Calculate the magnitude of the angular momentum of the Mars in a circular orbit around the sun. The Mars has mass 6.42×1023 kg , radius 3.40×106 m , and orbit radius 2.28×1011 m . The planet completes one rotation on its axis in 24.5 hours and one orbit in 687 days.

Parts made from blow molding are plastic, hollow, and thin-walled, such as bottles and containers that are available in a variety of shapes and sizes. Small products may include bottles for water, liquid soap, shampoo, motor oil, and milk, while larger containers include plastic drums, tubs, and storage tanks.

Final answer:

The buoyant force on a boat is the same in both salt water and fresh water, as it depends on the weight of the fluid displaced, not on the fluid's density.

Explanation:

According to the principle discovered by Archimedes, a boat experiences buoyant force which is equal to the weight of the water it displaces. This buoyant force is not dependent on the weight of the boat but rather on the density of the fluid and the volume of the fluid displaced.

As the density of salt water is higher than that of fresh water, a boat will displace less volume of salt water to float. Therefore, the weight of the salt water displaced is equal to the weight of fresh water displaced when the boat floats.

This means that the buoyant force on the boat is the same in both salt water and fresh water.

Answer:

Option B

Explanation:

According to quantum theory of radiation electromagnetic field or electromagnetic  radiation( like light) produce by accelerated charge object and the quantum of EM radiations is photon which has discrete energy. So, EM field can have only certain values of total energy and no other value due the discrete nature of the energy of photon. Hence option B is correct

Answer:

2b2t

Explanation:

2b2t

Answer:

1.2 x 10⁻¹¹ N

Explanation:

B = magnitude of magnetic field at the equator = 5 x 10⁻⁵ T

q = magnitude of charge acquired by the head = 4 x 10⁻⁹ C

v = speed of driving at the equator = 60 m/s

Magnitude of magnetic force on the head at the equator is given as

F = q v B

Inserting the above values in the equation

F = (4 x 10⁻⁹) (60) (5 x 10⁻⁵)

F = 1.2 x 10⁻¹¹ N

Answer:

B) 1218

Explanation:

N = Total number of turns in the solenoid

L = length of the solenoid = 34.00 cm = 0.34 m

B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T

i = current carried by the solenoid = 2.000 A

Magnetic field at the center of the solenoid is given as

[tex]B = \frac{\mu _{o}N i}{L}[/tex]

[tex]9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}[/tex]

N = 1218

The value of N is about B) 1218

[tex]\texttt{ }[/tex]

Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:

[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]

B = magnetic field strength from current carrying wire (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

d = distance (m)

[tex]\texttt{ }[/tex]

[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]

B = magnetic field strength at the center of the solenoid (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

N = number of turns

L = length of solenoid (m)

Let's tackle the problem now !

[tex]\texttt{ }[/tex]

Given:

Current = I = 2000 A

Length = L = 34.00 cm = 0.34 m

Magnetic field strength = B = 9000 mT = 9 T

Permeability of free space = μo = 4π × 10⁻⁷ T.m/A

Asked:

Number of turns = N = ?

Solution:

[tex]B = \mu_o \frac{I N}{L}}[/tex]

[tex]\frac{I N}{L} = B \div \mu_o[/tex]

[tex]IN = BL \div \mu_o[/tex]

[tex]N = BL \div (\mu_o I)[/tex]

[tex]N = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )[/tex]

[tex]\boxed {N \approx 1218}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Magnetic Field

Answer:

N/l = 104

Explanation:

Energy stored in the inductor is given by the formula

[tex]U = \frac{1}{2}Li^2[/tex]

now we have

[tex]6\times 10^{-6} = \frac{1}{2}L(0.400)^2[/tex]

now we have

[tex]L = 7.5 \times 10^{-5}[/tex]

now we have

[tex]L = \frac{\mu_0 N^2 \pi r^2}{l}[/tex]

[tex]7.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} N^2 \pi(0.05)^2}{0.7}[/tex]

[tex]N = 73 turns[/tex]

now winding density is turns per unit length

[tex]N/l = 104[/tex]

The winding density of the given solenoid is 104 turns per meter.

The formula for energy stored in the inductor can be used to determine the inductance of the solenoid as follows.

U = ¹/₂LI²

6 x 10⁻⁶ = ¹/₂ (L) x (0.4)²

6 x 10⁻⁶ = 0.0.8L

L = 7.5 x 10⁻⁵

The number of turns of the solenoid is calculated as follows;

[tex]L = \frac{\mu N^2\pi r^2}{l} \\\\7.5 \times 10^{-5} = \frac{(4\pi \times 10^{-7} ) \times N^2 \times \pi(0.05)^2}{0.7} \\\\N = 73 \ turns[/tex]

The winding density if the number of turns per length

N/l = 73/0.7

N/l = 104

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Answer:

N = 768.4 N

Explanation:

As per given FBD we can see that the person inside the elevator have two forces on it

1) Normal force upwards

2) weight downwards

Now from Newton's law of motion we can say

[tex]F_{net} = ma[/tex]

[tex]N - mg = ma[/tex]

[tex]N = mg + ma[/tex]

now plug in all values in it

[tex]N = 68(1.5) + 68(9.8)[/tex]

[tex]N = 768.4 N[/tex]

Answer:

180.04 nm

Explanation:

λ₀ = maximum wavelength for photoelectric emission in tungsten = 230 x 10⁻⁹ m

E₀ = maximum energy of ejected electron = 1.5 eV = 1.5 x 1.6 x 10⁻¹⁹ J

λ = wavelength of light used = ?

Using conservation of energy

Energy of the light used = Maximum energy required for photoelectric emission + Energy of ejected electron

[tex]\frac{hc}{\lambda }=\frac{hc}{\lambda_{o} } + E_{o}[/tex]

[tex]\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{\lambda }=\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{230 \times 10^{-9} } + 1.5 \times 1.6 \times 10^{-19}[/tex]

λ = 180.04 x 10⁻⁹ m

λ = 180.04 nm

Answer: [tex]11.17\ \text{ m/s}[/tex]

Explanation:

Given : The escape velocity : [tex]v=2.5\times10\text{ miles per hour}[/tex]

We know that 1 mile = 1609 meters  (approx)

and 1 hour= 3600 seconds

To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.

Thus, the escape velocity in m/s is given by :-

[tex]v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}[/tex]

Hence, the speed in m/s = 11.17

Answer:

Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.

Explanation:

Vertical motion of fish:

 Initial speed, u = 0

 Acceleration, a = 9.81 m/s²

 Displacement, s = 10 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 0² + 2 x 9.81 x 10 = 196.2

    v = 14 m/s

 Final vertical speed = 14 m/s

 Final horizontal speed = initial horizontal speed = 3 m/s

 Final velocity = 3 i - 14 j m/s

 Magnitude

     [tex]v=\sqrt{3^2+(-14)^2}=14.32m/s[/tex]

 Direction

      [tex]\theta =tan^{-1}\left ( \frac{-14}{3}\right )=-77.91^0[/tex]

 Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.

If ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.0 m/s .

A. The kayaker is 11 meters away from the wall.

B. The period of the kayaker's up and down motion is 5.5 seconds.

What is the distance?

A) The distance from the kayaker to the wall is:

Distance = 1/2 × Wavelength

Distance = 1/2 × 22 m

Distance = 11 m

Therefore the kayaker is 11 meters away from the wall.

B) The period can be calculated using the formula:

Period = 1 / Frequency

Use this formula:

Frequency = Velocity / Wavelength

Frequency = 4.0 m/s / 22 m

Frequency ≈ 0.182 Hz

Period:

Period = 1 / 0.182 Hz

Period ≈ 5.49 seconds

Period ≈ 5.5 seconds

Therefore, the period of the kayaker's up and down motion is approximately 5.5 seconds.

Answer:

4 m/s

0.82 s

Explanation:

h = height to which the fox jumps = 81 cm = 0.81 m

v₀ = speed at which the fox leaves the snow

v = speed of the fox at highest point = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

Using the kinematics equation

v² = v₀² + 2 a h

0² = v₀² + 2 (- 9.8) (0.81)

v₀ = 4 m/s

t = amount of time in air while going up

Using the equation

v = v₀ + a t

0 = 4 + (- 9.8) t

t = 0.41 s

T = Total time

Total time is given as

T = 2 t

T = 2 (0.41)

T  = 0.82 s

The speed at which the fox leaves the snow is approximately 3.987 m/s. The fox is in the air for approximately 0.407 seconds.

To calculate the speed at which the fox leaves the snow, we can use the concept of vertical motion and the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Since the fox jumps straight up, the initial velocity is 0 m/s and the displacement is 81 cm (or 0.81 m). Assuming the acceleration due to gravity is 9.8 m/s^2, we can now calculate the final velocity:

v^2 = u^2 + 2as
v^2 = 0^2 + 2(9.8)(0.81)
v^2 = 15.876
v = √15.876
v ≈ 3.987 m/s

The time the fox is in the air can be calculated using the equation v = u + at, where t is the time. Again, the initial velocity is 0 m/s and the acceleration due to gravity is 9.8 m/s^2. Plugging in these values, we have:

v = u + at
3.987 = 0 + (9.8)t
3.987 = 9.8t
t = 3.987/9.8
t ≈ 0.407 s

Answer:

Total displacement traveled = 298

Explanation:

According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).

So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.

Then after a little it can travel up another cm to reach the top.

Therefore, the total displacement traveled = 297 + 1 = 298 cm

"Displacement" means the distance and direction from start to finish, regardless of what happened in between.

The caterpillar's displacement is 99 cm straight up.

Answer:

The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex]  times.

Explanation:

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

[tex]\log K=\log A-\frac{Ea}{2.303\times RT}[/tex]

The expression used with catalyst and without catalyst is,

[tex]\log K_1=\log A-\frac{Ea_1}{2.303\times RT}[/tex]...(1)

[tex]\log K_2=\log A-\frac{Ea_2}{2.303\times RT}[/tex]...(2)

On subtracting (2) from (1)

[tex]\log \frac{K_2}{K_1}=\frac{Ea_1-Ea_2}{2.303RT}[/tex]

where,

[tex]K_2[/tex] = rate of reaction with catalyst

[tex]K_1[/tex] = rate of reaction without catalyst  

[tex]Ea_2[/tex] = activation energy with catalyst  = 29.3 kJ/mol = 29300 J/mol

[tex]Ea_1[/tex] = activation energy without catalyst  = 75.3 kJ/mol=75300 J/mol

R = gas constant =8.314 J /mol K

T = temperature = [tex]20^oC=273+20=293K[/tex]

Now on substituting all the values in the above formula, we get

[tex]\log \frac{K_2}{K_1}=\frac{75300 kJ/mol-29300 kJ/mol}{2.303\times 8.314 J/mol K\times 293}=1.58\times 10^{8}[/tex]

The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex]  times.

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is [tex]\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}[/tex]

Substituting values

    [tex]3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s[/tex]

Speed of block A after collision = 10 m/s

Option A is the correct answer.

Answer:

8m/s

Explanation:

The rate of heat generation in the rod is calculated by first determining the resistance using the formula R = ρL/A, then using that resistance value in the power formula P = V²/R. Using the provided values, the rate of heat generation in the rod under a potential difference of 0.50 V is 8.33 Watts.

The subject of this question is about the rate of heat generation in a rod under a potential difference, which is a topic in Physics. To solve for this, we first need to compute for the resistance of the rod using the formula R = ρL/A, where R is the resistance, ρ is the resistivity of the material, L is the length, and A is the cross-sectional area of the rod. Given that ρ = 6.0 × 10−8 Ω ⋅ m, L = 2.0 m, and A = (2.0 mm × 2.0 mm) = 4.0 × 10-6 m², we get R = (6.0 × 10−8 Ω ⋅ m * 2.0 m) / 4.0 × 10-6 m² = 0.03 Ω.

Next, we use the formula P = V²/R to calculate the rate of heat generation (power). Here, P is the power, V is the potential difference, and R is the resistance. With V = 0.50 V and R = 0.03 Ω, after substituting the values we get P = (0.50 V)² / 0.03 Ω = 8.33 Watts. Therefore, the rate at which heat is generated in the rod is 8.33 Watts.

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The rate at which heat is generated in the rod is found by calculating the current and using Joule's Law. The final rate of heat generation is 8.33 Watts.

Heat Generated in a Rod Due to Electric Current

To determine the rate at which heat is generated in the rod, we start by calculating the resistance of the rod. Given:

Length of the rod (L) = 2.0 m

Cross-sectional area (A) = (2.0 mm × 2.0 mm) = (2.0 × [tex]10^{-3}[/tex] m) × (2.0 × [tex]10^{-3}[/tex] m) = 4.0 × [tex]10^{-6}[/tex] m²

Resistivity of the material (ρ) = 6.0 × [tex]10^{-8}[/tex] Ω·m

Potential difference (V) = 0.50 V

The resistance (R) of the rod can be calculated using the formula:

R = ρ(L/A)

Substituting the given values:

R = (6.0 × [tex]10^{-8}[/tex] Ω·m) × (2.0 m / 4.0 × [tex]10^{-6}[/tex] m²) = 3.0 × [tex]10^{-2}[/tex] Ω

Next, we calculate the current (I) flowing through the rod using Ohm's Law:

I = V / R

Substituting the values:

I = 0.50 V / 3.0 × [tex]10^{-2}[/tex] Ω = 16.67 A

The rate of heat generated (P) in the rod is given by Joule's Law:

P = I²R

Substituting the calculated values:

P = (16.67 A)² × 3.0 ×[tex]10^{-2}[/tex] Ω = 8.33 W

Therefore, the rate at which heat is generated in the rod is 8.33 Watts.

Answer:

a) 7.35 x 10¹³ m/s²

b) 5.03 x 10⁻⁸ sec

c) 9.3 cm

d) 6.23 x 10⁻¹⁸ J

Explanation:

E = magnitude of electric field = 418 N/C

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of the electron = 9.1 x 10⁻³¹ kg

a)

acceleration of the electron is given as

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

[tex]a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}[/tex]

a = 7.35 x 10¹³ m/s²

b)

v = final velocity of the electron = 3.70 x 10⁶ m/s

v₀ = initial velocity of the electron = 0 m/s

t = time taken

Using the equation

v = v₀ + at

3.70 x 10⁶ = 0 + (7.35 x 10¹³) t

t = 5.03 x 10⁻⁸ sec

c)

d = distance traveled by the electron

using the equation

d = v₀ t + (0.5) at²

d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²

d = 0.093 m

d = 9.3 cm

d)

Kinetic energy of the electron is given as

KE = (0.5) m v²

KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²

KE = 6.23 x 10⁻¹⁸ J

Answer:

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She was the presenter of Nova feature Black hole Apocalypse and has writtenany science non-fiction books

www.jannalevin.com is her own page where u get her correct info and bio

Answer:

[tex]A = 5 \times 10^{32} m^2[/tex]

Explanation:

As we know that the energy stored in the capacitor is given as

[tex]Q = \frac{1}{2}CV^2[/tex]

here we know that

[tex]Q = 6 \times 10^{22} J[/tex]

also we know that

[tex]V = 5 Volts[/tex]

now we have

[tex]6 \times 10^{22} = \frac{1}{2}C(5^2)[/tex]

[tex]C = 4.8 \times 10^{21} F[/tex]

now we know the formula of capacitance

[tex]C = \frac{\epsilon_0 A}{d}[/tex]

[tex]4.8 \times 10^{21} = \frac{(8.85 \times 10^{-12})(A)}{1}[/tex]

[tex]A = 5 \times 10^{32} m^2[/tex]

Answer:

Torque = 7.07 N.m

Explanation:

It is given that,

Force acting on the wrench, F = 40 N

Length of wrench, l = 0.25 meters

It is attached perpendicular to the bolt such that the force is applied at an angle of 135 degrees to the wrench. The formula for torque is given by :

[tex]\tau=r\times F[/tex]

[tex]\tau=rF\ sin\theta[/tex]

[tex]\tau=0.25\ m\times 40\ N\ sin(135)[/tex]

[tex]\tau=7.07\ N.m[/tex]

So, the magnitude of torque applied to the wrench is 7.07 N-m. Hence, this is the required solution.

To calculate the magnitude of the torque, use the formula τ = r * F * sin(θ), where r is the lever arm length (0.25 m), F is the force applied (40 N), and θ is the angle between force and lever arm (135°). The sine of 135° provides the necessary component of the force that contributes to the torque.

The question deals with the concept of torque in physics, particularly how torque is influenced by the angle at which a force is applied. Torque (τ) is the product of the force (F) applied, the distance (r) from the pivot point to the point where the force is applied, and the sine of the angle (θ) between the force vector and the lever arm, which can be represented as τ = r * F * sin(θ). Given that a force of 40 N is applied to the 0.25-meter-long wrench at a 135-degree angle to the wrench, the magnitude of the torque can be calculated using this formula.

Using the provided equation:

Torque= 0.25 m * 40 N * sin(135°) = 0.25 m * 40 N * sin(135°)

Here, sin(135°) is a positive value since 135° is in the second quadrant where sine values are positive. It is important to note that the angle must be converted to radians or the correct sine value must be used if the calculator is set to degrees. The calculated torque will have the unit of Newton-meters (N.m).

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

[tex]Distance=Speed\times Time[/tex]

Thus distance covered in blinking of eye =

[tex]Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters[/tex]

Thus no of football fields=[tex]\frac{936}{91.4}=9.15Fields[/tex]

Answer:

217.28 m/s

Explanation:

u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

Let a be the acceleration.

Use third equation of motion.

v^2 = u^2 + 2 a s

47.1 x 47.1 = 0 + 2 a x 607

a = 1.83 m/s^2

For small plane

a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.

Use third equation of motion

28.2^2 = 0 + 2 x 1.83 x s

s = 217.28 m/s

Answer:

E= 3.4893 N/C

Explanation:

Given s=6.20 m , t=2.50μs, m=9.11*10^-31 Kg  , q= 1.6*10^-19 C

the distance traveled by the electron in time t is

s=ut+0.5at^2

here, u is the initial velocity of the electron, t is time taken and

a is acceleration.

Since the electron is initially at rest u=0

now s=0.5at^2

Therefore a=2s/t^2

also. we know that strength of electric field is

E=ma/q

[tex]E= \frac{2ma}{qt^2}[/tex]

now puting the values we get

[tex]E=\frac{9.11\times 10^-31\times 2\times 6.20}{1.6\times 10^-19\times (4.5\times 10^-6)^2}[/tex]

therefore, E= 3.4865 N/C

The magnitude of the electric field is calculated by first determining the acceleration of the electron and then using the electric force equation to find the electric field. The resulting electric field is 34.8 N/C.

To find the magnitude of the electric field, we first need to calculate the acceleration of the electron. Given that the electron travels a distance of 6.20 m in a time of 4.50 µs (4.50 × 10-6 s), we can use the equations of motion.

Initial velocity, u = 0 (since the electron is released from rest)

Time, t = 4.50 × 10-6 s

Distance, s = 6.20 m

Using the equation of motion: s = ut + 0.5at2

Substitute the values: 6.20 = 0 + 0.5a(4.50 × 10-6)2

6.20 = 0.5a(20.25 × 10-12)

a = 6.20 / (0.5 × 20.25 × 10-12)

a = 6.20 / (10.125 × 10-12)

a = 6.12 × 1011 m/s2

Now, we calculate the electric field using Newton's Second Law, F = ma, and the electric force equation, F = eE, where e is the charge of the electron (1.60 × 10-19 C) and E is the electric field.

ma = eE

(9.11 × 10-31 kg)(6.12 × 1011 m/s2) = (1.60 × 10-19 C)E

(5.57 × 10-19 N) = (1.60 × 10-19 C)E

E = 5.57 × 10-19 N / 1.60 × 10-19 C

E = 3.48 × 101 N/C

Thus, the magnitude of the electric field is 34.8 N/C.

Answer:

5.865 μs

Explanation:

t₀ = Time taken to decay a muon = 2.20 μs

c = Speed of Light in vacuum = 3×10⁸ m/s

v = Velocity of muon = 0.927 c

t = Lifetime observed

Time dilation

[tex]t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds[/tex]

∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs

Answer:

Velocity, V = 3t²- 28t+6

Displacement, s = t³ - 14t²+6t -8

At t = 5.8 s

        s = -249.05 m

        v = -55.48 m/s

At t = 12.7 s

        s = -141.48 m

        v = 134.27 m/s

Explanation:

We have acceleration of a particle is given by  a = 6t - 28

Velocity

      [tex]v=\int adt=\int (6t - 28)dt=3t^2-28t+C[/tex]

At t = 0 we have v₀ = 6 m/s

         v₀ = 6 =  3 x 0 ²-28 x 0+C

         C = 6

        So velocity, V = 3t² - 28t+6

Displacement

        [tex]s=\int vdt=\int (3t^2-28t+6)dt=t^3-14t^2+6t+C[/tex]

At t = 0 we have s₀ = -8 m

         s₀ = -8 =  0³ + 14 x 0² + 6 x 0 + C

         C = -8

        So displacement, s = t³ - 14t²+6t -8

At t = 5.8 s

        s = 5.8³ - 14 x 5.8²+6 x 5.8 - 8 = -249.05 m

        v =  3 x 5.8² - 28 x 5.8 + 6 = -55.48 m/s

At t = 12.7 s

        s = 12.7³ - 14 x 12.7²+6 x 12.7 - 8 = -141.48 m

        v =  3 x 12.7² - 28 x 12.7 + 6 = 134.27 m/s

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]

[tex]\dfrac{d\phi}{dt}[/tex] is the rate of change if magnetic flux.

[tex]\phi=BA\ cos\theta[/tex]

[tex]\theta[/tex] is the angle between the magnetic field and the normal to area vector.

[tex]\theta=90-60=30[/tex]

[tex]\epsilon=NA\dfrac{dB}{dt}\times cos30[/tex]

[tex]\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)[/tex]

[tex]\epsilon=0.0598\ T[/tex]

[tex]\epsilon=59.8\ mT[/tex]

or

EMF = 60 mT

So, the magnitude of  emf induced in the coil is 60 mT. Hence, this is the required solution.

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed [tex]\omega[/tex], since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

[tex]v=\omega r[/tex]

where

[tex]\omega[/tex] is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

Answer:

Explanation:

The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.

In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.

Here, In the balanced wheat stone bridge

R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.

Resistors in a Wheatstone bridge can be treated as combinations of series and/or parallel resistors for simplification in both balanced and unbalanced bridges. A balanced bridge allows separate treatment of two parallel branches, while unbalanced requires careful analysis. Not all resistor networks can be simplified into series or parallel models.

In a Wheatstone bridge, the resistors can indeed be treated as combinations of series and/or parallel resistors when aiming to simplify calculations or understand the behavior of the circuit. For an unbalanced bridge, resistors are not in simple series or parallel arrangements with respect to the entire circuit due to the bridge not being in equilibrium. However, within certain parts of the bridge, resistors may appear to be in series or parallel with each other. In a balanced bridge, where the bridge is in a state of equilibrium and the central voltmeter reads zero, the two arms of the bridge can be treated separately as two parallel voltage dividers, because no current flows through the meter, effectively decoupling the two parallel branches.

Attempting to simplify a complex resistor network encountered in bridges can indeed be done by identifying and replacing series and parallel resistor combinations step by step until a single resistance value is found. However, this approach cannot always be applied to any arbitrary combination of resistors. Some configurations might contain elements that cannot be reduced to mere parallel or series connections, usually because they form more intricate networks, such as bridges or loops not separable into simpler series or parallel sections.

In conclusion, while more complex connections of resistors in circuits like the Wheatstone bridge can often be broken down into combinations of series and parallel, this is not universally the case for all resistor networks. In certain scenarios, specific techniques or theorems such as Kirchhoff's laws might be required to analyze the circuit effectively.