2aS\\S=\frac{v+u}{2}t[/tex]

            [tex]v=u+gt=-1.53-(9.8*4)=-40.73 m/s^2[/tex]

           [tex]d_1=ut+\frac{1}{2}gt^2=(-1.53*4)+\frac{-9.8*16}{2} =-84.52 m\\d_2=vt=-40.73*4=-162.92 m\\d=-162,92+84.52=-78.4m[/tex]

                 [tex]v=u+gt=1.53-(9.8*4)=-37.67 m/s[/tex]

                 [tex]d_1=ut+\frac{1}{2}gt^2=(1.53*4)+\frac{-9.8*16}{2} =-72.28 m\\ d_2=vt=-37.67*4=-150.68 m\\d=-150.68+72.28=-78.4 m[/tex]

Thus, we can conclude that, the answers to the question are, a) after 4s, the speed of the mailbox will be, 40.73m/s downwards.

b)78.4m downwards.

c) v=37.67m/s downward. and d=78.4m downward.

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Answer:  When you cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.

Explanation: To find the correct answer, we need to know more about the traffic signal rules.

Thus, we can conclude that, the option C is correct.

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F = ma is the formula of Newton's Second Law of Motion. Newton's Second Law of Motion is described as Force is equal to the rate of change of momentum. For a constant mass, force equals mass duration acceleration.

Newton's Second Law of Motion F=ma is very significant because it indicates the relationship between forces and motion. It permits you to calculate the acceleration (and therefore velocity and position) of an object with known forces. This is extremely helpful for scientists, engineers, inventors, etc.

Therefore : F is directly proportional to m×a. This relation F is directly proportional to m×a can be turned into an equation by placing in a constant k.

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Answer: A satellite with a mass of 110 kg and a kinetic energy of 3.08×10^9 J must be moving at a speed of 7483 m/s.

Explanation: To find the answer we need to know about the kinetic energy of a body.

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

                                   [tex]m=110kg\\KE=3.08*10^9J\\[/tex]

                               [tex]v=\sqrt{\frac{2KE}{m} } =\sqrt{\frac{2*3.08*10^9}{110} } =7.483*10^3 m/s[/tex]

Thus, we can conclude that, the velocity of the satellite will be 7438m/s.

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In order to move at a speed of 7483 m/s, a satellite with a mass of 110 kg and a kinetic energy of 3.08× 10^-9 J must be travelling.

Understanding a body's kinetic energy is necessary in order to determine the solution.

                         [tex]KE=\frac{1}{2}mV^2[/tex]

               [tex]m=110kg\\KE=3.08*10^9J[/tex]

                        [tex]V=\sqrt{\frac{2KE}{m} }=7.48km/s[/tex]

Thus, we can infer that the satellite will move at a speed of 7438 m/s.

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If the strength of a magnetic field at B is three units the strength of the magnetic field at A, means that point B has three times more turns as compared to point A.

We can increase the strength of the magnetic field by increasing the number of turns which increases the strength of the current flow.

So we can conclude that point B has three times more turns as compared to point A.

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The new moment of inertia of the gymnast is 9kgm^2.

To find the answer, we need to know about the conservation of angular momentum.

                      I₁×w₁= I₂×w₂

                      I₂= I₁×w₁/w₂

                         = (6×12)/8

                         = 9kgm^2

Thus, we can conclude that the new moment of inertia of the gymnast is 9kgm^2.

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The angular speeds of the wheels is 1,647.28 rad/min

angular velocity is the speed at which the angle between two bodies changes as an item rotates or revolves around an axis.

The diameter of a bicycle, d = 20inches

Radius, r = 10 inches

The bicycle is travelling at 16 mi/h.

We need to find the angular speed of the wheels in rad/min.

We know that,

1 mi/h = 0.0166667 mi/min

It means,

16 mi/h = 0.26 mi/min

Also,

10 inch = 0.0001578356 miles

The angular speeds of the wheels is given by :

v = rω

ω = v/r

ω = 0.26/0.0001578356

ω = 1,647.28 rad/min

So, the angular speeds of the wheels is 1,647.28 rad/min

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The area of Piston A is 13.88 in^2

What is a hydraulic system?

The hydraulic system functions by using pressurized fluid.  The power of liquid fuel in hydraulics is significant and as a result, they are commonly used in heavy equipment.

Hydraulic system simply means a mechanical function that operates through the force of liquid pressure.

Here

A hydraulic system contains 3 pistons.

Pressure, P = 1440 psi

for piston A:

Area, A= 0.625 in^2

Force, F = 20,000 lb

We know that,

Force = Pressure x Area

F = P x A

Area ( piston A) = F / P

A = 20,000 / 1440

A = 13.88 in^2

Hence,

The area of Piston A is 13.88 in^2

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