2. A ball is tossed straight upwards with a velocity of 60m/s. Using an estimate of gravity of -10m/s², about how many seconds would it take to reach the peak (top of the flight). A.) 10 seconds B.) 3 seconds C.) 6 seconds D.) 12 seconds​

In the NEC Code, 210.8 indicates article 210, section 8 and is denoted as option A.

This is referred to the standards which are required for the safe installation of electric wires and equipment in USA.

This comprises of several articles covering different topics and sections which gives it a more elaborate meaning and understanding.

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The maximum height above the point of release it rises is 8.82 m

The force exerted by a spring on objects attached to its ends is proportional to the spring's change in length away from its equilibrium length.

F(spring) = k*x

here,

mass, m=0.240 kg

vertical spring constant, k = 5050 N/m

compression, x=0.090 m

When the spring is compressed, the spring potential energy is:

E = (k*x^2) / 2

substituting the values,

E = 5050* 0.090^2 / s

E = 20.91 J

This energy would be converted to kinetic, so the mass gains speed, which is then converted to gravitational potential energy once the mass reaches its highest point

E(p) = m*g*h

where, h is the maximum height

E(p) = E

m*g*h = 20.91

h = 20.91 / ( 9.8 * 0.248 )

h = 8.82 m

hence

The maximum height above the point of release is 8.82 m

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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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Richard Dyer written about stars that a star must be represented simultaneously as, just like us (the audience/regular people), but also at the same time possess something we do not have and something that makes them special, different, extraordinary.

According to the question

Richard Dyer is a British media theorist, who discussed a star’s special place in the audiences’ lives.

He (similarly to Hall) suggests that a star’s meta-narrative impacts on the consumption by the audience. He argued that the star image is manufactured and artificial and that individual stars have their own unique selling point (brand values) in order to grab and hold our attention – for

example – Justin Bieber’s hair, JLS’ colour codes...

Hence,  Richard Dyer written about stars that a star must be represented simultaneously as, just like us (the audience/regular people), but also at the same time possess something we do not have and something that makes them special, different, extraordinary.

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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

It looks like you're given

[tex]x(t) = -11\,\mathrm m + \left(1\dfrac{\rm m}{\rm s}\right) t - \left(5\dfrac{\rm m}{\mathrm s^3}\right) t^3[/tex]

[tex]y(t) = 19\,\mathrm m + \left(3\dfrac{\rm m}{\rm s}\right) t - \left(9\dfrac{\rm m}{\mathrm s^2}\right) t^2[/tex]

The particle's position vector at time [tex]t[/tex] is given by

[tex]\vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

Differentiate [tex]\vec r[/tex] twice to recover the velocity and acceleration vectors.

[tex]\vec a(t) = \dfrac{d\vec v(t)}{dt} = \dfrac{d^2\vec r}{dt^2} = x''(t)\,\vec\imath + y''(t)\,\vec\jmath \\\\ \implies \vec v(t) = \left(1\dfrac{\rm m}{\rm s} - \left(15\dfrac{\rm m}{\mathrm s^3}\right) t^2\right)\,\vec\imath + \left(3\dfrac{\rm m}{\rm s} - \left(18\dfrac{\rm m}{\mathrm s^2}\right) t\right) \,\vec\jmath \\\\ \implies \vec a(t) = -\left(30\dfrac{\rm m}{\mathrm s^3}\right)t \, \vec\imath - \left(18\dfrac{\rm m}{\mathrm s^2}\right) \,\vec\jmath[/tex]

At [tex]t=1.4\,\rm s[/tex], the particle has acceleration

[tex]\vec a(1.4\,\mathrm s) = \left(-42\,\vec\imath - 18\,\vec\jmath) \dfrac{\rm m}{\mathrm s^2}[/tex]

with magnitude

[tex]\|\vec a(1.4\,\mathrm s)\| = \sqrt{\left(-42\dfrac{\rm m}{\mathrm s^2}\right)^2 + \left(-18\dfrac{\rm m}{\mathrm s^2}\right)^2} \approx 45.695\dfrac{\rm m}{\mathrm s^2}[/tex]

and making an angle [tex]\theta[/tex] relative to the positive [tex]x[/tex]-axis such that

[tex]\tan(\theta) = \dfrac{-18}{-42} = \dfrac37[/tex]

Since both components of the acceleration vector have negative sign, the acceleration points into the third quadrant, so that we add a multiple of 180° after taking the inverse tangent of both sides, namely

[tex]\theta = \tan^{-1}\left(\dfrac37\right) - 180^\circ \approx -156.801^\circ[/tex]

Now, (a) the magnitude of the net force acting on the particle is, by Newton's second law,

[tex]F = (0.41\,\mathrm{kg})\|\vec a(1.4\,\mathrm s)\| \approx \boxed{18.735\,\mathrm N}[/tex]

and (b) makes the same angle as the acceleration vector, [tex]\theta \approx \boxed{-156.801^\circ}[/tex].

At this moment, its velocity vector is

[tex]\vec v(1.4\,\mathrm s) = \left(-28.4\vec\imath - 22.2\,\vec\jmath\right) \dfrac{\rm m}{\rm s}[/tex]

which (c) makes an angle [tex]\theta[/tex] such that

[tex]\tan(\theta) = \dfrac{-22.2}{-28.4} = \dfrac{111}{142}[/tex]

This vector also points in the third quadrant, so

[tex]\theta = \tan^{-1}\left(\dfrac{111}{142}\right) -180^\circ \approx \boxed{-142.986^\circ}[/tex]

The focal plane  is a plane perpendicular to the optic axis through the focal point.

What is  focal plane?

What is focal plane and lens?

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You need the atomic mass of calcium, oxygen and carbo, the balanced chemical equation and the molar ratio between Calcium carbonate and calcium oxide.

1. If you know the atomic mass of the elements, you can calculate the molar mass of Calcium Carbonate (CaCO3) and the mass of Calcium Oxide (CaO).

2. The balanced equation is CaCO3 -> CaO + CO2

3. The molar ratio between CaCO3 and CaO is 1:1, which means that the descomposing of one mole of calcium carbonate produce one mole of calcium oxide.

4- The last thing you should do is to conver the 4.7 kg of CaCO3 to moles and with the molar ratio calculate the mass of CaO

Explanation:

Any confusions you may ask!

An athlete produced 840 watts of power in 0.6 seconds. And will produce 504 joule of work did this individual perform during the exercise.

Solution:

we know that power is the work done completed in a given interval of time.

Power = [tex]\frac{work}{time}[/tex]

also, Work = Power × Time

Work = 840 × 0.6 = 504 joule

What is Power?

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The speed of Block A is 7.18 ft/s.

From energy conservation we know that the initial kinetic energy is equal to final kinetic energy.

Block A has weight of 60 lb.

Block B has weight of 10 lb.

In the question it is said that block A moves from rest at a distance of 5ft.

Relative motion constraint-

2sA + sB  =l

differentiating we get:

So,    2ΔsA + ΔsB  = 0

→  ΔsB = -2ΔsA

2 vA + vB  =0

→  vB = -2vA

Now applying principle of work and energy

TA0 +TBO + ΣwO-AB  =  TA+TB

1/2 mA v²A₀ + 1/2 mBv²B₀ + wAΔsA + wBΔsB = 1/2 mAv²A + 1/2 mBv²B

0 + 0 +6(5) - 10(10) = 1/2 (-60/32-2) v²A +1/2 (10/32.2) 2v²A

vA = 7.18 ft/s

So, the speed of block A is 7.18 ft/s.

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Head to tail rule helps us to find the resultant of forces because in this rule we join many vectors with each other forming resultant vector.

The joining of tail of the first vector with the head of the second vector represents the resultant vector whereas the direction of the resultant vector is from the tail of the first vector towards the head of the second.

So we can conclude that Head to tail rule helps us to find the resultant of forces because in this rule we join many vectors with each other forming resultant vector.

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The resultant magnitude and direction of the vectors are as follows:

The resultant magnitude and direction of the vectors are as follows:

A. A = √(5.7² + 3.4²)

A = 6.6

θ = tan⁻¹(3.4/5.7)

θ = 30.8° from the x-axis

B. B = √{(-10)² + (-3²)}

B 10.4

θ = tan⁻¹(3/10) + 180

θ = 16.7° + 180°

θ = 196.7° from the x-axis

C. C = √{(12)² + (-20²)}

C = 23.3

θ = 360° - tan⁻¹(20/12)

θ = 300° from the x-axis

D. D = √{(-30)² + (27²)}

D = 40.4

θ = 180 - tan⁻¹(27/30)

θ = 138° from the x-axis

E. E = √{(48)² + (12²)}

E = 49.5

θ = 360° - tan⁻¹(12/48)

θ = 346° from the x-axis

In conclusion, the resultant of the vectors and direction are obtained using the resultant vector formula.

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Mass of 1 mole of pennies is 1.506 × 10²¹ kg. 48.805 moles moles of pennies have a mass equal to the mass of the Moon.

Mass of one mole pennies= 6.023 × 10²³ × mass of one penny

= 6.023 × 10²³ ×0.0025 kg

= 1.506 × 10²¹ kg

= 48.805 moles

Thus, we can conclude that the mass of 1 mole of pennies is 1.506 × 10²¹ kg and 48.805 moles moles of pennies have a mass equal to the mass of the Moon.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: A penny has a mass of 2.50g and the Moon has a mass of 7.35 X 10^22kg. Use this information to answer the questions below.

A) What is the mass of 1 moIe of pennies? Round your answer to 3 significant digits.

B) How many moles of pennies have a mass equal to the mass of the Moon? Round your answer to 3 significant digits.

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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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Answer:

Explanation:

Formula

E = I * R                                      Substitute the givens into the formula

Givens

E = 120 volts

I = ?

R = 4800 0hms

Solution

Ohms law is a 3 letter equation that relates volts ohms (resistance) * current (in amperes)

E = i * R

120 V = I * 4800 ohms               Divide both sides by 4800  

120 V/4800 = I * 4800/4800     Divide

0.025 amps = I

Answer

I = 0.025 amps

The net gain in potential energy by the crate is 120 J.

Potential energy is the energy that an object has stored in it as a result of its position or condition. Potential energy exists in a stretched spring, a book held above your head, and a bicycle on top of a hill. The joule, which is represented by the letter "J," is the standard unit for calculating potential energy.

We use the idea of potential energy, which is given by the equation,

PE = mgh

where,

m = mass of the crate

g = gravitational acceleration

h = height

Given,

mg = 40 N

L = 5 m

angle of inclination, ∅ = 37°

h = L sin∅

h = 5 (sin 37°)

h = 5 (3/5)

h = 3 m

Put the values in the above formula,

PE = mgh

PE = 40 (3)

PE = 120 J

Therefore, the net gain in potential energy by the crate is 120 J.

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The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

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

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

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Inspiratory reserve volume is the additional amount of air that can be inhaled at the end of quiet inspiration.

Inhalation : The act of drawing air into the lungs during inspiration, also known as inhalation.

By establishing a pressure differential between the lungs and the atmosphere, it is accomplished.

The diaphragm contracts toward the abdominal cavity as air enters the lungs, creating more room in the thoracic cavity for the air to fit.

Inspiratory reserve volume is the additional volume of air that can be maximally inspired after finishing a typical, quiet inspiration.

Or the volume of air that can be inhaled following a regular breath is known as the inspiratory reserve volume.

It is denoted by IRV.

In men average IRV is 3.3 L whereas in women it is about 1.9 L

Hence, Inspiratory reserve volume is the additional amount of air that can be inhaled at the end of quiet inspiration.

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A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 [tex]m^{3}[/tex]/sec .

Discharge rate = velocity * area

                 = velocity * depth * width

                 = 2 * 2 * 10 = 40 [tex]m^{3}[/tex]/sec

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 [tex]m^{3}[/tex]/sec .

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Using the left hand rule, if current points up and the field is toward you,  the  motion points up.

The left hand rule states that if the thumb, the fore finger and the middle finger are held mutually perpendicular to each other, the thumb points in the direction motion, the middle finger points in the direction of current while the fore finger points in the direction of the magnetic field.

Thus using the left hand rule, if current points up and the field is toward you,  the  motion points up.

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The hottest celebrity is this one.

The majority of stars fall under a select few spectral categories. The hottest to the coldest stars are listed in the spectral types of the Henry Draper and Bright Star Catalogues (see stellar classification). The letters O, B, A, F, G, K, and M are used to indicate these categories in decreasing order of temperature.

Using spectral types, we can categorise stars based on their colour or the spectral lines that are visible in their light. The spectral types are OBAFGKM, with the hottest stars being O and the coldest being M. To humans, cooler stars appear redder and hotter stars appear bluer.

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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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Hudson Bay is the  restricted basin that has the coolest temperatures

Hudson Bay is a restricted basin which  remains frozen or is dominated by ice over the summer solstice and through- out much of the high-sun season. This basin experiences a harsh continental climate.

The average annual temperature in almost the entire bay is around 0 °C (32 °F) or below. In the extreme northeast, winter temperatures average as low as −29 °C or −20.2 °F. The region of this basin has very low year-round average temperatures.

This basin starts freezing up by early November, and the northern part of the basin is typically entirely iced over by the end of the month.

correct answer is Hudson bay

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The reason for which the jovian planets lost very little of their original atmosphere is due to their large mass.

The jovian planets are celestial bodies formed by gaseous huge masses of dust instead of rock planes such as, among others, the Earth planet.

These types of planets (jovian planets)  include four major celestial bodies, i.e., Jupiter, Saturn, Uranus, and Neptune.

The jovian planets are characterized to have a thick atmosphere that is composed of different elements such as hydrogen, and methane.

In conclusion, the reason for which the jovian planets lost very little of their original atmosphere is due to their large mass.

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Work done by the force is 6.802 Joule

Given:

Therefore displacement of coin along x axis,x=3m

displacement of coin along y-axis,y=4m

force has magnitude = 2.0 N

To Find:

Work done by the force

Solution: Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

component of force along x-axis, F = 2.0Ncos100

F = 2x(-.173)

F = -(.346)N

component of force in y direction,F₁ = 2.0sin100

F = 2x(.98)

F = 1.96N

work is done by the force on the coin during the displacement will be given as

W = Fx*x + Fy*y

W = -(.346) x 3m + 1.96 x 4

W = 6.802 Joule

Hence, Work done by the force is 6.802 Joule

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The object will drop 5 cm if released  before coming to rest.

We know from Hooke's law that the work that is done when the spring is stretched is given by the formula;

W = 1/2 kx^2

k= 100 N/m

x =  1.00 cm or 0.01 m

W = 0.5 * 100 * 0.01

W = 0.5 J

Now;

W = mgh

h = W/mg

m =  1.00 kg

g = 9.8 m/s^2

h =  0.5 J/1.00 kg  * 9.8 m/s^2

h = 0.05 m or 5 cm

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The decreasing order of the momenta are b>a=d=e>c.

Solution: The momentum of all the options are mentioned below according to the formula:

a. p = 20 × 1 = 20gm/s

b. p = 20 × 2 = 40gm/s

c. p =10 × 1 = 10gm/s

d. p = 10 × 2 = 20 gm/s

e. p =200 × 0.1 = 20gm/s

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Question:

I hope you are trying to ask the question below:

Rank in order, from largest to smallest, the momenta a to e of the objects

a. 20 g & 1 m/s

b. 20 g & 2 m/s

c. 10 g & 1 m/s

d. 10 g & 2 m/s

e. 200 g & 0.1 m/s

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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The SI unit of current is the ampere, which is defined as the flow of charge in units of coulomb over a period of time measured in seconds.

Electrical charge carriers, often electrons or atoms deficient in electrons, travel as current. The capital letter I is a typical way to represent the current. The ampere, denoted by the letter A, is the common unit. An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor. One coulomb of electrical charge (6.24 x 1018 charge carriers) traveling by a given place in one second is represented by one ampere of current.

Conventional current, also known as Franklin current, is thought by physicists to flow from relatively positive points to comparatively negative locations. The most prevalent charge carriers, electrons, are negatively charged. From somewhat negative to relatively good points, they move.

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The student's answer is wrong because the unit of his answer is meter - seconds which is not unit of distance (meter).

The total distance traveled by an object projected upward is calculated from third equation of motion a shown below.

v² = u² - 2gh

where;

at maximum height, the final velocity is zero, v = 0

0 = u² - 2gh

2gh = u²

h = u²/2g

The equation for the total distance an object has traveled during upward projection is "h = u²/2g".

d = (2vh)/g = (m/s x m) / (m/s²)  = (m²/s) / (m/s²) = m.s

Thus, the student's answer is wrong because the unit of his answer is meter - seconds which is not unit of distance (meter).

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