An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the oscillator repeats its motion every 0.412 s. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.

Answer:

c. The astronaut does not need to worry: the charge will remain on the outside surface.

Explanation:

The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.

Answer:

497.6 N

Explanation:

From the question,

The net force on the skydiver = weight of the skydiver- the resistive force of air

F' = W-F...................... Equation 1

Where W = weight of the skydiver, F = resistive force of air.

But,

W = mg................ Equation 2

Where m = mass of the skydiver, g = acceleration due to gravity.

Substitute equation 2 into equation 1

F' = mg-F............ Equation 3

Given: m = 87 kg, F = 355 N, g = 9.8 m/s²

Substitute these values into equation 3

F' = 87(9.8)-355

F' = 852.6-355

F' = 487.6 N

Answer:

498 N down

Explanation:

Answer:

-17.8 V

Explanation:

The induced emf in a coil is given as:

[tex]E = \frac{-NdB\pi r^2}{dt}[/tex]

where N = number of loops

dB = change in magnetic field

r = radius of coil

dt = elapsed time

From the question:

N = 50

dB = final magnetic field - initial magnetic field

dB = 0.35 - 0.10 = 0.25 T

r = 3 cm

dt = 2 ms = 0.002 secs

Therefore, the induced emf is:

[tex]E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V[/tex]

Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.

Answer:

Explanation:

each grid corresponding  0.1s⁻¹.

0.2grid unit = 0.2×0.1 =0.02s⁻¹

distance of the star from telescope

d = 1/p

d= 1/0.02= 50 par sec

1par sec = 3.26 light year

1 light year = 9.5×10¹²km

3.26ly=3.084×10¹³km

d= 50×3.084×10¹³=1.55×10¹⁵km

Answer:

a.   vmax = 0.445 m/s

b.   amx = 1.65 m/s^2

c.    v = 0.32 m/s

d.    1.12 m/s^2

e.     t = 0.21 s

Explanation:

a. The maximum speed of the object is given by the following formula:

[tex]v_{max}=\omega A[/tex]    (1)

w: angular frequency of the object

A: amplitude of the motion = 12.2cm = 0.122m

The angular frequency is calculated by using:

[tex]\omega=\sqrt{\frac{k}{m}}[/tex]     (2)

k: spring constant = 8.00 N/m

m: mass of the object = 0.600kg

you replace the equation (2) into the equation (1) and replace the values of the other parameters:

[tex]v_{max}=\sqrt{\frac{k}{m}}A=(\sqrt{\frac{8.00N/m}{0.600kg}})(0.122m)=0.445\frac{m}{s}[/tex]

The maximum speed is 0.445m/s

b. The maximum acceleration is:

[tex]a_{max}=\omega^2 A\\\omega=\sqrt{\frac{8.00N/m}{0.600kg}}=3.65\frac{rad}{s}\\\\a_{max}=(3.65rad/s)^2(0.122m)=1.625\frac{m}{s^2}[/tex]

The maximum acceleration is 1.65m/s^2

c. To calculate the value of the speed for x = 8.60cm you first find the time t by using the following equation of motion for a simple harmonic motion:

[tex]x=Asin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{x}{A})\\\\t=\frac{1}{3.65rad/s}sin^{-1}(\frac{8.60cm}{12.2cm})=0.21s[/tex]

You use this value of t in the following equation for v:

[tex]v=\omega Acos(\omega t)\\\\v=(3.65rad/s)(0.122m)cos((3.65rad/s)(0.21s))=0.32\frac{m}{s}[/tex]

The speed of the object when it is at x = 8.60cm is 0.32m/s

d. The acceleration is:

[tex]a=-\omega^2 A sin(\omega t)\\\\a=-(3.65rad/s)^2(0.122m)sin((3.65rad/s)(0.21))=1.12\frac{m}{s^2}[/tex]

The acceleration is 1.12 m/s^2

e. The time is 0.21s

Answer:

A) 12752.55 J

B) 12507.75 J

C) 12262.95 J

Explanation:

We are given;

Mass;m = 85 kg

Vertical distance; d = 15 m

From change in kinetic energy, the work done by the applied force to pull the spelunker is given by;

Change in kinetic energy = Wa + Wg

Where Wg = -mgd

A) In the first stage;the the work done is given by;

Wa = mgd + ½m(v_f)² - ½m(v_i)²

Since initially stationary, v_i = 0

So, we have;

Wa = mgd + ½m(v_f)²

v_f = 2.4 m/s

So,

Wa = (85 × 9.81 × 15) + ((1/2) × 85 × 2.4²)

Wa = 12752.55 J

B) for the second stage, there is a constant speed of 2.4 m/s

So, v_f = v_i

Thus; Wa = mgd

Wa = (85 × 9.81 × 15)

Wa = 12507.75 J

C) he finally decelerated to zero.

So v_f = 0 while v_i is 2.4 m/s

Thus;

Wa = mgd - ½m(v_i)²

Wa = (85 × 9.81 × 15) - (½ × 85 × 2.4²)

Wa = 12507.75 - 244.8

Wa = 12262.95 J

Answer:

This increases the intensity of the light wave.

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.

To find the force, we need to know about the mathematical formulation of force.

Mathematically, it is written as

F= m×a= m×(∆V/∆t)

Here, initially the velocity of the person is 30m/s. But after applying the force, he came to rest. So his final velocity is 0 m/s. ∆V= 30m/s

When ∆t=5 seconds, F= 60×(30/5)=360N

When ∆t=0.5 seconds, F= 60×(30/0.5)=3600N

When ∆t=0.05 seconds, F= 60×(30/0.05)=36000N

Thus, we can conclude that 360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.

Learn more about force here:

brainly.com/question/12785175

#SPJ2

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

[tex]\Delta P \propto \rho \cdot \Delta h[/tex]

[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]

Where:

[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.

[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.

[tex]\Delta h[/tex] - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]

[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]

Where:

[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.

[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.

If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:

[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]

[tex]\Delta h_{air} = 158.140\,m[/tex]

The height of the building is 158.140 meters.

158.13m

Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e

P = ρhg

Let;

Pressure measured at the roof top =  ([tex]P_{R}[/tex])

Pressure measured at the ground level =  ([tex]P_{G}[/tex])

Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P               ---------------(i)

(a) P = ρhg             -----------(***)

But;

ρ = density of air = 1.29kg/m³  

h = height of column of air = height of building

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (***)

P = 1.29 x h x 10

P = 12.9h Pa

(b) [tex]P_{G}[/tex] =  ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (*)

[tex]P_{G}[/tex] =  13600 x 0.76 x 10

[tex]P_{G}[/tex] = 103360 Pa

(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g       --------------(**)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (**)

[tex]P_{R}[/tex]  =  13600 x 0.745 x 10

[tex]P_{R}[/tex]  = 101320 Pa

(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P  

103360 = 101320 + 12.9h

Solve for h;

12.9h = 103360 - 101320

12.9h = 2040

h = [tex]\frac{2040}{12.9}[/tex]

h = 158.13m

Therefore, the height of the building is 158.13m

Answer:

E = (1 / 4π ε₀ )  q r / R³

Explanation:

Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law

       F = K q₁ q₂ / r²

we used law Gauss

Ф = ∫ E .dA = q_{int} /ε₀

E 4π r² = q_{int} /ε₀  

E = q_{int} / 4π ε₀ r²

we replace the charge inside  

E = (1 / 4π ε₀ r²) ρ 4/3 π r³  

E = ρ r / 3 ε₀

the density for the entire atom is  

ρ = Q / V  

V = 4/3 π R³  

we substitute  

E = (r / 3ε₀ ) Q 3/4π R³  

E = (1 / 4π ε₀ ) q r / R³

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]

The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]

[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]

Therefore,  the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

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

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

The acceleration of the train is 1.581 m/s² inward.

Explanation:

Given;

initial velocity of the train, u = 86.0 km/h = 23.889 m/s

final velocity of the train, v = 56.0 km/h = 15.556 m/s

change in time, Δt = 18 s

The total acceleration of particles moving along a curved path is given as vector sum of the tangential acceleration and radial acceleration

[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]

where;

[tex]a_t[/tex] is the tangential acceleration

[tex]a_r[/tex] is radial acceleration

[tex]a_t = \frac{v-u}{t} \\a_t = \frac{15.556-23.889}{18} \\\\a_t = -0.463 \ m/s^2 \\\\a_t = 0.463 \ m/s^2 \ \ (inward)[/tex]

[tex]a_r = \frac{v^2}{r} \\\\a_r = \frac{15.556^2}{160} \\\\a_r = 1.512 \ m/s^2[/tex]

[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{(-0.463)^2+(1.512)^2} \\\\a = \sqrt{2.5005} \\\\a = 1.581 \ m/s^2[/tex]

Therefore, the acceleration at the moment the train speed reaches 56.0 km/h is 1.581 m/s² inward.

Answer:

Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.

Explanation:

In this case now, the bulb will serve as the load on the battery (resistance).

For the current to last longer, the least amount of energy must be drawn.

The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.

Let us take the resistance of each bulb as 'R'

If we arrange the bulbs in series, then, the total resistance will be

Rt = R + R = 2R

at a EMF of V from the battery, current I through the battery will be

I = V/2R

If we arrange the bulbs in parallel, then , the total resistance will be

1/Rt = 1/R + 1/R

1/Rt = 2/R

therefore

Rt = R/2

at an EMF of V from the battery, the current I that will be drawn through the battery will be

I = 2V/R

we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series

From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery

Answer:

The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].

Explanation:

Given the absence of external forces, the final angular speed of the merry-go-round can be determined with the resource of the Principle of Angular Momentum Conservation, which is described in this case as:

[tex]I_{g, m} \cdot \omega_{o,m} + I_{g, p}\cdot \omega_{o,p} = (I_{g,m} + I_{g, p})\cdot \omega_{f}[/tex]

Where:

[tex]I_{g,m}[/tex] - Moment of inertia of the merry-go-round with respect to its axis of rotation, measured in [tex]kg\cdot m^{2}[/tex].

[tex]I_{g,p}[/tex] - Moment of inertia of the person with respect to the axis of rotation of the merry-go-round, measured in [tex]kg\cdot m^{2}[/tex].

[tex]\omega_{o, m}[/tex] - Initial angular speed of the merry-go-round with respect to its axis of rotation, measured in radians per second.

[tex]\omega_{o,p}[/tex] - Initial angular speed of the merry-go-round with respect to the axis of rotation of the merry-go-round, measured in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed of the merry-go-round-person system, measured in radians per second.

The final angular speed is cleared:

[tex]\omega_{f} = \frac{I_{g,m}\cdot \omega_{o,m}+I_{g,p}\cdot \omega_{o,p}}{I_{g,m}+I_{g,p}}[/tex]

Merry-go-round is modelled as uniform disk-like rigid body, whereas the person can be modelled as a particle. The expressions for their moments of inertia are, respectively:

Merry-go-round

[tex]I_{g,m} = \frac{1}{2}\cdot M \cdot R^{2}[/tex]

Where:

[tex]M[/tex] - The mass of the merry-go-round, measured in kilograms.

[tex]R[/tex] - Radius of the merry-go-round, measured in meters.

Person

[tex]I_{g,p} = m\cdot r^{2}[/tex]

Where:

[tex]m[/tex] - The mass of the person, measured in kilograms.

[tex]r[/tex] - Distance of the person with respect to the axis of rotation of the merry-go-round, measured in meters.

If [tex]M = 185\,kg[/tex], [tex]m = 63.4\,kg[/tex], [tex]R = r = 2.83\,m[/tex], the moments of inertia are, respectively:

[tex]I_{g,m} = \frac{1}{2}\cdot (185\,kg)\cdot (2.83\,m)^{2}[/tex]

[tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex]

[tex]I_{g,p} = (63.4\,kg)\cdot (2.83\,m)^{2}[/tex]

[tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex]

The angular speed experimented by the person with respect to the axis of rotation of the merry-go-round is:

[tex]\omega_{o,p} = \frac{v_{p}}{r}[/tex]

[tex]\omega_{o,p} = \frac{3.51\,\frac{m}{s} }{2.83\,m}[/tex]

[tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex]

Given that [tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex], [tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex], [tex]\omega_{o,m} = 4.405\,\frac{rad}{s}[/tex] and [tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex], the final angular speed of the merry-go-round is:

[tex]\omega_{f} = \frac{(740.823\,kg\cdot m^{2})\cdot \left(4.405\,\frac{rad}{s} \right)+(507.764\,kg\cdot m^{2})\cdot \left(1.240\,\frac{rad}{s} \right)}{740.823\,kg\cdot m^{2}+507.764\,kg\cdot m^{2}}[/tex]

[tex]\omega_{f} = 3.118\,\frac{rad}{s}[/tex]

[tex]\omega_{f} = 0.496\,\frac{rad}{s}[/tex]

The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].

Answer:

Explanation:

Image distance and focal length of a convex mirror is negative.

a) Using the mirror formula to determine the image distance;

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

f = focal length of the mirror = -5.40m

u = object distance from the mirror = 3.00m

v = image distance from the mirror

[tex]\frac{1}{-5.40} = \frac{1}{3} + \frac{1}{v}[/tex]

[tex]\frac{1}{v} = \frac{-1}{5.4} - \frac{1}{3}[/tex]

[tex]\frac{1}{v} = \frac{-1}{5.4} - \frac{1}{3}\\\frac{1}{v} = -0.1852 -0.3333\\\frac{1}{v} = - 0.5185\\v = \frac{-1}{0.5185} \\v = -1.93m[/tex]

b) Magnification = image distance/object distance

Magnification = 1.93/3

Magnification = +0.643

Note that we do not use the negative value of the image distance to calculate the magnification of the object

a. The  Image distance of the truck should be -1.93m.

b, The magnification of the object distance should be +0.643.

here the mirror formula should be used

a.

1/f = 1/u + 1/v

where,

f = focal length of the mirror = -5.40m

u = object distance from the mirror = 3.00m

v = image distance from the mirror

So,

1/5.40 = 1/3 + 1/v

1/v = -1/5.4 - 1/3

So, v = -1.93m

b. Now the Magnification = image distance/object distance

= 1.93/3

= +0.643

learn more about distance here: https://brainly.com/question/21586135

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

[tex]V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}[/tex]

Where;

d is the separation or distance between the two parallel plates;

[tex]d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm[/tex]

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

Answer:

(a) σ = 3.41*10⁻7C/m^2

(b) E = 38,530.1 N/C

Explanation:

(a) In order to calculate the resulting surface charge density, you use the following formula:

[tex]\sigma=\frac{Q}{S}[/tex]     (1)

σ: surface charge density

Q: charge of the satellite = 3.1 µC = 3.1*10^-6C

S: surface area of the satellite

The satellite has a spherical form, then, the area of the surface is given by:

[tex]S=4\pi r^2[/tex]     (2)

r: radius of the satellite = d/2 = 1.7m/2 = 0.85m

You replace the equation (2) into the equation (1) and solve for the surface charge density:

[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]

The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2

(b) The electric field just outside the surface is calculate d by using the following formula:

[tex]E=k\frac{Q}{R^2}[/tex]      (3)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the satellite = 0.85m

[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]

The magnitude of the electric field just outside the sphere is 38,530.1 N/C

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

[tex]L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}[/tex]

Where

[tex]I_{i}\ and \ \omega_{i}[/tex] initial moment of inertia and angular velocity

[tex]I_{f}\ and \ \omega_{f}[/tex] is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

[tex]I_{i} = \frac{2}{5}m_{s}r^{2}[/tex]

Where [tex]m_{s}[/tex] is the satellite mass

r is the  radus of the sphere

Substititute 1900kg for m and 4.6m for r

[tex]I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}[/tex]

The final moment of inertia of the satellite about the centre of mass

[tex]I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}[/tex]

Where [tex]m_{x}[/tex] is the antenna's mass and

I is the length of the antenna

[tex]I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}[/tex]

So, the Final rotation rate of the satellite is:

[tex]I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s[/tex]

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Answer:

1.35 kJ  

Explanation:

KE = ½mv² = ½ × 0.030  kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

Given:-

To Find: Kinetic energy of the bullet.

We know,

Eₖ = ½mv²

where,

thus,

Eₖ = ½(30 g)(300 m/s)²

= (15 g)(90000 m²/s²)

= 1350000 g m²/s²

= 1350 kg m²/s²

= 1350 J

= 1.35 kJ (Ans.)

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

Answer:

force on larger piston = [tex]\frac{fA}{a}[/tex]

Explanation:

we label the pistons as piston A and piston B

small piston A:

area = a

force = f

large piston B:

area = A

force  = ?

Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.

we know that pressure = force ÷ area

pressure of piston A = [tex]\frac{f}{a}[/tex]

pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]

obeying Pascal's law, the system pressures must be equal. Therefore

[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]

force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]

Answer:

N' = 1.17*10^21 molecules/m^3

Explanation:

In order to calculate the surface density in Titan's atmosphere, in molecules per cubic meter, you use the following formula for ideal gases:

[tex]PV=nRT[/tex]     (1)

P: pressure = 1.5 Earth-atmospheres = 1.5 atm

R: ideal gas constant = 8.205 m^3.atm.mol^-1.K^-1

T: temperature in Titan = 94K

n: number of moles

To obtain the number of moles per cubic meter you write the equation (1) as follow:

[tex]N=\frac{n}{V}=\frac{P}{RT}[/tex]      (2)

N: moles per cubic meter

You use the Avogadro's number for the number of molecules:

[tex]N'=NA=\frac{AP}{RT}[/tex]    (3)

A: Avogadro's number = 6.022*109^23 molecules/mol

N': molecules per cubic meter

You replace the values of all parameters in the equation (3):

[tex]N'=\frac{(6.022*10^{23}molecules/mol)(1.5atm)}{(8.205m^3.atm.K^{-1}mol^{-1})(94K)}\\\\N'=1.17*10^{21}\frac{molecules}{m^3}[/tex]

The surface density of molecules in Titan is 1.17*10^21 molecules/m^3

Answer:

1.2×1026molecules/m3

Explanation:

given

Answer:

Differences between regular and irregular objects are:

Hope this helps...

Good luck on your assignment..

Answer:

The total internal energy change for the entire process is  -0.94 kJ

Explanation:

Process A to B is an isothermal process, therefore, [tex]u_A[/tex] - [tex]u_B[/tex] = 0

Process B to C

P = -mV + C

When P = 12, V = 0.12

When P = 4, V = 0.135

Therefore, we have;

12 = -m·0.12 + C

4 = -m·0.135 + C

Solving gives

m = 533.33

C = 76

[tex]T = \dfrac{1}{nR} \times (-533.33 \times V^2 + 76 \times V)[/tex]

p₂ = p₁V₁/V₂ = 12*0.1/0.12 = 10 kPa

The work done = 0.5*(0.135 - 0.12)*(4 - 10.0) = -0.045 kJ = -45 J

For heat supplied

Assuming an approximate polytropic process, we have;

Work done = (p₃×v₃ - p₂×v₂)/(n - 1)

Which gives;

-45 = (4*0.135 - 10*0.12)/(n -1)

∴ n -1 = (4*0.135 - 10*0.12)/-45 =   14.67

n = 15.67

Q = W×(n - γ)/(γ - 1)

Q = -45*(15.67 - 1.4)/(1.4 - 1) = -1,605.375 J

u₃ - u₂ = Q + W = -1,605.375 J - 45 J = -1650 J = -1.65 kJ

For the constant pressure process D to C, we have;

[tex]Q = c_p \times \dfrac{p}{R} \times (V_4 -V_3) = \dfrac{5}{2} \times p \times (V_4 -V_3)[/tex]

Q₄₋₃ = (0.1 - 0.135) * 4*5/2 = -0.35 kJ

W₄₋₃ = 4*(0.1 - 0.135) = -0.14 kJ

u₄ - u₃ = Q₄₋₃ + W₄₋₃ = -0.14 kJ + -0.35 kJ = -0.49 kJ

For the process D to A, we have a constant volume process

[tex]Q_{1-4} = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]

W₁₋₄ = 0 for constant volume process, therefore, u₁ - u₄ = 1.2 kJ

The total internal energy change Δ[tex]u_{process}[/tex] for the entire process is therefore;

Δ[tex]u_{process}[/tex] = u₂ - u₁ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄ = 0  - 1.65 - 0.49 + 1.2 = -0.94 kJ.

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer: Acceleration of the car at time = 10 sec is 108 [tex]m/s^{2}[/tex] and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

[tex]M\frac{dv}{dt} = u\frac{dM}{dt}[/tex]

[tex]\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt[/tex]

       = [tex]u\int_{M_{o}}^{M_{f}} \frac{dM}{M}[/tex]

[tex]v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}[/tex]

[tex]v_{o} = 0[/tex]

As, [tex]v_{f} = u ln (\frac{M_{f}}{M_{o}})[/tex]

u = -2900 m/s

[tex]M_{f} = M_{o} - m \times t_{f}[/tex]

           = [tex]2500 kg + 1000 kg - 95 kg \times t_{f}s[/tex]

           = [tex](3500 - 95t_{f})s[/tex]

[tex]v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s[/tex]

Also, we know that

     a = [tex]\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}[/tex]

        = [tex]\frac{u}{3500 - 95 t} \times (-95) m/s^{2}[/tex]

        = [tex]\frac{95 \times 2900}{3500 - 95t} m/s^{2}[/tex]

At t = 10 sec,

[tex]v_{f}[/tex] = 918.34 m/s

and,   a = 108 [tex]m/s^{2}[/tex]