You are driving down the highway at 65 m p h, which is 29 m/s. Your tires have a radius of 0.30 m. a. How many times per second does each tire rotate? b. What is the speed of a point at the top of a tire, relative to the ground?

Answer:

Force between two identical conducting ball will be [tex]2.65\times 10^{-5}N[/tex]                        

Explanation:

We have given there is two identical conducting small spheres having charge [tex]q_1=12nC=12\times 10^{-9}C[/tex] and [tex]q_2=-21nC=-21\times 10^{-9}C[/tex]

Distance between centers of the ball is 0.290 m

So r = 0.290 m

We have to fond the electric force between the balls

According to Coulomb's law force between two charges is given by

[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex]

So force [tex]F=\frac{9\times 10^{9}\times 12\times 10^{-9}\times 21\times 10^{-9}}{0.293^2}=2.65\times 10^{-5}N[/tex]

So force between two identical conducting ball will be [tex]2.65\times 10^{-5}N[/tex]

To solve this problem we will use Henry's law. This law states that at a constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure exerted by that gas on the liquid. Mathematically it is formulated as follows:

[tex]C = K_H*P[/tex]

Where,

[tex]K_H[/tex] = Henry's constant for C02 at 25°C is equal to [tex]3.6*10^{-2}M/atm[/tex]

C = Gas concentration is 0.19M

Replacing we have,

[tex]0.19 M = (3.6*10^-2 M/atm)*P[/tex]

[tex]P = 5.277 atm[/tex]

Therefore the pressure of carbon dioxide is 5.277 atm

To calculate the pressure of CO2 needed to maintain a 0.19 M concentration in a can of soda, you apply Henry's law, which results in the pressure equalling 0.00684 atm.

The subject of this question is Henry's law, which is used in Chemistry to relate the solubility of a gas in a liquid to the pressure of that gas above the liquid. In this case, we're asked to calculate the pressure of the CO2 gas needed to maintain a certain concentration in a can of soda. Henry's law is defined as: P = kH × C, where P is the gas pressure, kH is Henry's law constant, and C is the concentration.

To solve this problem, we rearrange Henry's law to find the pressure: P = kH × C. Given that the Henry's law constant for CO2 is 3.6 × 10−2 M/atm and the CO2 concentration is 0.19 M, we can apply these values and get P = (3.6 × 10−2 M/atm) × (0.19 M) which equals 0.00684 atm.

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Answer:

Coefficients are 1/2, -z and +1

Explanation:

The step by step explanation is in the attachment given below.

Answer:

Power will be 356.90 watt  

Explanation:

We have given total number of steps in the stair = 1600

And height id the stair h = 320 m

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

Mass is given m = 75 kg

So work done in climbing the stairs [tex]W=mgh=75\times 9.8\times 320=235200J[/tex]

Time is given t = 10 min 59 sec

So time [tex]t=10\times 60+59=659sec[/tex]

We know that power is rate of doing work

So power [tex]P=\frac{Work}{time}=\frac{235200}{659}=356.90watt[/tex]

So power will be 356.90 watt

The man generated approximately 357.26 watts or 0.479 horsepower while climbing the stairs.

To calculate the power generated by the man while climbing the stairs, we can use the work-energy principle. The work done by the man in climbing to a height  against the force of gravity is equal to the change in his gravitational potential energy. The power is then the work done divided by the time taken to do it.

 The gravitational potential energy (PE) is given by:

[tex]\[ PE = mgh \][/tex]

where:

-[tex]\( m \)[/tex] is the mass of the man (75 kg),

-[tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \) on Earth),[/tex]

- [tex]\( h \)[/tex] is the height climbed (320 m).

 First, we calculate the work done (PE):

[tex]\[ PE = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 320 \, \text{m} \][/tex]

[tex]\[ PE = 235,440 \, \text{J} \][/tex]

 Now, we convert the time from minutes to seconds because power is typically calculated in joules per second (watts). The time taken is 10 minutes and 59 seconds, which is:

[tex]\[ 10 \times 60 \, \text{s} + 59 \, \text{s} = 659 \, \text{s} \][/tex]

 Power (P) is the work done divided by the time taken:

[tex]\[ P = \frac{PE}{t} \][/tex]

[tex]\[ P = \frac{235,440 \, \text{J}}{659 \, \text{s}} \][/tex]

[tex]\[ P \approx 357.26 \, \text{W} \][/tex]

 To convert watts to horsepower, we use the conversion factor [tex]\( 1 \, \text{hp} = 746 \, \text{W} \):[/tex]

[tex]\[ P_{\text{hp}} = \frac{P}{746} \][/tex]

[tex]\[ P_{\text{hp}} \approx \frac{357.26 \, \text{W}}{746} \][/tex]

[tex]\[ P_{\text{hp}} \approx 0.479 \, \text{hp} \][/tex]

 Therefore, the man generated approximately 357.26 watts or 0.479 horsepower while climbing the stairs.

 The final answer in the correct format is:

[tex]\[ \boxed{357.26 \, \text{W}} \][/tex]

[tex]\[ \boxed{0.479 \, \text{hp}} \][/tex]

 The answer is:[tex]0.479 \, \text{hp}.[/tex]

Answer: A would experience a greater attraction to a point charge a distance away.

Explanation: An atom that is more easily polarizable, the separation of charge in that atom will be greater. The larger the dipole moment, the larger is the force.

Answer: Atom A

atom A would experience a greater attraction to a point charge a distance away because it is easily polarized.

Explanation:

Polarization can be defined as the process of separating opposite charges within an object. The positive charge becomes separated from the negative charge, when this happens the atom is said to have been polarized and have an electric dipole moment. So, the polarized molecules are attracted more toward the charged object because the field increases in the direction of the charged object. Therefore, atom A would experience a greater attraction to a point charge a distance away because it is easily polarized.

Answer:

ms^(-3)

Explanation:

V = kt^2

m/s = k (s)^2

k = m/ s^(3)

k = ms^(-3)

Explanation of finding the net electric force on a charge Q using Coulomb's Law in an arrangement with charges on the square's corners.

To find the net electric force on a charge Q located at the center of a square with charges on its corners, we must consider the forces exerted by each charge using Coulomb's Law.

The net electric force on charge Q can be calculated by adding the individual force vectors from the charges Q₁ and Q₂.

By determining the force vectors and summing them up, we can express the net electric force on charge Q in terms of q, Q, a, and appropriate constants.

The magnitude of the net electric force exerted on the charge Q at the center of the square is [tex]\(\frac{4kqQ}{a^2}\).[/tex]

To find the magnitude of the net electric force exerted on the charge Q at the center of the square, we can calculate the electric force exerted by each individual charge and then sum them up vectorially.

Let q be the magnitude of each corner charge and a be the side length of the square.

The electric force between two charges q separated by a distance r is given by Coulomb's law:

[tex]\[F = \frac{k \cdot q \cdot Q}{r^2}\][/tex]

Considering symmetry, the horizontal components of the electric forces from the charges on the left and right sides cancel out, as do the vertical components from the top and bottom charges.

Thus, the net electric force acting on Q is the sum of the forces from the top and bottom charges, which are equal in magnitude:

[tex]\[F_{\text{net}} = 2 \cdot F_{\text{top}}\][/tex]

[tex]\[F_{\text{net}} = 2 \cdot \frac{k \cdot q \cdot Q}{(\frac{a}{\sqrt{2}})^2}\][/tex]

[tex]\[F_{\text{net}} = 2 \cdot \frac{k \cdot q \cdot Q}{\frac{a^2}{2}}\][/tex]

[tex]\[F_{\text{net}} = \frac{4kqQ}{a^2}\][/tex]

Therefore, the magnitude of the net electric force exerted on the charge Q at the center of the square is [tex]\(\frac{4kqQ}{a^2}\).[/tex]

To solve this problem we will apply the concept designed to generate the conversion from one unit to another. Basically, what is sought is to eliminate the units of the denominator and the numerator of the conversion factor and the unit to be converted respectively, leaving the units of the new unit. In mathematical terms this is,

[tex]1 in = 2.54cm[/tex]

If we want to convert 28.4in then the conversion factor versus the unit would be

[tex]x = 28.4 in (\frac{2.54cm}{1in})[/tex]

[tex]x = 72.136 cm[/tex]

Therefore the factor of conversion will be 2.54 and the final units for the value given is 72.136cm

Answer:

DC = 3 div. AC= 0

Explanation:

When the input is directed to the input circuitry, and the "A input" switch is set to DC, the pure 15 V DC signal will be showed on the screen, so, if the vertical setting is at 5 V /div, the trace will deflect exactly 3 div.

If the switch is on AC, as this setting inserts a capacitor in series (which is located here to block any unwanted DC component superimposed to an AC signal) the DC signal will be blocked, so no trace will be deflected on the screen (after completed the transient period).

In this case, there are 3 divisions if the switch is on direct current (DC), whereas there are 0 divisions if the switch is on alternating current (AC).

In conclusion, there are 3 divisions if the switch is on direct current, whereas there are 0 divisions if the switch is on alternating current.

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Answer:

The tension in the string is 43.9 N.

Explanation:

Given that,

Mass of rock = 5.0 kg

Density of rock = 4800 kg/m³

We need to calculate the volume of rock

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

[tex]V=\dfrac{m}{\rho}[/tex]

Put the value into the formula

[tex]V=\dfrac{5.0}{4800}[/tex]

[tex]V=0.001041\ m^3[/tex]

We need to calculate the volume of water

[tex]V_{w}=\dfrac{V}{2}[/tex]

Put the value of volume

[tex]V_{w}=\dfrac{0.001041}{2}[/tex]

[tex]V_{w}=0.0005205\ m^3[/tex]

We need to calculate the mass of water displaced

Using formula of mass

[tex]m = 1000\times0.0005205[/tex]

[tex]m=0.5205\ kg[/tex]

We need to calculate weight of water displaced

Using formula of weight of water

[tex]W=0.5205\times9.8[/tex]

[tex]W=5.1009\ N[/tex]

Weight of rock is

[tex]W_{r}=5.0\times9.8[/tex]

[tex]W_{r}=49\ N[/tex]

We need to calculate the tension in the string

Using formula of tension

[tex]T=\text{weight of rock - weight of water displaced}[/tex]

Put the value into the formula

[tex]T=49-5.1009[/tex]

[tex]T=43.9\ N[/tex]

Hence, The tension in the string is 43.9 N.

The tension in the string is mathematically given as

T=43.9 N

Question Parameter(s):

A 5.0kg rock whose density is 4800 kg/m3 suspended bya string

Generally, the equation for the Volume is mathematically given as

[tex]V=\frac{m}{\rho}[/tex]

Therefore

V=5.04/800

V=0.001041 m^3

for water

Vw=0.0005205 m^3

Weight of water

W=0.520*9.8

W=5.1009 N

Weight of rock

Wr=49N

In conclusion, Tension on string

T=49-5.1009

T=43.9 N

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Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²[tex]\sqrt{((17.4)^{2} +(16.9)^{2}}[/tex] = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

Answer:hello

Explanation:

Answer:

Fx = (783.451 , 373.063 , 240.124 , 240.124) N

Fy = (470.745 , 597.026 , 318.66)

Fres =768.216

Theta = 77.2 degrees from + x

Explanation:

F1 = 914N; F2 = 704N ; F3 = 399N

Part A

Fx1 = F1*cos (A) = 914*cos (31) = +783.451 N

Fx2 = F2*sin (B) = 704*sin (32) = -373.063 N

Fx3 = F3*cos (C) = 399*cos (53) = -240.124 N

                                   Sum of Fx = +170.26 N

Part B

Fy1 = F1*sin (A) = 914*sin (31) = +470.745 N

Fy2 = F2*cos (B) = 704*cos (32) = +597.026 N

Fy3 = F3*sin (C) = 399*sin (53) = -318.66 N

                                     Sum of Fy = + 749.111 N

Part C

Fres = sqrt (Fx^2 + Fy^2 )

Fres = sqrt (170.26^2 + 749.111^2)

Fres = 768.216 N

Part D

Angle = arctan (Fy/Fx)

Angle = arctan (749.111/170.26)

Angle = 77.2 degrees

To find the x and y components of each pull, we use trigonometry. The magnitude of the resultant pull is found by summing up the squares of the x and y components and taking the square root. The direction of the resultant pull is determined using the inverse tangent function.

To find the x components of each pull, we can use trigonometry. The x component of a force is given by the magnitude of the force multiplied by the cosine of the angle it makes with the x-axis. So, the x components of the three pulls are: F1x = F1 × cos(theta1), F2x = F2 × cos(theta2), F3x = F3 × cos(theta3).

To find the y components of each pull, we use the same approach but with the sine function. So, the y components of the three pulls are: F1y = F1 × sin(theta1), F2y = F2 × sin(theta2), F3y = F3 × sin(theta3).

Once we have the x and y components of each pull, we can find the magnitude of the resultant pull by summing up the squares of the x and y components and taking the square root: F = √((F1x)² + (F2x)² + (F3x)² + (F1y)² + (F2y)² + (F3y)²).

To find the direction of the resultant pull, we can use the inverse tangent function. The angle counted from the +x axis in the counterclockwise direction is given by: theta = atan(F1y + F2y + F3y / F1x + F2x + F3x).

So, the answers to the question are: F1x, F2x, F3x = 914N × cos(theta1), 704N × cos(theta2), 399N × cos(theta3); F1y, F2y, F3y = 914N × sin(theta1), 704N × sin(theta2), 399N × sin(theta3); F = √((F1x)² + (F2x)² + (F3x)² + (F1y)² + (F2y)² + (F3y)²); theta = atan(F1y + F2y + F3y / F1x + F2x + F3x).

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Answer:

a)The acceleration of the airplane is 2.5 m/s².

b)The distance AB is 1125 m.

Explanation:

Hi there!

a)The equation of velocity of an object moving in a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time "t".

v0 = initial velocity.

a = acceleration.

t = time

We have the following information:

The airplane starts with zero velocity (v0 = 0) and its velocity after 30 s is 270 km/h (converted into m/s: 270 km/h · 1000 m/1 km · 1 h /3600 s = 75 m/s). Then, we can solve the equation to obtain the acceleration:

75 m/s = a · 30 s

75 m/s / 30 s = a

a = 2.5 m/s²

The acceleration of the airplane is 2.5 m/s².

b)The distance AB can be calculated using the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time "t".

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at A, then, x0 = 0. Since the airplane is initially at rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

Let´s find the position of the airplane after 30 s:

x = 1/2 · 2.5 m/s² · (30 s)²

x = 1125 m

The position of B is 1125 m away from A (the origin), then, the distance AB is 1125 m.

(a) the acceleration of the plane is 2.5 m/s²

(b) The distance AB is 1125m

(a) Given that the initial velocity of the plane at point A is u = 0

and the final velocity of the plane at point B of take-off is v = 270 km/h

[tex]v=\frac{270\times1000}{3600}m/s=75m/s[/tex]

the time taken to teach point B is t = 30s

So from the first equation of motion, we get:

[tex]v=u+at[/tex]

where a is the acceleration.

[tex]75=a\times30\\\\a=2.5\;m/s^2[/tex]

(b) From the second equation of motion we get,

[tex]s=ut+\frac{1}{2} at^2[/tex]

where s is the distance

[tex]s = \frac{1}{2}\times2.5\times30^2\\\\s=1125m[/tex]

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Answer

70 m/s,  140 m/s, 28 m/s²

Explanation:

Average velocity = total distance travel / time = 350 / 5 = 70 m/s

average velocity = (initial velocity + final velocity) / 2

70 × 2 = 140 m/s

change in velocity = final velocity - initial velocity = 140 m/s - 0 = 140 m/s

acceleration = (final velocity - initial velocity) / t = 140 m/s / 5 = 28 m/s²

Answer:

1. Final velocity is 140m/s

2. Change in velocity is 140m/s

3. Acceleration is28m/s^2

Explanation:

Initial velocity u=0,

Distance travelled S=350m

Time taken t=5 seconds

The car move constant acceleration, then we can use any of equations of motion

v=u+at

v^2=u^2+2as

s=ut+(at^2)/2

Using equation 3

S=ut+(at^2)/2

350= 0×5+ (a ×5^2)/2

350= 0+(a×25)/2

350=25a/2

350×2=25a

700=25a

a=700/25

a=28m/s^2. Answer

1. Now, using equation 1

v=u+at

v=0+28×5

v=0+140

v=140m/s. Answer

2. The changed in the velocity of the car is final velocity minus the initial velocity

Change in velocity = v-u

Change in velocity = 140-0

Therefore,

Change in velocity is 140m/s. Answer

3. The acceleration has been answered before solving and it is

a= 28m/s^2

Answer:

A. 10^12 meters which is bigger than the distance between Sun and Earth

Explanation:

In a case where we take all the atoms in a single drop of water and put them on a single line as closely packed as they can be, the total length of the line would be a function of the diameter of an atom.

Total length L = Diameter of an atom d × number of atom in a droplet of water N

L = dN

N = 10^22

d ~= 0.1nm = 10^-10m

L = 10^22 × 10^-10 m

L = 10^12 m

So, the length would be approximately 10^12 m

distance between the sun and earth is 147.34million km

D= 1.47 × 10^11 m

L > D

Therefore, the length would be approximately 10^12 m

Which is greater than the distance between the earth and the sun.

The force on an ion interacting with an induced dipole can be found using Coulomb's law. The force can be represented as Fionondipole = qEinduced, where q is the charge of the ion and Einduced is the induced electric field. The magnitude of the force can be expressed as (1/(4πε0)) * (2αEext)/(r3) * q, where α is the polarizability of the molecule, Eext is the external electric field, and r is the distance between the ion and the molecule.

The force experienced by an ion interacting with an induced dipole can be found using Coulomb's law. The force can be represented as Fionondipole = qEinduced, where q is the charge of the ion and Einduced is the induced electric field. The induced electric field can be expressed as Einduced = (1/(4πε0)) * (2αEext)/(r3), where α is the polarizability of the molecule, Eext is the external electric field, and r is the distance between the ion and the molecule. Combining these expressions, the magnitude of the force Fionondipole can be written as:

Fionondipole = (1/(4πε0)) * (2αEext)/(r3) * q

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Answer:

f=qB/2[tex]\pi[/tex]m

Explanation:

The cyclotron frequency is the number of revolutions the particle makes in one second. calculate the cyclotron frequency for this particle

solution

The Lorentz force  [tex]F_{lorentz} =F_{centripetal}[/tex]

is the centripetal force  

and makes the particles path to revolve in a circle:

qvB=mv^2/r

radius=r

m=mas of the particle

B=magnetic flux

q=quantity of charge

v=velocity of the particle

v=qBr/m

where v is the velocity of the particle

recall the velocity v=rω

v=2*Pi*f*r

[tex]2\pi *f*r[/tex]=qBr/m

f=qB/2[tex]\pi[/tex]m

the cyclotron frequency is therefore f=qB/2[tex]\pi[/tex]m

the period(the time it take to make a complete oscillation) will be the inverse of frequency

T=2[tex]\pi[/tex]m/qB

Answer:

[tex]\mu_k=\dfrac{gsin\theta -a}{gcos\theta}[/tex]

Explanation:

g = Acceleration due to gravity = 9.80 m/s²

a = Acceleration= 3.6 m/s²

[tex]\theta[/tex] = Angle = 27°

The equation is

[tex]\mu_kmgcos\theta=mgsin\theta -ma[/tex]

Mass gets cancelled

[tex]\\\Rightarrow \mu_kgcos\theta=gsin\theta -a[/tex]

Rearranging for [tex]\mu_k[/tex]

[tex]\\\Rightarrow \mu_k=\dfrac{gsin\theta -a}{gcos\theta}[/tex]

The simplified expression is

[tex]\mathbf{\mu_k=\dfrac{gsin\theta -a}{gcos\theta}}[/tex]

*the options are incomplete. The above answer is the required solution

The simplified expression for coefficient of kinetic friction is 0.097.

The given parameters:

The simplified expression for coefficient of kinetic friction is calculated as follows;

μkmgcosθ = mgsinθ − ma

m(μkgcosθ) = m(gsinθ − a)

μkgcosθ = gsinθ − a

[tex]\mu_k = \frac{gsin\theta - a}{gcos\theta} \\\\\mu_k = \frac{(9.8 \times sin27) - 3.6}{9.8 \times cos(27)} \\\\\mu_k = 0.097[/tex]

Thus, the simplified expression for coefficient of kinetic friction is 0.097.

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Answer:

When the ball is held motionless above the floor, the ball possesses only GPE  energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.

Explanation:

If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.

If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.

[tex]K = \frac{1}{2}mv^2\\U = mgh[/tex]

Answer:

Explanation:

A. GPE

B. GPE

C. KE

Answer:

(a) 0.017m/s^2

(b) 17/100,000

(c) 0.17m, 0.558ft

Explanation:

(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s

Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2

(b) g = 9.8m/s^2, a = 0.0017m/s^2

a/g = 0.0017/9.8 = 0.00017 = 17/100,000

(c) Distance = speed × time = 0.017m/s × 10s = 0.17m

Distance in foot = 0.17 × 3.2808ft = 0.558ft

Answer:

a) acceleration a = 2.68 m/s^2

b) fraction of g = 0.273 or 27.3%

c) distance travelled d = 134m (in SI unit)

d = 439.5ft (in feet)

Explanation:

a) converting 60mph (miles per hour) to m/s

∆v = 60 miles/hour × 1609.344meters/mile × 1/3600 seconds/hour

∆v = 26.82m/s

t = 10s

acceleration = change in velocity/time = ∆v/t

Acceleration a = 26.82/10 = 2.68m/s^2

b) acceleration due to gravity is g = 9.8m/s^2

Fraction of g = a/g = 2.68/9.8 = 0.273

Fraction = 0.273 or 27.3%

c) distance travelled can be calculated using the equation below.

d = ut + 0.5at^2 ....1

Initial speed u = 0

time t = 10s

Acceleration a = 2.68m/s^2

d = distance covered.

From equation 1, putting u = 0

d = 0.5at^2

d = 0.5(2.68 × 10^2)

d = 134m

Converting meters to feet

d = 134m × 3.28ft/m

d = 439.5ft

Final answer:

The rotational inertia of an object depends on its shape and mass distribution. More information about the construction and its dimensions is needed to calculate the rotational inertia about the desired axis.

Explanation:

The rotational inertia of an object, also known as its moment of inertia, depends on its shape and mass distribution. To determine the rotational inertia of the construction about an axis perpendicular to the picture and passing through point B, we need more information about the construction and its dimensions. The moment of inertia can be calculated using various formulas depending on the shape of the object.

Answer:

M=4.736×10³⁰kg

Explanation:

Given Data

Diameter of planet= 1.86×10⁷m

Radius R=Diameter/2=1.86×10⁷/2⇒0.93×10⁷m

Period of rotation of planet t=22.3 hr⇒80280 seconds

Period of revolution around star T=422 earth days⇒3.646×10⁷ seconds

Radius of orbit r=2.2×10¹¹m

Acceleration of gravity on surface of planet g=12.2 m/s²

To Find

Mass of star

Solution

To find mass of star.For that we will consider the revolution of planet around the star with given period

[tex]T^{2}=\frac{4\pi ^{2}r^{3}  }{GM}\\  M=\frac{4\pi ^{2}r^{3}  }{GT^{2}}\\M=\frac{4\pi ^{2}(2.2*10^{11} )^{3}  }{(6.67*10^{-11} )(3.646*10^{7} )^{2}}\\M=4.736*10^{30} kg[/tex]

Answer:

Explanation:

Answer is in the attachment below:

To find the electric field strength that would store 12.5 J of energy in every 6.00 mm3 of space, one must use the formula for energy density related to electric field strength, considering the permittivity of free space.

To solve for the electric field strength that would store 12.5 J of energy in every 6.00 mm3 of space, we need to use the relationship between the electric field (E), the energy density (u), and the permittivity of free space (ε0). The energy density in an electric field is given by u = ½ε0E2. Rearranging for E, and substituting in the given values allows us to solve for the electric field strength.

Assuming ε0 = 8.85 x 10-12 C2/N·m2, and converting the volume from mm3 to m3 (6.00 mm3 = 6.00 x 10-9 m3), we can substitute these values into the rearranged formula E = √(2u/ε0) to find the electric field strength needed to store 12.5 J of energy.

Final answer:

Scientific notation is essential in handling large numbers in physics. The gravitational constant is crucial for calculating gravitational forces. Understanding Earth's mass involves utilizing known values such as its radius and the gravitational constant.

Explanation:

Scientific Notation: Scientific notation is commonly used in astronomy and other sciences to represent very large or small numbers efficiently.

Gravitational Constant: The gravitational constant, denoted by G, is a fundamental constant used to calculate gravitational forces.

Earth's Mass Calculation: Earth's mass, represented as 5.97 x 10²⁴ kg, can be calculated using known values like its radius and the gravitational constant.

Answer:

a) 12.01 W

b) 63.506 N

c) 11.82 s

Explanation:

Given data:

mass = 75 kg

slope of inclination = 3 degree

speed of skier is 2.00 m/s

air Resistance = 25 N

a) From the given information we have following relation

[tex]F_n = mgsin\theta + F_a[/tex]

[tex]F_n = 75\times 9.81 sin 3 + 25 = 63.506 N[/tex]

Power output

[tex]P =F_n \times v[/tex]

[tex]P = 63.506 \times 2 = 127.01 W[/tex]

b) average force can be calculate as

[tex]F_n = mgsin\theta + F_a[/tex]

[tex]F_n = 75 \times 9.81 sin 3 + 25[/tex]

F_n = 63.506 s

c)

force of acceleration is f = ma

solving for acceleration

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{63.506}{75}[/tex]

a = 0.846 m/s^2

we know acceleration is given as [tex]a =\frac{v -u}{t}[/tex]

solving for time t

[tex]t = \frac{v-u}{a}[/tex]

[tex]t =\frac{10-0}{0.846}[/tex]

t = 11.82 s

Answer:Kobe

Explanation:

The small difference at the time of reporting the length should be the random uncertainty.

The following information related to the random certainty is:

There should be a difference in the length as compared to the real one for various members.

Therefore we can conclude that The small difference at the time of reporting the length should be the random uncertainty.

Learn more about the string here: brainly.com/question/4087119

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

[tex]v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s[/tex]

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

[tex]\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m[/tex]

The locations are +5.9 cm or -5.9 cm

Final answer:

The maximum speed of the object is approximately 1.06 m/s. When the object's velocity is one-third of the maximum speed, the object is located approximately 0.060 m from the equilibrium position.

Explanation:

To calculate the maximum speed of the object, the equation for the period of oscillation can be used. The period, T, can be calculated using the formula T = 2π√(m/k), where m is the mass of the object and k is the force constant of the spring. Substituting the given values, we have T = 2π√(0.225/74.5) = 0.942 s. The maximum speed, vmax, can be determined using the formula vmax = 2πA/T, where A is the amplitude of the oscillation. In this case, the amplitude is given as 6.25 cm, which goes to 0.0625 m. Substituting the values, we have vmax = 2π(0.0625)/0.942 = 1.06 m/s.

To find the locations of the object when its velocity is one-third of the maximum speed, we can use the equation for the displacement of an object undergoing simple harmonic motion. The equation is given as x = Acos(ωt + φ), where x is the displacement, A is the amplitude, t is the time, ω is the angular frequency, and φ is the phase constant. Since the object is released from rest, the phase constant is 0. The angular frequency can be calculated using the formula ω = 2π/T, where T is the period. Substituting the given values, we have ω = 2π/0.942 = 6.69 rad/s. The time when the velocity is one-third of the maximum speed can be found by rearranging the formula for velocity, v = ωAsin(ωt + φ), to T/6.69 = Asin(ωt + φ). Solving for t, we find that t ≈ 0.303 s. Substituting this value into the equation for displacement, we have x = Acos(ωt + φ) = 0.0625cos(6.69(0.303)) ≈ 0.060 m.

Answer:The horizontal area enclosed by the waterline of the ship is 12,797.88m2

Explanation: Mass of the ship=29000kg

Total mass of 50 ships= 50×29000

Total Mass=1450000kg

Volume of water with mass= mass/ density

Assuming density of ocean water =1030kgm3

Volume=1450000/1030

Volume=1407.77m3

Area enclosed by the ship =A=V/h

h=11cm=0.11m

Area= 1407.77/0.11

Area=12,797.88m2