A vacuum-filled parallel plate capacitor has an energy density of 0.1 J/m^3 and the plates are separated by 0.2 mm.
An electron is released from rest near negativity charge plate of the capacitor.

How fast is the electron moving when it reaches the positively charged plate?

Answer:

C

Explanation:

The electric field inside a conductor is always zero if the charges inside the conductor are not moving.

Since the electron are not moving then they must be in electrostatic equilibrium which means the electric field inside the conductor is zero. if the electric field existed inside the conductor then there will be net force on all the electrons and the electrons will accelerate.

The electric field inside a conductor is always zero if charges inside the conductor are not moving. Under electrostatic conditions, free electrons in the conductor rearrange to cancel any internal electric fields. Thus, the correct answer is Option C.

1. In a conductor in electrostatic equilibrium, the electric field inside the conductor is zero. This is because any free electrons within the conductor will move in response to any electric field until they reach a state where there is no net force acting on them. This movement of electrons cancels out any existing electric field.

2. Properties of conductors in electrostatic equilibrium include that any excess charge resides on the surface of the conductor, and the electric field just outside the surface is perpendicular to the surface.

3. Therefore, in the absence of moving charges (static conditions), the electric field inside a conductor must be zero.

Answer:

Temperature of air at exit = 24.32 C, After reducing hot air the temperature of the exit air becomes = 20.11 C

Explanation:

ρ = P/R(Ti) where ρ is the density of air at the entry, P is pressure of air at entrance, R is the gas constant, Ti is the temperature at entry

ρ = (2.07 x 10⁵)/(287)(473) = 1.525 kg/m³

Calculate the mass flow rate given by

m (flow rate) = (ρ x u(i) x A(i)) where u(i) is the speed of air, A(i) is the area of the tube (πr²) of the tube

m (flow rate) = 1.525 x (π x 0.0125²) x 6 = 4.491 x 10⁻³ kg/s

The Reynold's Number for the air inside the tube is given by

R(i) = (ρ x u(i) x d)/μ where d is the inner diameter of the tube and μ is the dynamic viscosity of air (found from the table at Temp = 473 K)

R(i) = (1.525) x (6) x 0.025/2.58 x 10⁻⁵ = 8866

Calculate the convection heat transfer Coefficient as

h(i) = (k/d)(R(i)^0.8)(Pr^0.3) where k is the thermal conductivity constant known from table and Pr is the Prandtl's Number which can also be found from the table at Temperature = 473 K

h(i) = (0.0383/0.025) x (8866^0.8) x (0.681^0.3) = 1965.1 W/m². C

The fluid temperature is given by T(f) = (T(i) + T(o))/2 where T(i) is the temperature of entry and T(o) is the temperature of air at exit

T(f) = (200 + 20)/2 = 110 C = 383 K

Now calculate the Reynold's Number and the Convection heat transfer Coefficient for the outside

R(o) = (μ∞ x do)/V(f)  where μ∞ is the speed of the air outside, do is the outer diameter of the tube and V(f) is the kinematic viscosity which can be known from the table at temperature = 383 K

R(o) = (12 x 0.0266)/(25.15 x 10⁻⁶) = 12692

h(o) = K(f)/d(o)(0.193 x Ro^0.618)(∛Pr) where K(f) is the Thermal conductivity of air on the outside known from the table along with the Prandtl's Number (Pr) from the table at temperature = 383 K

h(o) = (0.0324/0.0266) x (0.193 x 12692^0.618) x (0.69^1/3) = 71.36 W/m². C

Calculate the overall heat transfer coefficient given by

U = 1/{(1/h(i)) + A(i)/(A(o) x h(o))} simplifying the equation we get

U = 1/{(1/h(i) + (πd(i)L)/(πd(o)L) x h(o)} = 1/{(1/h(i) + di/(d(o) x h(o))}

U = 1/{(1/1965.1) + 0.025/(0.0266 x 71.36)} = 73.1 W/m². C

Find out the minimum capacity rate by

C(min) = m (flow rate) x C(a) where C(a) is the specific heat of air known from the table at temperature = 473 K

C(min) = (4.491 x 10⁻³) x (1030) = 4.626 W/ C

hence the Number of Units Transferred may be calculated by

NTU = U x A(i)/C(min) = (73.1 x π x 0.025 x 3)/4.626 = 3.723

Calculate the effectiveness of heat ex-changer using

∈ = 1 - е^(-NTU) = 1 - e^(-3.723) = 0.976

Use the following equation to find the exit temperature of the air

(Ti - Te) = ∈(Ti - To) where Te is the exit temperature

(200 - Te) = (0.976) x (200 - 20)

Te = 24.32 C

The effect of reducing the hot air flow by half, we need to calculate a new value of Number of Units transferred followed by the new Effectiveness of heat ex-changer and finally the exit temperature under these new conditions.

Since the new NTU is half of the previous NTU we can say that

NTU (new) = 2 x NTU = 2 x 3.723 = 7.446

∈(new) = 1 - e^(-7.446) = 0.999

(200 - Te (new)) = (0.999) x (200 - 20)

Te (new) = 20.11 C

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

[tex]\Delta x = 6.17 cm = 0.0617 m[/tex]

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

[tex]2mg = k'\Delta x[/tex] (1)

where

m = 76.2 kg is the mass of each children

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

[tex]k'[/tex] is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

[tex]k'=k+k=2k[/tex]

Substituting into (1) and solving for k, we find:

[tex]2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m[/tex]

b)

The period of the oscillating system is given by

[tex]T=2\pi \sqrt{\frac{m}{k'}}[/tex]

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

[tex]k'=2k=2(12,103)=24,206 N/m[/tex] is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

[tex]m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg[/tex]

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

[tex]f=\frac{1}{T}[/tex]

where

f is the frequency

T is the period

In  this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

[tex]f=\frac{1}{2.09}=0.478 Hz[/tex]

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

[tex]t=10T=10(2.09)=20.9 s[/tex]

#LearnwithBrainly

Answer:

(I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]

Explanation:

Given that,

Radius of aorta = 10 mm

Speed = 300 mm/s

Radius of capillary [tex]r=4\times10^{-3}\ mm[/tex]

Speed of blood [tex]v=5\times10^{-4}\ m/s[/tex]

(I). We need to calculate the effective cross sectional area of the capillaries

Using continuity equation

[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]

Where. v₁ = speed of blood in capillarity

A₂ = area of cross section of aorta

v₂ =speed of blood in aorta

Put the value into the formula

[tex]A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}[/tex]

[tex]A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}[/tex]

[tex]A_{1}=0.188\ m^2[/tex]

(II). We need to calculate the approximate number of capillaries

Using formula of area of cross section

[tex]A_{1}=N\pi r_{c}^2[/tex]

[tex]N=\dfrac{A_{1}}{\pi\times r_{c}^2}[/tex]

Put the value into the formula

[tex]N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}[/tex]

[tex]N=3.74\times10^{9}[/tex]

Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

[tex]\rho(r) = ar^2 - br^3[/tex]

r is the radius of the spherical shell

dr is the thickness

volume of shell

[tex]dV = 4 \pi r^2 dr[/tex]

mass of shell

[tex]dM = \rho(r)dV[/tex]

[tex]\rho = \rho_0 - br[/tex]

now,

[tex]dM = (\rho_0 - br)(4 \pi r^2)dr[/tex]

integrating both side

[tex]M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr[/tex]

[tex]M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})[/tex]

[tex]M = \pi R^3(\dfrac{\rho_0}{3}+\rho)[/tex]

we know,

[tex]a = \dfrac{GM}{R^2}[/tex]

[tex]a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}[/tex]

[tex]a =\pi RG(\dfrac{\rho_0}{3}+\rho)[/tex]

[tex]a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)[/tex]

a = 9.94 m/s²

Answer:

L = 0.0319 H

Explanation:

Given that,

Number of loops in the solenoid, N = 900

Radius of the wire, r = 3 cm = 0.03 m

Length of the rod, l = 9 cm = 0.09 m

To find,

Self inductance in the solenoid

Solution,

The expression for the self inductance of the solenoid is given by :

[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]

[tex]L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}[/tex]

L = 0.0319 H

So, the self inductance of the solenoid is 0.0319 henries.

Answer:

$0.662772

Explanation:

v = Volume of can = 16 fl oz.

[tex]1\ floz.=1.805\ in^3[/tex]

r = Radius of can

h = Height of can = 3r

Volume of cylinder is given by

[tex]\pi r^2h=16\times 1.805\\\Rightarrow \pi r^23r=16\times 1.805\\\Rightarrow 3\pi r^3=16\times 1.805\\\Rightarrow r=\left(\frac{16\times 1.805}{3\times 3.14}\right)^{\frac{1}{3}}\\\Rightarrow r=1.45247\ in[/tex]

h=3r\\\Rightarrow h=3\times 1.45247\\\Rightarrow h=4.35741\ in[/tex]

Surface area of sides is given by

[tex]2\pi rh\\ =2\times 3.14\times 1.45247\times 4.35741\\ =39.76632\ in^2[/tex]

Surface area of top and bottom is given by

[tex]2\pi r^2\\ =2\times 3.14\times 1.45247^2\\ =13.25544\ in^2[/tex]

Cost of making the can will be

[tex]39.76632\times 0.01+13.25544\times 0.02=\$0.662772[/tex]

The cost to make the can is $0.662772

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   [tex]V = \epsilon - Ir[/tex]

now,

   [tex]IR = \epsilon - Ir[/tex]

   [tex]I(R+r) = \epsilon[/tex]

   [tex]I= \dfrac{\epsilon}{R+r}[/tex]

inserting the values

   [tex]I= \dfrac{12}{0.05+0.01}[/tex]

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 [tex]I= \dfrac{\epsilon}{R+r}[/tex]

   [tex]I= \dfrac{12}{0.14+0.01}[/tex]

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W

Answer:

Explanation:

Resistance of motor, R = 0.05 ohm

internal resistance of battery, r = 0.01 ohm

Voltage of battery, V = 12 V

(a) Total resistance, R' = R + r = 0.05 + 0.01 = 0.06 ohm

Let the current be i.

use Ohm's law

i = V / R'

i = 12 / 0.06 = 200 A

(b) Voltage across motor, V' = i x R = 200 x 0.05 = 10 V

(c) Power, P = i²R = 200 x 200 x 0.05 = 2000 Watt.

(d) Total resistance, R' = 0.05 + 0.1 + 0.09 = 0.15 ohm

i = V / R' = 12 / 0.15 = 80 A

V' = i x R = 80 x 0.05 = 4 V

P' = i²R = 80 x 80 x 0.05 = 320 Watt

Answer:

The activity is 811.77 Bq

Solution:

As per the question:

Half life of Thorium, [tex]t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs[/tex]

Mass of Thorium, m = 200 mg = 0.2 g

M = 232 g/mol

Now,

No. of nuclei of Thorium in 200 mg of Thorium:

[tex]N = \frac{N_{o}m}{M}[/tex]

where

[tex]N_{o }[/tex] = Avagadro's number

Thus

[tex]N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}[/tex]

Also,

Activity is given by:

[tex]\frac{0.693}{t_{\frac{1}{2}}}\times N[/tex]

[tex]A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr[/tex]

[tex]A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq[/tex]

Answer:

Explanation:

Given

inclination [tex]=\beta [/tex]

Assuming radius of sphere is r

Now from Free Body Diagram

[tex]Mg\sin \theta -f_r=Ma[/tex]

where [tex]f_r=friction\ force[/tex]

[tex]a=acceleration\ of\ system[/tex]

Now friction will Provide the Torque

[tex]f_r\times r=I\cdot \alpha [/tex]

where [tex]I=moment\ of\ inertia[/tex]

[tex]\alpha =angular\ acceleration [/tex]

[tex]f_r\times r=\frac{2}{3}Mr^2\times \frac{a}{r}[/tex]

in pure rolling [tex]a=\alpha r[/tex]

[tex]f_r=\frac{2}{3}Ma[/tex]

[tex]mg\sin \beta -\frac{2}{3}Ma=Ma[/tex]

[tex]Mg\sin \beta =\frac{5}{3}Ma[/tex]

[tex]a=\frac{3g\sin \beta }{5}[/tex]

Answer:

10.28571 N

Explanation:

m = Mass of toboggan = 15 kg

u = Initial velocity = 4.8 m/s

v = Final velocity = 0

t = Time taken = 7 seconds

Friction force is given by the change in momentum over time

[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{15(0-4.8)}{7}\\\Rightarrow F=-10.28571\ N[/tex]

The magnitude of the average friction force acting on the toboggan while it was moving is 10.28571 N

Using the principle of conservation of momentum, the magnitude of the average friction force acting on the toboggan can be found. The initial momentum of the toboggan is equal to the change in momentum, which is equal to the mass of the toboggan multiplied by the change in velocity. Dividing the change in momentum by the time interval gives the magnitude of the average friction force as 10.29 N.

To find the magnitude of the average friction force acting on the toboggan, we can use the principle of conservation of momentum. The initial momentum of the toboggan is given by P = m * v, where m is the mass (15.0 kg) and v is the velocity (4.80 m/s). The final momentum is zero, as the toboggan comes to a stop. Therefore, the change in momentum is equal to the initial momentum.

The change in momentum is given by δP = m * δv, where δv is the change in velocity. Since the velocity changes from 4.80 m/s to 0 m/s, the change in velocity is -4.80 m/s. Therefore, the change in momentum is -15.0 kg * 4.80 m/s = -72.0 kg*m/s.

The average friction force is equal to the change in momentum divided by the time interval. The time interval is given as 7.00 s. Therefore, the magnitude of the average friction force is |-72.0 kg*m/s / 7.00 s| = 10.29 N.

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#SPJ11

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

Answer:

The radius of the second coil will be 7.5 cm

Explanation:

We have given that radius of the first coil [tex]r_1=5.1cm=0.051m[/tex]

Peak emf of the first generator [tex]e_1=1.8Volt[/tex]

Peak emf of the second generator [tex]e_2=3.9Volt[/tex]

We know that induced emf is rate of change of magnetic flux , that is equal to magnetic field multiply by change in area

So emf will be proportional to the change in area

So [tex]\frac{e_1}{e_2}=\frac{A_1}{A_2}[/tex]

[tex]\frac{1.8}{3.9}=\frac{0.051^2}{r_2^2}[/tex]

[tex]r_2=0.075m = 7.5cm[/tex]

So the radius of the second coil will be 7.5 cm

Final answer:

To calculate the change in volume of a diamond exposed to atmospheric pressure, we can use the formula for bulk modulus. The increase in the diamond's radius can be found using the formula for increase in volume.

Explanation:

(a) To calculate the change in volume, we can use the formula for bulk modulus:

ΔV = V * (Pf - Pi) / B

Where ΔV is the change in volume, V is the initial volume, Pf is the final pressure, Pi is the initial pressure, and B is the bulk modulus.

Substituting the given values, we get:

ΔV = (4/3) * π * r^3 * (Pf - Pi) / B

Since the sphere is symmetrical, the change in radius (Δr) is the same in all directions. So, we can calculate it as:

Δr = ΔV / ((4/3) * π * (r^2) * ΔP)

(b) To find the increase in the diamond's radius, we can use the formula for increase in volume:

ΔV = (4/3) * π * (Rf^3 - Ri^3)

Substituting the given values, we get:

Rf = (3 * ΔV + 4 * π * Ri^3) / (4 * π * Ri^2)

The diamond's volume expands by 2.08 x 10⁻⁹ m³ and its radius increases by 10 micrometers upon being exposed to atmospheric pressure. These calculations utilize the mass, density, and bulk modulus of the diamond.

First, we convert the mass of the diamond into kilograms:

2.8 carats * 0.200 g/carat = 0.56 g = 0.00056 kg

Next, calculate the initial volume ([tex]V_i[/tex]) of the diamond using the density formula:

Density = Mass / Volume

So, the initial volume:

[tex]V_i[/tex] = Mass / Density  

[tex]V_i[/tex] = 0.56 g / 3.52 g/cm³

[tex]V_i[/tex] = 0.1591 cm³ = 1.591 × 10⁻⁷ m³

The bulk modulus (B) of diamond is given as 4.43 × 10¹¹ N/m² and the pressure change (ΔP) is:

ΔP = 58 × 10⁹ N/m²

Using the formula for volume change (ΔV), where:

ΔV / V = -ΔP / B

we get:

ΔV = -[tex]V_i[/tex] * ΔP / B

ΔV = -1.591 × 10⁻⁷ m³ * 58 × 109 N/m² / 4.43 × 10¹¹ N/m²

ΔV ≈ -2.08 × 10⁻⁹ m³

Therefore, the volume expansion when exposed to atmospheric pressure is:

ΔV = 2.08 × 10⁻⁹ m³

(b) For the increase in radius (Δr), use the formula for the volume of a sphere:

V = (4/3)πr³

The new volume ([tex]V_f[/tex]) is:

[tex]V_f[/tex] = Vi + ΔV

[tex]V_f[/tex] ≈ 1.591 × 10⁻⁷ m³ + 2.08 × 10⁻⁹ m³

[tex]V_f[/tex] ≈ 1.611 × 10⁻⁷ m³

Setting the initial and final volumes equal to the sphere volume formula, we solve for the radii:

[tex]r_{i[/tex] = (3[tex]V_i[/tex] / 4π)^(1/3)

[tex]r_{i[/tex] ≈ (3 * 1.591 × 10⁻⁷ m³ / 4π)^(1/3)

[tex]r_{i[/tex] ≈ 3.37 × 10-3 m

[tex]r_f[/tex] = (3[tex]V_f[/tex] / 4π)^(1/3)

[tex]r_f[/tex] ≈ (3 * 1.611 × 10⁻⁷ m³ / 4π)^(1/3)

[tex]r_f[/tex] ≈ 3.38 × 10⁻³ m

Thus the increase in radius Δr is:

Δr ≈ [tex]r_f[/tex] - [tex]r_{i[/tex]

Δr ≈ 3.38 × 10⁻³ m - 3.37 × 10⁻³ m

Δr = 0.01 × 10⁻³ m

Δr = 10 × 10⁻⁶ m = 10 μm

Answer:

The approximate combined sound  intensity is [tex]I_{T}=1.1\times10^{-4}W/m^{2}[/tex]

Explanation:

The decibel  scale intensity for busy traffic is 80 dB. so intensity will be

[tex]10log(\frac{I_{1}}{I_{0}} )=80[/tex], therefore [tex]I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}[/tex]

In the same way for the loud conversation having a decibel intensity of 70 dB.

[tex]10log(\frac{I_{2}}{I_{0}} )=70[/tex], therefore [tex]I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}[/tex]

Finally we add both of them [tex]I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}[/tex], is the approximate combined sound  intensity.

Answer:

The loop penetrate 4 cm into the magnetic field.

Explanation:

Given that,

Width w= 5 cm

Length L= 10 cm

mass m = 40 g

Resistance R = 20 mΩ

Initial velocity = 1 m/s

Magnetic field = 2 T

We need to calculate the induced emf

Using formula of emf

[tex]\epsilon=v_{0}Bw[/tex]

Put the value into the formula

[tex]\epsilon =1\times2\times5\times10^{-2}[/tex]

[tex]\epsilon =10\times10^{-2}\ volt[/tex]

We need to calculate the current

Using Lenz's formula

[tex]i=\dfrac{\epsilon}{R}[/tex]

[tex]i=\dfrac{10\times10^{-2}}{20\times10^{-3}}[/tex]

[tex]i=5\ A[/tex]

We need to calculate the force

Using formula of force

[tex]F=i(\vec{w}\times\vec{B})[/tex]

[tex]F=iwB[/tex]

Put the value into the formula

[tex]F=5\times5\times10^{-2}\times2[/tex]

[tex]F=0.5\ N[/tex]

We need to calculate the acceleration

Using formula of acceleration

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

Put the value in to the formula

[tex]a=\dfrac{0.5}{40\times10^{-3}}[/tex]

[tex]a=12.5\ m/s^2[/tex]

We need to calculate the distance

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0-1^2}{2\times(-12.5)}[/tex]

[tex]s=0.04\ m[/tex]

[tex]s=4\ cm[/tex]

Hence, The loop penetrate 4 cm into the magnetic field.

Answer:

3 cm

Explanation:

To answer this question the equation of continuity can be used

For us to use the equaition of continuity, we will make a few assumptions:

[tex]A_{1}v_{1} = A_{2} v_{2} \\\pi (3cm)^{2} * 5m/s = \pi r^{2} *20m/s\\\\r = 1.5 cm\\D = 2r = 2*1.5cm=3 cm[/tex]

Answer:

0.256 hours

Explanation:

Vectors in the plane

We know Office A is walking at 5 mph directly south. Let [tex]X_A[/tex] be its distance. In t hours he has walked

[tex]X_A=5t\ \text{miles}[/tex]

Office B is walking at 6 mph directly west. In t hours his distance is

[tex]X_B=6t\ \text{miles}[/tex]

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

[tex]D=\sqrt{X_A^2+X_B^2}[/tex]

[tex]D=\sqrt{(5t)^2+(6t)^2}[/tex]

[tex]D=\sqrt{61}t[/tex]

This distance is known to be 2 miles, so

[tex]\sqrt{61}t=2[/tex]

[tex]t =\frac{2}{\sqrt{61}}=0.256\ hours[/tex]

t is approximately 15 minutes

Answer:

The amplitude of the resultant wave are

(a). 0.0772 m

(b). 0.0692 m

Explanation:

Given that,

Frequency = 135 Hz

Wavelength = 2 cm

Amplitude = 0.04 m

We need to calculate the angular frequency

[tex]\omega=2\pi f[/tex]

[tex]\omega=2\times\pi\times135[/tex]

[tex]\omega=848.23\ rad/s[/tex]

As the two waves are identical except in their phase,

The amplitude of the resultant wave is given by

[tex]y+y=A\sin(kx-\omega t)+Asin(kx-\omega t+\phi)[/tex]

[tex]y+y=A[2\sin(kx-\omega t+\dfrac{\phi}{2})\cos\phi\dfrac{\phi}{2}[/tex]

[tex]y'=2A\cos(\dfrac{\phi}{2})\sin(kx-\omega t+\dfrac{\phi}{2})[/tex]

(a). We need to calculate the amplitude of the resultant wave

For [tex]\phi =\dfrac{\pi}{6}[/tex]

The amplitude of the resultant wave is

[tex]A'=2A\cos(\dfrac{\phi}{2})[/tex]

Put the value into the formula

[tex]A'=2\times0.04\cos(\dfrac{\pi}{12})[/tex]

[tex]A'=0.0772\ m[/tex]

(b), We need to calculate the amplitude of the resultant wave

For [tex]\phi =\dfrac{\pi}{3}[/tex]

[tex]A'=2\times0.04\cos(\dfrac{\pi}{6})[/tex]

[tex]A'=0.0692\ m[/tex]

Hence, The amplitude of the resultant wave are

(a). 0.0772 m

(b). 0.0692 m

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

The average velocity of the sled, given by the formula v_avg = s / t, evaluates the distance covered over a certain duration of time. The sled has moved a distance 's' during a time interval 't', and so its average velocity over this interval is simply 's' divided by 't'. The presence of friction does not affect the calculation of average velocity.

The average velocity of the sled is simply the distance traveled over a certain time interval. In this case, the sled starts from rest and is being pulled by a horizontal force F against a frictional force. Average Velocity, v_avg, is given by the formula:

v_avg = s / t

In physics, velocity is a measure of the rate of change of position concerning time, so the average velocity is simply the total distance (or displacement) divided by the total time. Given the fact that the sled has moved a distance, s, during the time interval t, the average velocity is simply s divided by t.

With this definition, the sled's average velocity can be determined without needing to tabulate its speed or direction throughout the time interval even in the presence of friction.

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Answer:

Angular displacement will be [tex]\frac{1}{4}\Theta[/tex]

So option (b) will be the correct option

Explanation:

We have given that firstly object is at rest

So [tex]\omega _i=0rad/sec[/tex]

From law of motion we know that angular displacement is given by

[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2[/tex]

Now angular displacement by the object in [tex]\frac{t}{2}sec[/tex]

[tex]\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta[/tex]

So option (b) will be the correct option

The angle the object rotate through in the time [tex]\frac{1}{2} t[/tex] is [tex]\frac{1}{4} (\theta)[/tex]

Given the following data:

To determine the angle the object rotate through in the time [tex]\frac{1}{2} t[/tex]:

Mathematically, angular displacement is given by this formula:

[tex]\theta = \omega_i t +\frac{1}{2} \alpha t^2[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\theta = 0( t )+\frac{1}{2} \alpha t^2\\\\\theta = \frac{1}{2} \alpha t^2[/tex]

when t = [tex]\frac{1}{2} t[/tex]:

[tex]\theta = \frac{1}{2} \alpha (\frac{t}{2} )^2\\\\\theta = \frac{1}{2} \alpha (\frac{t^2}{4} )\\\\\theta =\frac{1}{4} (\frac{1}{2} \alpha t^2)\\\\\theta =\frac{1}{4} (\theta)[/tex]

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Answer:

[tex]M=6.4243\ g[/tex]

Explanation:

Given:

To stop this balloon from rising up we need to counter the buoyant force.

mass of balloon after inflation:

[tex]m=m_h+m_b[/tex]

[tex]m=0.0084\times 0.180+0.0029[/tex]

[tex]m=0.004412\ kg[/tex]

Now the density of inflated balloon:

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

[tex]\rho_b=\frac{0.004412}{0.0084}[/tex]

[tex]\rho_b=0.5252\ kg.m^{-3}[/tex]

Now the buoyant force on balloon

[tex]F_B=V(\rho_a-\rho_b).g[/tex]

[tex]F_B=0.0084(1.29-0.5252)\times 9.8[/tex]

[tex]F_B=0.063\ N[/tex]

∴Mass to be hung:

[tex]M=\frac{F_B}{g}[/tex]

[tex]M=0.00642432\ kg[/tex]

[tex]M=6.4243\ g[/tex]

Answer:

0.25938 W

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]\nu[/tex] = Frequency = 3 GHz

d = Diameter of lossless antenna = 1 m

r = Radius = [tex]\frac{d}{2}=\frac{1}{2}=0.5\ m[/tex]

[tex]A_t[/tex] = Area of transmitter

[tex]A_r[/tex] = Area of receiver

R = Distance between the antennae = 40 km

[tex]P_r[/tex] = Power of receiver = [tex]10\times 10^{-9}\ W[/tex]

[tex]P_t[/tex] = Power of Transmitter

Wavelength

[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{3\times 10^9}\\\Rightarrow \lambda=0.1\ m[/tex]

From Friis transmission formula we have

[tex]\frac{P_t}{P_r}=\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=P_r\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=10\times 10^{-9}\frac{0.1^2\times (40\times 10^3)^2}{\pi 0.5^2\times \pi 0.5^2}\\\Rightarrow P_t=0.25938\ W[/tex]

The power that should be transmitted is 0.25938 W

Answer:

[tex]V_{ft}= 317 cm/s[/tex]

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

[tex]P_i = P_f[/tex]

Where:

[tex]P_i=M_cV_{ic} + M_tV_{it}[/tex]

[tex]P_f = M_cV_{fc} + M_tV_{ft}[/tex]

Now:

[tex]M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}[/tex]

Where [tex]M_c[/tex] is the mass of the car, [tex]V_{ic}[/tex] is the initial velocity of the car, [tex]M_t[/tex] is the mass of train, [tex]V_{fc}[/tex] is the final velocity of the car and [tex]V_{ft}[/tex] is the final velocity of the train.

Replacing data:

[tex](1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}[/tex]

Solving for [tex]V_{ft}[/tex]:

[tex]V_{ft}= 3.17 m/s[/tex]

Changed to cm/s, we get:

[tex]V_{ft}= 3.17*100 = 317 cm/s[/tex]

b) The kinetic energy K is calculated as:

K = [tex]\frac{1}{2}MV^2[/tex]

where M is the mass and V is the velocity.

So, the initial K is:

[tex]K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2[/tex]

[tex]K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2[/tex]

[tex]K_i = 22.06 J[/tex]

And the final K is:

[tex]K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2[/tex]

[tex]K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2[/tex]

[tex]K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2[/tex]

[tex]K_f = 19.61 J[/tex]

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

Answer:

[tex]v=3.564\ m.s^{-1}[/tex]

[tex]\Delta v =2.16\ m.s^{-1}[/tex]

Explanation:

Given:

(A)

height between John and William, [tex]h=1.8\ m[/tex]

Using the equation of motion:

[tex]v_J^2=u_J^2+2 (g.sin\theta).l[/tex]

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

[tex]v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3[/tex]

[tex]v_J=5.94\ m.s^{-1}[/tex]

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

[tex]m_J.v_J+m_w.v_w=(m_J+m_w).v[/tex]

where: v = velocity with which they move together after collision

[tex]30\times 5.94+0=(30+20)v[/tex]

[tex]v=3.564\ m.s^{-1}[/tex] is the velocity with which they leave the slide.

(B)

Now we find the force along the slide due to the body weight:

[tex]F=m_J.g.sin\theta[/tex]

[tex]F=30\times 9.8\times \frac{1.8}{3}[/tex]

[tex]F=176.4\ N[/tex]

Hence the net force along the slide:

[tex]F_R=71.4\ N[/tex]

Now the acceleration of John:

[tex]a_j=\frac{F_R}{m_J}[/tex]

[tex]a_j=\frac{71.4}{30}[/tex]

[tex]a_j=2.38\ m.s^{-2}[/tex]

Now the new velocity:

[tex]v_J_n^2=u_J^2+2.(a_j).l[/tex]

[tex]v_J_n^2=0^2+2\times 2.38\times 3[/tex]

[tex]v_J_n=3.78\ m.s^{-1}[/tex]

Hence the new velocity is slower by

[tex]\Delta v =(v_J-v_J_n)[/tex]

[tex]\Delta v =5.94-3.78= 2.16\ m.s^{-1}[/tex]

Based on the motion after the collision, cart A is concluded to be hollow.

Let's analyze the situation step by step.

Now, let's consider the possible scenarios:

Therefore, based on the information provided, the conclusion is: (A) Cart A is hollow.

Answer:

Frequency, f = 3.35 Hz

Explanation:

It is given that,

Mass of the object, m = 8 kg

Stretching in the spring, x = 2.2 cm

When the spring is hanged up in the Earth's gravitational field, its weight is balanced by the force in the spring. So,

[tex]mg=kx[/tex]

k is the spring constant

[tex]k=\dfrac{mg}{x}[/tex]

[tex]k=\dfrac{8\times 9.8}{2.2\times 10^{-2}}[/tex]            

k = 3563.63 N/m

Let f is the frequency of oscillation. Its expression is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{3563.63}{8}}[/tex]        

f = 3.35 Hz

So, the frequency of oscillation by the spring is 3.35 Hz. Hence, this is the required solution.                                                                                                                                                                                                            

Answer:

a.) 5.24 10⁻³ N . b) 0.013 N

Explanation:

a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.

We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.

In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.

The only force that has horizontal and vertical components, is the tension in the string.

We can apply Newton's 2nd Law to both directions, as follows:

T cos 24º - mg = 0

-T sin 24º + Fe = 0

where T= Tension in the string.

Fe = Electrical Force

mg = Fg = gravity force

⇒ T = mg/ cos 24º

Replacing in the horizontal forces equation:

-mg/cos 24º . sin 24º = -Fe

∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N

b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:

T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N

Final answer:

To find the magnitude of the electrical force (F⃗ elec) exerted on the plastic ball, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The electrical force is the only horizontal force acting on the ball, so it must be balanced by the horizontal component of the tension in the string.

Explanation:

The horizontal component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the horizontal component of the tension is:

T_horizontal = T * cos(24.0 degrees)

Since the ball is in equilibrium, the magnitude of the electrical force is equal to the horizontal component of the tension:

|F⃗ elec| = T_horizontal

To find the tension in the string, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The gravitational force (F⃗ g) acting on the ball is balanced by the vertical component of the tension in the string.

The vertical component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the vertical component of the tension is:

T_vertical = T * sin(24.0 degrees)

Since the ball is in equilibrium, the magnitude of the gravitational force is equal to the vertical component of the tension:

|F⃗ g| = T_vertical

The gravitational force can be calculated using the mass of the ball (m) and the acceleration due to gravity (g):

|F⃗ g| = m * g

Substituting the expression for the vertical component of the tension into the equation for the gravitational force, we get:

m * g = T * sin(24.0 degrees)

Solving for the tension (T), we get:

T = (m * g) / sin(24.0 degrees)

Substituting the given values for the mass of the ball (m) and the acceleration due to gravity (g), we get:

T = (1.20 g * 9.8 m/s^2) / sin(24.0 degrees)

T ≈ 5.10 N

b. The tension in the string is approximately 5.10 N.

Answer:[tex]a=\frac{g\sin \theta }{2}[/tex]

Explanation:

Given

inclination is [tex]\theta [/tex]

let M be the mass and r be the radius of uniform circular ring

Moment of Inertia of ring [tex]I=mr^2[/tex]

Friction will Provide the Torque to ring

[tex]f_r\times r=I\times \alpha [/tex]

[tex]f_r\times r=mr^2\times \alpha [/tex]

in pure Rolling [tex]a=\alpha r[/tex]

[tex]\alpha =\frac{a}{r}[/tex]

[tex]f_r=ma[/tex]

Form FBD [tex]mg\sin \theta -f_r=ma[/tex]

[tex]mg\sin \theta =ma+ma[/tex]

[tex]2ma=mg\sin \theta [/tex]

[tex]a=\frac{g\sin \theta }{2}[/tex]

The magnitude of the component in the xy plane of the angular momentum around the origin is 7.68 kg・m²/s. This is calculated using the formula for angular momentum, with the velocity determined from the product of the radius and angular speed.

The magnitude of the component in the xy plane of the angular momentum around the origin can be calculated using the formula for angular momentum, L = mvr, where m is the mass, v is the velocity (which can be obtained from v = rw where r is the radius, and w is the angular speed), and r is the perpendicular distance from the center of the circular path to the origin. In this case, m = 2.0 kg, r = 0.60 m (the distance from the z-axis, not the radius of the circle), and the velocity v = (0.40 m)(16 rad/s) = 6.4 m/s.

Plugging these values into the angular momentum formula gives us, L = mvr = (2.0 kg)(6.4 m/s)(0.60 m) = 7.68 kg・m²/s as the magnitude of the component in the xy plane of the angular momentum around the origin.

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