A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a long, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).

Answer:

Distance = 0.738 m

Explanation:

Solution:  

First convert Km/h into m/s.  

75 km/h * 1000 m/km * 1 hr/3600 sec = 20.8333 m/s  

According to third equation of motion:

[tex]Vf^{2}[/tex] – [tex]Vi^{2}[/tex] = 2 * acceleration * distance

Vf= final velocity

Vi= initial velocity

putting values in third equation of motion....

[tex]0.2^{2}[/tex] – [tex]20.8333^{2}[/tex] = 2 * (-30 * 9.8) * distance

note:

negative sign is due to deceleration

Distance = 0.738 m

Given an initial speed of 75km/h and a maximum deceleration of 30 G's, the front end of the car must be designed to collapse over a distance of approximately 0.73m to prevent injuries that could be fatal in an accident.

The problem is asking for the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest, given a certain initial speed and maximum deceleration. This is a problem of kinematics, which is a branch of physics. To solve it, we can use the equation d = (v_f^2 - v_i^2) / (2*a), where:

So, substituting these values into our equation, we find that the front end of the car must be designed to collapse over a distance of approximately 0.73 meters to prevent injuries that could be fatal.

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Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

Answer:b and e

Explanation:

When electron and proton are placed in the same Electric field then the force experienced by electron and proton is the same as the charge possessed by them is the same . The direction of force is different on them but the magnitude is the same.

The electric field force is the product of charge and strength of the electric field.

We know the mass of an electron is less than the mass of proton that is why the acceleration of electron is more as compared to proton for the same Electric force.

thus option b and e are correct.                                  

The statements which are true of a "free" electron and a "free" proton are placed in an identical electric field are: Choice b and Choice e.

According to the question:

In essence, the magnitude of the Coulumb's force of attraction or repulsion is the same for the free electron and free proton.

However, the direction of the forces are opposite as the protons and electron have opposing signs.

Additionally, since the mass of an electron is relatively infinitesimal compared to a proton, The acceleration of the electron is greater than that of the proton.

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Answer:

0.64 m from the first charge

Explanation:

Force is given by

[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}[/tex]

[tex]F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}[/tex]

These forces are equal

[tex]\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m[/tex]

The distance that charge should be placed is 0.64 m from the first charge

The charge +3.70 x [tex]10^{-9[/tex]C should be placed at approximately 0.96 meters from the origin for the net electric force to be zero. The solution involves balancing the Coulomb forces from two charges placed at different points. The calculations involve setting up the equation for the forces and solving it step-by-step.

Finding the Point Where the Net Electric Force is Zero

To determine the point where the charge +3.70 x [tex]10^{-9[/tex] C should be placed so that the net electric force on it is zero, we need to consider the forces exerted by both charges +2.00 x [tex]10^{-9[/tex] C (at the origin) and +4.50 x [tex]10^{-9[/tex] C (at x = 1.6 m).

Step-by-Step Explanation:

Let the position where the net force is zero be at distance x from the origin (charge +2.00 x [tex]10^{-9[/tex] C).

The distance from the charge +4.50 x [tex]10^{-9[/tex] C to this point will then be (1.6 - x) meters.

According to Coulomb's Law, the force due to a charge is given by F = k * |q1 * q2| / [tex]r^2[/tex], where k is the Coulomb constant (8.98755 x[tex]10^{9[/tex] N m²/C²).

For the net force on the charge +3.70 x [tex]10^{-9[/tex] C to be zero, the magnitudes of the forces due to the two other charges must be equal:

Set these forces equal to each other:

(8.98755 x[tex]10^{9[/tex] N m²/C²) * (2.00 x [tex]10^{-9[/tex]C) * (3.70 x [tex]10^{-9[/tex] C) / x² = (8.98755 x [tex]10^{9[/tex] N m²/C²) * (4.50 x [tex]10^{-9[/tex] C) * (3.70 x [tex]10^{-9[/tex] C) / (1.6 - x)²

Simplify and solve for x:

(2.00 x [tex]10^{-9[/tex]) / x² = (4.50 x [tex]10^{-9[/tex]) / (1.6 - x)²

2 / x² = 4.5 / (1.6 - x)²

x² / (1.6 - x)² = 2 / 4.5

x² / (1.6 - x)² = 0.444

To solve for x, take the square root of both sides:

Finally, solve the equation x = √0.444 * (1.6 - x):

Thus, the charge +3.70 x [tex]10^{-9[/tex] C should be placed at approximately 0.96 meters from the origin to have the net electric force on it be zero.

Answer:

The answer to your question is v₁ = 5.74 m/s

Explanation:

Data

v₁ = ?

h₁ = 0 m

v₂ = 0.75 m/s

h₂ = 1.65 m

g = 9.81 m/s²

Formula

                mgh₁   +  1/2mv₁²   =   mgh₂  +  1/2mv₂²

mass is not consider (if we factor mass, it is cancelled)

                        gh₁ + 1/2v₁²  =   gh₂  +  1/2v₂²

Substitution

                        (9.81)(0) + 1/2v₁² = (9.81)(1.65) + 1/2(0.75)²

Simplification

                                 0    +  1/2v₁² = 16.19 + 0.28

Solve for v₁

                                            1/2v₁² = 16.47

                                                 v₁² = 2(16.47)

                                                 v₁² = 32.94

Result

                                                v₁ = 5.74 m/s

Minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

Given here,

v₁ - initial velocity = ?

h₁ - initial height = 0 m

v₂ - final velocity = 0.75 m/s

h₂ - final height = 1.65 m

g - gravitational acceleration = 9.81 m/s²

The speed can be calculated by using the formula,

[tex]\bold { mgh_1 + \dfrac 12 mv_1^2 = mgh_2 + \dfrac 12mv_2^2}[/tex]            

factor the mass,

 [tex]\bold { gh_1 + \dfrac 12 v_1^2 = gh_2 + \dfrac 12v_2^2}[/tex]

put the values in the formula, and solve it for V1

[tex]\bold { (9.81)(0) + \dfrac 12v_1^2 = (9.81)(1.65) + \dfrac 12(0.75)^2}\\\\\bold { \dfrac 12v_1^2 = 16.19 + 0.28}\\\\\bold { \dfrac 12v_1^2 = 16.47}\\\\ \bold {v_1^2 = 2(16.47)}\\\\\bold {v_1^2= 32.94}\\\\\bold { v_1 = 5.74\ m/s}[/tex]

Therefore,  minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

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Answer:

IGNEOUS ROCKS

Explanation: Igneous rocks are those rocks that solidify from magma.

Igneous rock is divided into two ,they are:

1. Intrusive

Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.

2 Extrusive igneous rock is also known as known as volcanic rocks

Answer:

[tex]F_C=20800\ N[/tex]

Explanation:

Given:

During the motion of the car on a flat circular track there acts a force which pushes the car inwards to comply in the circular motion. This force is called centripetal force which is generated due to the frictional force between the road and the tyres of the car.

This centripetal force is given as:

[tex]F_C=m.\frac{v^2}{r}[/tex]

here: r = radius of the track = [tex]\frac{d}{2}[/tex]

[tex]F_C=1300\times \frac{40^2}{100}[/tex]

[tex]F_C=20800\ N[/tex] is the force which makes the car turn inwards going with the given uniform speed.

Final answer:

The net force on a 1300 kg car driving around a 200-m diameter track at 40 m/s is calculated using the centripetal force formula, resulting in a force of 20,800 N directed towards the center of the track.

Explanation:

The question asks to calculate the magnitude of the net force acting on a 1300 kg car driving around a 200-m diameter circular track at a speed of 40 m/s.

First, we convert the diameter to radius by dividing by 2, giving a radius (r) of 100 m. To find the net force (Fnet), we use the formula for centripetal force: Fnet = m * (v² / r), where m is mass, v is velocity, and r is radius.

Substituting the given values:

Mass (m) = 1300 kg

Velocity (v) = 40 m/s

Radius (r) = 100 m

This yields:

Fnet = 1300 kg * (40 m/s)² / 100 m

Calculating this, we get:

Fnet = 1300 kg * 1600 m²/s² / 100 m

Fnet = 1300 kg * 16 m/s²

The net force is 20,800 N directed towards the center of the circular track.

To determine the heaviest weight a weight lifter can lift without breaking his legs, we need to calculate the stress on the bones. By using the dimensions of the tibia and the breaking stress of bone, we can calculate the maximum weight the lifter can lift.

To determine the heaviest weight a weight lifter can lift without breaking his legs, we need to calculate the stress on the bones. Stress is defined as the force applied per unit area. In this case, we can calculate the stress on the tibia using the formula stress = force/area. The area of the cross section of the tibia can be found by subtracting the area of the inner circle from the area of the outer circle. Once we have the stress, we can use it to determine the maximum weight the lifter can lift without breaking his legs.

Using the given dimensions of the tibia, we can calculate the area and stress:

Outer radius = 3.6 cm / 2 = 1.8 cm = 0.018 m

Inner radius = 2.3 cm / 2 = 1.15 cm = 0.0115 m

Area of outer circle = π * (0.018 m)^2 = 0.001018 m^2

Area of inner circle = π * (0.0115 m)^2 = 0.000415 m^2

Area of tibia = Area of outer circle - Area of inner circle = 0.001018 m^2 - 0.000415 m^2 = 0.000603 m^2

Force on the tibia = weight lifted = mass * acceleration due to gravity = 90.0 kg * 9.8 m/s^2 = 882 N

Stress on the tibia = force/area = 882 N / 0.000603 m^2 = 1,460,070 Pa (or 1.46 MPa)

Therefore, the lifter can lift a maximum weight without breaking his legs if the stress on the tibia is less than or equal to the breaking stress of bone. In this case, the lifter can lift:

Maximum weight = breaking stress * area = 150 MPa * 0.000603 m^2 = 90.45 N

So, the lifter can lift a maximum weight of approximately 90.45 N without breaking his legs.

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Answer:

(A)  111.77m

(B) 9.07m

(C)55.88m/s

(D)20.54m/s

Explanation:

step 1 " we have to calculate the time it took the projectile to get to its maximum height

(a)  t = usinθ/g

= 49sin 33/9.81

49×0.5446/9.81

=2.72s

the horizontal distance =  ucosθ×t  , because the projectile horizontal motion is unaffected by the force of graavity

= 49cos33 ×2.72

=111.77m

(B) with the same projectile fired the same way , the horizontal distance = 55.8m

55.8 = ucosθ×t

55.8 = 49cos33 ×t

t = 55.8/49cos33

t= 1.36s

height of the projectile = 1/2 ×g×t²

=1/2 ×9.81×1.36²

= 9.07m

(c) Velocity of the projectile when it hits the wall

V₀ = ucosθ×t

49cos 33 × 1.36

=55.88 m/s

(D) speed = distance / time

distance /2×t   ; total time of flight

= 55.88/ 2.72

=20.54m/s

Answer:

a friction force that opposes its movement

All this force has a direction opposite to that of the car

Explanation:

A car traveling at 83 km / h with constant speed has a friction force that opposes its movement, this force comes from several sources:

.- The air resistance that is proportional to the speed

/ - the resistance between the rubbers and the pavement that is constant

.- The internal friction of the different engine components

.- The formation of eddies due to the lack of aerodynamic shape of the car

All these forces must be counteracted the motor force

All this force has a direction opposite to that of the car

Answer:

Yes, a frictional force would be required parallel to the bank in the direction downwards the bank to prevent the car from moving off the bank.

Question:

A 1050-kg car rounds a curve of radius 72 m banked at an angle of 14°. If the car is traveling at 83 kmh, will a friction force be required? If so, in what direction?

Explanation:

When a car is to move round a curve with a particular radius, the centrifugal force due to rounding the curve would act on it, which would try to push it off the bank.

As a result of the inclined bank, a force as a result of gravity will act on it to prevent it from moving off the bank.

Fc = mv^2/r

Fw = mgsin14

Where;

Fc is the centrifugal force.

m = mass

Fw = force as a result of weight.

v = velocity of car

r = radius of circular path

g = acceleration due to gravity.

Looking at the two forces, due to the high speed of the car, low angle of inclination of the bank and relatively low radius of round path, the centrifugal force will be high than the force resulting from the weight to keep the car on the track.

Fc > Fw

Fc is directed out of the bank

Fw is try to keep the car on track

Therefore, when Fc > Fw and without frictional force acting inward(down the bank) to neutralise the centrifugal force the car would move off the bank and out of the round curve track. So frictional force is needed in the direction downwards the bank to keep the car on track.

Answer:

the electron jump goes from the negative terminal to the positive terminal

Explanation:

In a battery the energy is produced by a chemical reaction, which accumulates electrons in a terminal, called a negative terminal, but by convention the current goes from the positive to the negative terminal.

   We can see from this explanation that the negative wire is the one with an excess is electrons, so the electron jump goes from the negative terminal to the positive terminal

The final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).

Given equations:

[tex]\[ Q_c = Q_a \][/tex]

[tex]\[ m_c \cdot c_w \cdot (T_{ic} - T_f) = m_a \cdot c_a \cdot (T_f - T_{ia}) \][/tex]

Substitute the given values:

[tex]\[ 0.5 \cdot 1000 \cdot (85 - T_f) = 0.2 \cdot 900 \cdot (T_f - (-20)) \][/tex]

Now, distribute and simplify:

[tex]\[ 42500 - 500 \cdot T_f = 180 \cdot T_f + 3600 \][/tex]

Combine like terms:

[tex]\[ 500 \cdot T_f + 180 \cdot T_f = 42500 - 3600 \][/tex]

[tex]\[ 680 \cdot T_f = 38900 \][/tex]

Now, solve for A [tex]\( T_f \):[/tex]

[tex]\[ T_f = \frac{38900}{680} \][/tex]

[tex]\[ T_f \approx 57.21 \, ^\circ C \][/tex]

So, the final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).

Answer:

1.5 Hz

0.67 s

4 m

[tex]\pi[/tex]

2.82842 m

Explanation:

The equation is

[tex]x=4cos(3\pi t+\pi)[/tex]

It is of the form

[tex]x=Acos(2\pi ft+\phi)[/tex]

Comparing the equations we get

[tex]3\pi=2\pi f\\\Rightarrow f=\dfrac{3}{2}\\\Rightarrow f=1.5\ Hz[/tex]

Frequency is 1.5 Hz

Time period is given by

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1.5}\\\Rightarrow T=0.67\ s[/tex]

The period of the motion is 0.67 s

Amplitude

[tex]A=4\ m[/tex]

Amplitude is 4 m

Phase constant

[tex]\phi=\pi[/tex]

The phase constant is [tex]\pi[/tex]

At t = 0.25 s

[tex]x=4cos(3\pi t+\pi)\\\Rightarrow x=4cos(3\pi\times 0.25+\pi)\\\Rightarrow x=2.82842\ m[/tex]

The position of the particle is 2.82842 m

Answer:

[tex]\Delta x=1-0=1\ m[/tex]

[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]

[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]

Explanation:

The equation governing the position of the particle moving along x-axis is given as:

[tex]x=5\times t^2-4\times t^3[/tex]

we know that the time derivative of position gives us the velocity:

[tex]\frac{d}{dt} x=v[/tex]

[tex]v=10\ t-12\ t^2[/tex]

and the time derivative of of velocity gives us the acceleration:

[tex]\frac{d}{dt} v=a[/tex]

[tex]a=10-24\ t[/tex]

Now, when t = 0

[tex]x=0\ m[/tex]

[tex]v=0\ m.s^{-1}[/tex]

[tex]a=10\ m.s^{-2}[/tex]

When t=1 s

[tex]x_1=5\times 1^2-4\times 1^3=1\ m[/tex]

[tex]v_1=10\times 1-12\times 1^2=-2\ m.s^{-1}[/tex]

[tex]a_1=10-24\times 1=-14\ m.s^{-2}[/tex]

Hence,

Displacement between the stipulated time:

[tex]\Delta x=x_1-x[/tex]

[tex]\Delta x=1-0=1\ m[/tex]

Velocity between the stipulated time:

[tex]\Delta v=v_1-v[/tex]

[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]

Acceleration between the stipulated time:

[tex]\Delta a=a_1-a[/tex]

[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]

Here negative sign indicates that the vectors are in negative x direction.

Answer:[tex]44.1\times 10^6[/tex] atoms

Explanation:

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = [tex]1.83\times 10^{-7}N/m^2=1.81\times 10^{-12}atm[/tex]     [tex]1N/m^2=9.87\times 10^{-6}atm[/tex]

V= Volume of the gas = [tex]1cm^3=1ml=0.001L[/tex]       (1L=1000ml)

T= Temperature of the gas = 28°C = 301 K      [tex]0^0C=273K[/tex]  

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= ?

[tex]n=\frac{PV}{RT}=\frac{1.81\times 10^{-12}atm\times 0.001L}0.0821Latm/Kmol\times 301K}=7.32\times 10^{-17}moles[/tex]

Number of atoms =[tex]moles\times {\text {avogadro's number}}=7.32\times 10^{-17}mol\times 6.023\times 10^{23}mol^{-1}=44.1\times 10^6atoms[/tex]

Thus there are [tex]44.1\times 10^6[/tex] atoms  in a cubic centimeter at this pressure and temperature.

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

The magnitude of the vector sum of the electric forces exerted on particle 3 is 0.598 N, with the direction angle being 90 degrees, as both forces act along the positive y-axis due to symmetry.

To solve the student's question, we'll have to calculate the forces exerted on particle 3 by particle 1 and particle 2, respectively, and then find the vector sum of these forces. Since both particle 1 and particle 2 are on the x-axis at the same distance from the origin and carry the same charge, the forces they exert on particle 3 will be equal in magnitude and opposite in direction along the y-axis. Therefore, the resulting force on particle 3 will be twice the force exerted by one particle along the y-axis.

Using the formula:
F = k × |q₁ × q₂| / r²
For particle 1 or 2 exerting force on particle 3:
F = (8.9875 × 10⁹) × (3.0 × 10⁻⁹ C × 9.0 × 10⁻⁶ C) / (0.09 m)²
F = 8.9875 × 10⁹ × 2.7 × 10⁻¹⁴ C / 0.0081 m²
F = 0.299 N (Force exerted by one particle)
The total force on particle 3 will be twice this value: 0.598 N.

The direction angle of the vector sum will be 90 degrees since both forces act along the y-axis and hence the resultant force is directed straight up along the positive y-axis when measured counterclockwise from the positive x-axis.

Magnitude is √2 N, Direction is 45° counterclockwise from positive x-axis.

Coulomb's Law: formula for electric force between point charges.

[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]

[tex]\( q_1 = q_2 = 3.0 \, \text{nC} = 3.0 \times 10^{-9} \, \text{C} \),[/tex]

[tex]\( q_3 = 9.0 \, \text{μC} = 9.0 \times 10^{-6} \, \text{C} \),[/tex]

The distance between particle 3 and particles 1 and 2 is [tex]\( 90 \, \text{mm} = 0.09 \, \text{m} \)[/tex],

All charges are located on the x-axis.

The electric force exerted by particle 1 on particle 3:

[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q_3|}{r^2} \][/tex]

The electric force exerted by particle 2 on particle 3:

[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q_3|}{r^2} \][/tex]

Particles 1 and 2, equidistant from 3, have equal electric forces. The magnitude of each force:

[tex]\[ F_1 = \frac{(8.99 \times 10^9) \cdot |3.0 \times 10^{-9} \cdot 9.0 \times 10^{-6}|}{(0.09)^2} \][/tex]

[tex]\[ F_1 \approx \frac{8.09 \times 10^{-5}}{0.0081} \][/tex]

[tex]\[ F_1 \approx 1.00 \, \text{N} \][/tex]

Similarly, [tex]\( F_2 \approx 1.00 \, \text{N} \)[/tex].

Now, the total electric force [tex]\( F_{\text{total}} \)[/tex] exerted on particle 3 is the vector sum of [tex]\( F_1 \) and \( F_2 \)[/tex].

[tex]\[ F_{\text{total}} = \sqrt{F_1^2 + F_2^2} \][/tex]

[tex]\[ F_{\text{total}} = \sqrt{(1.00)^2 + (1.00)^2} \][/tex]

[tex]\[ F_{\text{total}} = \sqrt{2} \, \text{N} \][/tex]

Magnitude of total electric force on particle 3 is [tex]\( \sqrt{2} \, \text{N} \)[/tex].

[tex]\( \theta \)[/tex] is the angle from x-axis to [tex]\( F_{\text{total}} \)[/tex].

[tex]\[ \theta = \arctan\left(\frac{F_2}{F_1}\right) \][/tex]

[tex]\[ \theta = \arctan\left(\frac{1.00}{1.00}\right) \][/tex]

[tex]\[ \theta = \arctan(1) \][/tex]

[tex]\[ \theta \approx 45^\circ \][/tex]

To calculate the angular velocity of the solid metal disk, we can use the equation for rotational motion: τ = Iα. We are given the torque applied to the disk (4.28 N) and the angle it is turned (3.32°). Using the formulas for moment of inertia and angular acceleration, we can calculate the final angular velocity.

To calculate the angular velocity of the solid metal disk, we can use the equation for rotational motion:

τ = Iα

Where τ is the torque applied to the disk, I is the moment of inertia of the disk, and α is the angular acceleration of the disk.

We are given the torque applied to the disk (4.28 N) and the angle it is turned (3.32°). We can calculate the moment of inertia of the disk using the formula:

I = (1/2)mr^2

Where m is the mass of the disk and r is the radius of the disk.

Using the given diameter (29.0 cm), we can calculate the radius (14.5 cm or 0.145 m).

Plugging in the values into the formulas, we can solve for the angular acceleration:

α = τ / I

Finally, we can use the equation for angular motion to find the angular velocity:

ω = ω0 + αt

Since the disk is released from rest, the initial angular velocity (ω0) is 0. Plugging in the values, we can solve for the final angular velocity.

Answer:

[tex]L=6.21m[/tex]

Explanation:

For the simple pendulum problem we need to remember that:

[tex]\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0[/tex],

where [tex]\theta[/tex] is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

[tex]\omega^{2}=\frac{g}{L}[/tex],

where [tex]\omega[/tex] is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

[tex]\omega=\frac{2\pi}{T}[/tex],

where T is the oscillation period. Now, we can easily solve for L:

[tex](\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m[/tex]

Answer:

The time difference is 8.33 ms.

The phase difference between them is 60°

Explanation:

Given that,

Frequency = 10 Hz

Angle = 30°

We need to calculate the time difference

Using formula of time difference

[tex]\Delta t=\dfrac{\phi}{360^{\circ}\times f}[/tex]

Put the value into the formula

[tex]\Delta t=\dfrac{30}{360\times10}[/tex]

[tex]\Delta t=8.33\ ms[/tex]

If the frequency of these sine waves doubles, but the time difference stays the same,

[tex]f=20\ Hz[/tex]

We need to calculate the phase difference between them

Using formula of phase difference

[tex]\Delta \phi=\Delta t\times360\times f[/tex]

Put the value in to the formula

[tex]\Delta=8.33\times10^{-3}\times360\times20[/tex]

[tex]\Delta \phi=60^{\circ}[/tex]

Hence, The time difference is 8.33 ms.

The phase difference between them is 60°

Answer:

The resultant velocity of the boat is 3.6 mil/h

Explanation:

given information:

flowing rate to the east (positive x-axis), v₁ = 2 mil/h (0.2) = 2i

boat's speed to the north (positive y-axis), v₂ = 3 mil/h (3,0) = 3j

the resultant velocity of the boat:

[tex]v_{R}[/tex] = √3²+2²

    = √13

    = 3.6 mil/h

The resultant velocity of the boat is approximately 3.61 mph, making an angle of 56.31° north of east. This is calculated using the Pythagorean theorem for magnitude and the inverse tangent function for direction.

To find the resultant velocity of the boat when it heads north across a river at a rate of 3 miles per hour with an eastward current at 2 miles per hour, we can use vector addition. The velocity of the boat heading north (Vboat) is perpendicular to the velocity of the river's current (Vriver).

The northward velocity can be considered as the y-component (Vy = 3 mph), and the eastward current as the x-component (Vx = 2 mph). The resultant velocity (Vtot) is the vector sum of these two components and can be calculated using the Pythagorean theorem:

Vtot = √(Vx2 + Vy2)

Plugging in the values we get:

Vtot = √(22 + 32)

Vtot = √(4 + 9)

Vtot = √13

Vtot = 3.61 mph (approximately)

To find the direction of the resultant velocity, we can use the inverse tangent function (tan-1) to calculate the angle (θ) the resultant velocity vector makes with the positive x-axis (eastward direction).

θ = tan-1(Vy/Vx)

θ = tan-1(3/2)

θ = 56.31° (approximately)

Since the boat is heading north and the current flows east, the resultant velocity makes an angle of 56.31° north of east.

Answer:

The density of the liquid = 1278.95 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of  a body to its volume. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of object/Density of liquid = Weight of object in air/Upthrust in liquid.

D₁/D₂ = W/U .......................... Equation 1

D₂ = D₁(U/W)........................ Equation 2

Where D₁ = Density of aluminum, D₂ = Density of liquid, W = weight of aluminum, U = upthrust.

m₁ = 3.8 kg, m₂ = 2.00 kg, g = 9.8 m/s²

W = m₁g = 3.8(9.8) = 37.24 N.

U = lost in weight of the aluminum = (m₁ - m₂)g = (3.8-2.0)9.8

U = 1.8(9.8) = 17.64 N.

Constant: D₁ = 2700 kg/m³

Substituting these values into equation 2

D₂ = 2700(17.64)/37.24

D₂ = 1278.95 kg/m³

Thus the density of the liquid = 1278.95 kg/m³

Answer:The process that prevents the bodies of dead plant from filling the earth is decomposition of their remains.

Explanation:

1. Fungi and bacteria in the soil help decompose the remains of dead plant and uses them to enrich and nourish other plants that are alive.

2. Insects, worms and other invertebrates feed on the too

Answer:

2 and 3

Explanation:

The right answer is the option 2 and 3,

This is all about the electrons transfer from one material to the other material.

For example if the electrons in the valence shell of one material are loosely attached then the other material's atoms try to take those electron to complete their shells and that is how the charges transfers from one another.

And it could also be happen as in option 2.

Answer: 0.07meter

Pointing to the left.

Explanation:

The Electric potential V of the proton is 3.8v. while the Electric Field strength is 56V/m. We first calculate the distance r the proton would travel

E= V/r

r = V/E

r = 3.8V/(56V/m)

r =0.07m.

The proton which is positively charged will move in the direction of the Electric Field that is to the left.

When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The force on the proton can be calculated using the formula F = q*E. The distance the proton will get before coming to a stop can be calculated using the formula d = (1/2)at^2.

When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The magnitude of the force can be determined using the formula:

F = q*E

Where F is the force, q is the charge of the proton, and E is the magnitude of the electric field. In this case, the force on the proton is given by:

F = (1.6 x 10^-19 C)(56 N/C) = -8.96 x 10^-18 N

The negative sign indicates that the force is in the opposite direction of the electric field. Since the proton is moving to the right, it will slow down and eventually come to a stop. The distance the proton will get can be calculated using the formula:

d = (1/2)at^2

Where d is the distance, a is the acceleration, and t is the time. Since the proton starts from rest, its initial velocity is zero. The acceleration can be calculated using the formula:

a = F/m

Where a is the acceleration, F is the force, and m is the mass of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg. Plugging in the values, we get:

a = (-8.96 x 10^-18 N) / (1.67 x 10^-27 kg) = -5.38 x 10^8 m/s^2

Substituting this value of acceleration and the given initial velocity into the distance formula, we can calculate the distance the proton will get:

d = (1/2)(-5.38 x 10^8 m/s^2)(t^2)

Where t is the time for which the proton travels. The time can be calculated using the formula:

t = v/a

Where t is the time, v is the initial velocity, and a is the acceleration. Plugging in the values, we get:

t = (3.8 m/s) / (-5.38 x 10^8 m/s^2) = -7.08 x 10^-9 s

Since the time cannot be negative, we take the magnitude of the time:

t = 7.08 x 10^-9 s

Substituting this value of time into the distance formula, we get:

d = (1/2)(-5.38 x 10^8 m/s^2)((7.08 x 10^-9 s)^2) = 1.05 x 10^-15 m

Therefore, the proton will get a distance of 1.05 x 10^-15 meters before coming to a stop.

Answer:

497.35 m.

Explanation:

Echo: This is the sound heard after the reflection of sound wave on a plane surface.

v = 2x/t.......................................... Equation 1

Where v = speed of sound in air, x = length of the lake, t = time taken to hear the echo.

Making x the subject of the equation,

x = vt/2.......................................... Equation 2

Given: v = 343 m/s, t = 2.9 s.

Substitute into equation 2

x = 343×2.9/2

x = 497.35 m

Thus the length of the lake  = 497.35 m.

Final answer:

To determine the magnitude of the electric field through which an electron narrowly misses the upper plate, apply principles of kinematics and the formula for electric force. Calculate the time of flight using the horizontal motion, and use this to establish the vertical acceleration due to electric force. From the electron's charge and acceleration, derive the electric field magnitude.

Explanation:

To calculate the magnitude of the electric field through which an electron passes and just misses the upper plate upon exit, we need to apply the concepts of kinematics and electric forces. Since the electrical force is the only vertical force acting on the electron, it will cause a vertical acceleration according to Newton's second law (F = ma), where the force F is the electric force (F = qE) and the acceleration a is due to this force.

The vertical distance traveled by the electron just before it exits the field is equal to half the distance between the plates, which is 0.5 cm (or 0.005 m). The time it takes for the electron to travel this vertical distance can be calculated by using the initial horizontal speed and the horizontal distance (2 cm or 0.02 m) it travels. Using the formula t = d/v (where d is the horizontal distance and v is the horizontal velocity), we find t = 0.02 m / 5.35×10¶6 m/s = 3.74×10¶8 seconds.

The vertical acceleration a can be found by using the vertical distance (s), initial vertical speed (u which is 0 in this case), and time (t), using the equation s = ut + 0.5at². From this, we find a = 2s/t². Knowing the charge of an electron (e = 1.6×10¶19 C) and its mass (m = 9.11×10¶31 kg), the electric field E can then be calculated as E = F/q = ma/q. Plugging in the values, we get the magnitude of the electric field.

Answer:

Please refer to the attachment below.

Explanation:

Please refer to the attachment below for explanation.

Answer:

After [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Explanation:

Let initially voltage across capacitor is [tex]v_0[/tex]

After discharging the voltage across the capacitor is .85% of its initial voltage

So final voltage [tex]v=0.0085v_0[/tex]

We know that voltage across capacitor is given by [tex]v=v_0e^{\frac{-t}{\tau }}[/tex]

So [tex]0.85v_0=v_0e^{\frac{-t}{\tau }}[/tex]

[tex]e^{\frac{-t}{\tau }}=0.0085[/tex]

[tex]{\frac{-t}{\tau }}=ln0.0085[/tex]

[tex]-t=-4.77\tau[/tex]

[tex]t=4.77\tau[/tex]

So after [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor

Time constant of capacitive circuit is equal to the ratio of resistance and capacitance

The number of time constants the voltage across a discharging cap has decayed to 0.85% of its' initial value is;

4.768τ

In RC time constants, the formula for voltage across capacitor when discharging is given by;

v = v₀e^(-t/τ)

where;

v₀ = initial voltage on the capacitor

v = the voltage after time t

t = time in seconds

τ = time constant

We are told that the capacitor decayed to 0.85% of its' initial vale. Thus;

v = 0.0085v₀

Thus;

0.0085v₀ = v₀e^(-t/τ)

v₀ will cancel out to give;

0.0085 = e^(-t/τ)

In 0.0085 = -t/τ

-4.768 = -t/τ

Since we want to find number of time constants, then let us make t the subject to get;

t = 4.768τ

Read more at; https://brainly.com/question/15905489

Answer:

3.72 cm

Explanation:

the coordiante of the negative charge =( 0,0)

the  coordinate of the positive charge = (0,3.75)

the distance simply is  3.72-0 = 3.72 cm