A charge of +2.00 x 10^-9 C is placed at the origin, and another charge of +4.50 x 10^-9 C is placed at x = 1.6 m. The Coulomb constant is 8.98755 x 10^9 N m2/C2. Find the point (coordinate) between these two charges where a charge of +3.70 x 10^-9 C should be placed so that the net electric force on it is zero.

Answer:

A. T = 0.4358s and f = 2.29hz

B. A = 15.67m

C. amax = 3258.71m/s

D. amax = 22601J

E. Ek = 3616.16J

Explanation:

A. The period of the motion, T = 2pi*(sqrt(m/k))

Where m is the mass of the body in motion = 885g = 0.885kg

k = the spring constant = 184N/m2

T = 2pi*(sqrt(0.885/184))

= 0.4358s

Frequency of the motion, f = 1/T

T = 0.4358s

f = 2.2949hz

B. Maximum speed, Vmax = A*(sqrt(k/m))

Where A = amplitude of the motion

Making amplitude subject of formula,

A = Vmax(sqrt(m/k))

= 226*(sqrt(0.885/184))

= 15.6739m

C. Maximum acceleration, amax = A*(k/m)

= 15.6739*(184/0.885)

= 3258.71m/s

D. Total energy, Etotal = 1/2*(m * Vmax)2

= 1/2 * 0.885 * (226)2

= 22601J

E. Kinetic energy, Ek = Etotal - mechanical energy

Ek = 1/2*(k*A2) - 1/2*(k*x2)

Where x = 0.40A

Ek = 1/2*((k*A2) - (k*0.40A)2)

= 1/2*k*A2*(1 - 0.16)

= 1/2*k*A2*0.16

But 1/2*k*A2 = 22601J

Therefore, Ek = 22601*0.16

= 3616.16J

To develop this problem we will apply the energy conservation theorem and the principle of work. Basically we will have that the kinetic, potential and work energy must be equivalent to the kinetic energy just before the impact. We will start converting the units given to the British system and then proceed with the Calculations.

Remember that according to the energy balance in this case it would be balanced like this

[tex]T_1 +\sum U_{1-2} = T_2[/tex]

[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]

Here

m = mass

[tex]v_{1,2}[/tex]= Velocity at each moment

[tex]W_{drag}[/tex]= Work by drag

h = Height

g = Acceleration due to gravity

Mass

[tex]m =255000 lb (\frac{1slug}{32.174lb})[/tex]

[tex]m = 7925.654slug[/tex]

Initial Velocity

[tex]v_1 = 550 \frac{mi}{h} (\frac{5280ft}{1mi})(\frac{1hr}{3600s})[/tex]

[tex]v_1 = 806.667ft/s[/tex]

Work by drag

[tex]W_{drag} = (2.96*10^6BTU)(\frac{778.169lb\cdot ft}{1BTU})[/tex]

[tex]W_{drag} = 2303380240lb\cdot ft[/tex]

By work energy principle

[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]

Replacing,

[tex]\frac{1}{2} (7925.654slug)(806.667ft/s)^2 +(7925.654slug)(32.08ft/s^2)(30000ft) -2303380240lb\cdot ft = \frac{1}{2} (7925.654slug)v_2^2[/tex]

Solving for [tex]v_2[/tex], we have that

[tex]v_2 = 1412.2 ft/s[/tex]

Converting this value,

[tex]v_2 = 1412.2 ft/s (\frac{1mi}{5280ft})(\frac{3600s}{1h})[/tex]

[tex]v_2 = 962.85mi/h[/tex]

Therefore the velocity of the aircraft at the time of impact is 962.85mi/h

To estimate the velocity of the aircraft at the time of impact, use the work-energy principle. The estimated velocity is 256.15 mi/h.

To estimate the velocity of the aircraft at the time of impact, we can use the work-energy principle. The work done by the drag force is equal to the change in kinetic energy of the aircraft. The work done by the drag force is given as 2.96x10^6 Btu. We need to convert this energy into foot-pounds, and then use the kinetic energy equation to find the velocity. The equation is:

K = (1/2) mv^2

Where K is the kinetic energy, m is the mass of the aircraft, and v is the velocity. Rearranging the equation, we have:

v = √((2K) / m)

Plugging in the given values, we get:

v = √((2(2.96x10^6 Btu) * (3.968x10^8 ft-lbf/Btu)) / (255,000 lb * 32.08 ft/s^2)) = 256.15 mi/h

Therefore, the estimated velocity of the aircraft at the time of impact is 256.15 mi/h.

To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.

The velocity from the uniform circular motion can be described as

[tex]v = \frac{2 \pi r}{T}[/tex]

Here,

r = Radius

T = Period

Replacing,

[tex]v = \frac{2\pi (8.4)}{6}[/tex]

[tex]v =8.7964 m/s[/tex]

From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore

[tex]F_f = F_c[/tex]

[tex]\mu N = \frac{mv^2}{r}[/tex]

Here,

[tex]\mu =[/tex] Coefficient of friction

N = Normal Force

m = mass

v = Velocity

r = Radius

The value of the Normal force is equal to the Weight, then

[tex]\mu(mg) = \frac{mv^2}{r}[/tex]

Rearranging to find the coefficient of friction

[tex]\mu = \frac{v^2}{gr}[/tex]

Replacing,

[tex]\mu = \frac{(8.7964)^2}{(9.8)(8.4)}[/tex]

[tex]\mu =0.9399[/tex]

Therefore the minimum coefficient of friction to prevent the cat from sliding off is  0.9399

Answer:

Tension T = 13.14N

Explanation:

Given:

Mass of rock m = 1.50kg

Density of rock p = 4700kg/m^3

Volume of rock V = mass/density = m/p

V = 1.50kg/4700kg/m3 = 3.19×10^-4m3

Taking the summation of forces acting on the rock;

T-W+Fb = 0

T = W - Fb .....1

T = tension

W = weight of rock

Fb = buoyant force

Fb = pw(0.5V)g = density of water × Volume under water×™ acceleration due to gravity

g = 9.8m/s^2

T = mg - pw(0.5V)g

T = 1.50×9.8 - 1000kg/m^3 ×0.5(3.19 × 10^-4) × 9.8

T = 13.14N

Final answer:

To find the tension in the string, we can use the concept of buoyancy. By equating the buoyant force and weight of the rock, we can solve for the tension. The tension will be equal to the weight of the rock minus the buoyant force.

Explanation:

In order to find the tension in the string, we can use the concept of buoyancy. Since half of the rock's volume is under water, it experiences an upward buoyant force equal to the weight of the water displaced by that volume. The buoyant force can be found using the following equation:

Buoyant force = density of water * volume of water displaced * acceleration due to gravity

The weight of the rock is equal to its mass multiplied by the acceleration due to gravity.

By equating the buoyant force and weight of the rock, we can solve for the tension in the string. The tension will be equal to the weight of the rock minus the buoyant force.

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

[tex]E = \frac{mg}{q}[/tex]

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

[tex]q = \frac{mg}{E}[/tex]

Replacing,

[tex]q = \frac{(3.37*10^{-9})(9.8)}{11000}[/tex]

[tex]q = 3.002*10^{-12}C[/tex]

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

[tex]q = ne[/tex]

Here,

n = Number of electrons

e = Charge of each electron

[tex]n = \frac{q}{e}[/tex]

Replacing,

[tex]n = \frac{3.002*10^{-12}}{1.6*10^{-19}}[/tex]

[tex]n = 2.44*10^7[/tex]

Therefore the number of electrons that reside on the drop is [tex]2.44*10^7[/tex]

To determine the number of excess electrons or protons residing on the water drop, we can use the principle of electrostatics. When an electric field is applied to a charged object, the electrostatic force acting on it can be calculated using the equation:

[tex]F = qE[/tex]

Where:

F is the electrostatic force

q is the charge on the object

E is the electric field strength

In this case, the electrostatic force acting on the water drop is balanced by the gravitational force, so we have:

F = mg

Where:

m is the mass of the water drop

g is the acceleration due to gravity

We can equate these two forces and solve for the charge q:

qE = mg

From this equation, we can isolate the charge q:

q = mg / E

Now we can calculate the charge on the water drop:

m = 3.37 × 10^(-9) kg

g = 9.8 m/s^2

E = 11000 N/C

Substituting the values into the equation:

q = (3.37 × 10^(-9) kg * 9.8 m/s^2) / 11000 N/C

Calculating this expression:

q = 3.037 × 10^(-15) C

The elementary charge of an electron or proton is approximately 1.602 × 10^(-19) C. To find the number of excess electrons or protons, we can divide the calculated charge by the elementary charge:

Number of excess electrons or protons = q / elementary charge

Number of excess electrons or protons = (3.037 × 10^(-15) C) / (1.602 × 10^(-19) C)

Calculating this expression:

Number of excess electrons or protons ≈ 1.895 × 10^(4)

Therefore, the water drop has approximately 18,950 excess electrons or protons.

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Answer:

[tex]\Delta f=f_{Lr}-f_{Se}[/tex]

147.45 Hz

Explanation:

v = Speed of sound in water = 1482 m/s

[tex]v_w[/tex] = Speed of whale = 4.95 m/s

Frequency of the wave in stationary condition

[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]

Ship's frequency which is reflected back

[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]

The difference in frequency is given by

[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]

[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]

[tex]f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz[/tex]

[tex]f_{Se}=22\ kHz[/tex]

[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]

The difference in wavelength is 147.45 Hz

Answer:

[tex]410.2 nm[/tex]

Explanation:

We are given that

[tex]n_1=2,n_2=6[/tex]

We have to find the wavelength of laser should you used.

We know that

[tex]\frac{1}{\lambda}=R(\frac{1}{n^2_1}-\frac{1}{n^2_2})[/tex]

Where [tex]R=1.097\times 10^7/m[/tex]=Rydberg constant

[tex]\lambda[/tex]=Wavelength

Using the formula

[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{2^2}-\frac{1}{6^2})[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{4}-\frac{1}{36})[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{9-1}{36}=1.097\times 10^7\times \frac{8}{36}[/tex]

[tex]\frac{1}{\lambda}=\frac{1.097\times 10^7}{4}[/tex]

Using identity:[tex]\frac{1}{a^x}=a^{-x}[/tex]

[tex]\lambda=\frac{4}{1.097}\times 10^{-7}[/tex]=[tex]4.102\times 10^{-7} m[/tex]

1 nm=[tex]10^{-9} m[/tex]

[tex]\lambda=4.102\times 100 \times 10^{-9}=410.2\times 10^{-9} [/tex] m=410.2 nm

Hence, the wavelength of laser=[tex]410.2 nm[/tex]

Answer

given,

speed of the fastest runner = 6.2 m/s

speed of the slower runner = 5.5 m/s

Assume, we need to calculate  time when both the runner meet for the first time.

distance cover by the fast runner = distance cover by the slow runner

  6.2 t = 5.5 t + 200

   0.7 t = 200

      t = 285.71 s

time after which both the runner meet.

Distance they covered after starting

D = s x t

D = 6.2 x 285.71

D = 1771.43 m.

now, calculating the second time both will meet.

they will take double time to meet again

 t' = 2 x 285.71

t '= 571.42 s

The ideal thermal efficiency for the steam power plant operating on the reheat Rankine cycle is approximately 57.0%.

To determine the thermal efficiency of the steam power plant operating on the ideal reheat Rankine cycle, we will follow the steps in the Problem-Solving Strategies for Thermodynamics:

Identify the Temperatures and Pressures

Boiler Pressure (P1): 17.5 MPa (High-pressure turbine inlet)

Reheater Pressure (P2): 2 MPa (Low-pressure turbine inlet)

Condenser Pressure (P3): 50 kPa

High-Pressure Turbine Inlet Temperature (T1): 550°C = 823 K

Low-Pressure Turbine Inlet Temperature (T2): 300°C = 573 K

Calculate Maximum Efficiency (Carnot Efficiency)

The maximum theoretical efficiency (Carnot efficiency) for a heat engine operating between two temperatures is given by the formula:
[tex]\eta_{max} = 1 - \frac{T_c}{T_h}[/tex]

where:

[tex]T_h[/tex] = Highest temperature (at the boiler, T1) = 823 K

[tex]T_c[/tex] = Lowest temperature (at the condenser) = Convert 50 kPa to temperature

Using steam tables or Mollier diagrams, the saturation temperature corresponding to 50 kPa is approximately 81°C = 354 K.

Now substituting:
[tex]\eta_{max} = 1 - \frac{354}{823} \approx 0.570 \text{ or } 57.0\%[/tex]

Answer:

Down

       F1ₓ = 219.6N

       [tex]F1_{y}[/tex]  = 1038.8 N

Top

       F2ₓ = 219.6 N

       [tex]F2_{y}[/tex] = 0  

Explanation:

For this exercise we must make a free body diagram of the ladder, see attached, then use the balance equations on each axis

Transnational Balance

X axis

        F1ₓ -F2ₓ = 0

        F1ₓ = F2ₓ

Y Axis  

         [tex]F1_{y}[/tex] -  [tex]F2_{y}[/tex] - W - W_man = 0           (1)

Rotational balance

The reference system is placed at the bottom of the stairs and we can turn the anti-clockwise direction of rotation as positive

           F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0

Let us write the data they give, the masses of the ladder (m = 14.0 kg), the mass of man (m_man = 92 kg), the center of mass of the ladder that is 2m from the bottom (the height) and the position of the man which is 3 m high

Let's look with trigonometry for distances

The angle of the stairs is

           cos θ = x / L

           θ = cos⁻¹ x / L

           θ = cos⁻¹ 2 / 5.6

           θ = 69⁰

Height y

          tan 69 = y / x

          y = x tan 69

          y = 2 tan 69

          y = 5.21 m

Distance x

          tan 69 = 2 / x

          x = 2 / tan 69

          x = 0.7677 m

The distance x_man

          x_man = 3 / tan 69

          x_man = 1,152 m

They indicate that between the scalars and the support there is no friction so the vertical force at the top is zero

          [tex]F2_{y}[/tex] = 0

Let's replace in the translational equilibrium equation

         F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0

         F2ₓ 5.21 -0 - 14.0 9.8 0.7677 - 92.0 9.8 1,152 = 0

         F2ₓ = 1143.97 / 5.21

         F2ₓ = 219.6 N

We use equation 1

         [tex]F1_{y}[/tex] + 0 - W - W_man = 0

        [tex]F1_{y}[/tex] = W + W_man

        [tex]F1_{y}[/tex]  = (m + m_man) g

         [tex]F1_{y}[/tex]  = (14 +92) 9.8

         [tex]F1_{y}[/tex]  = 1038.8 N

We can write the force on each part of the ladder

Down

       F1ₓ = 219.6N

       [tex]F1_{y}[/tex]  = 1038.8 N

Top

       F2ₓ = 219.6 N

       [tex]F2_{y}[/tex] = 0  

Answer:

The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

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

Put the value into the formula

[tex]a=\dfrac{160-0}{2.10}[/tex]

[tex]a=76.19\ m/s^2[/tex]

We need to calculate the magnitude of the electric field

Using formula of electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

[tex]E=\dfrac{a+g}{\dfrac{q}{m}}[/tex]

Put the value into the formula

[tex]E=\dfrac{76.19+9.8}{0.100}[/tex]

[tex]E=8.6\times10^{2}\ N/C[/tex]

The direction is upward.

Hence, The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]

Answer:

E = 8.6 x 10² N/C

Explanation:

given,

initial speed of charge,u = 0 m/s

final speed of charge,v = 160 m/s

time,t = 2.1 s

charge-to-mass ratio = 0.100 C/kg

Electric field of the region = ?

Acceleration of the charge

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

[tex]a = \dfrac{160 - 0}{2.1}[/tex]

a = 76.19 m/s²

specific charge = [tex]\dfrac{q}{m} = 0.1[/tex]

now,

Electric field,

[tex]E = \dfrac{F}{q}[/tex]

charge is moving upwards so,

[tex]E = \dfrac{(a + g)}{\dfrac{q}{m}}[/tex]

[tex]E = \dfrac{(76.19+9.8)}{0.1}[/tex]

E = 860 N/C

electric field , E = 8.6 x 10² N/C

hence, the magnitude of electric field is equal to E = 8.6 x 10² N/C

Answer:

2. VA

Explanation:

The valency electron or outer electron of a neutral atom of an element determines the group at which an element belong. The electronic configuration of an atom that have valency of 5 can be represented as 2 5 ,  2 8 5 etc.

The electronic configuration 2 5 represent Nitrogen atom while 2 8 5 represent phosphorus atom. The valency 5 depicts the element belongs to group 5A(VA).

This means atoms of valency electron of 3 belong to group IIIA, valency electrons of 4 belongs to group IVA, valency electron of 6 belong to group VIA  and valency electrons of  7 belong to group VIIA.  

The electric potential energy of a system comprising two protons 2.1 x 10^-15 m apart can be calculated using Coulomb's law, which gives us a result of approximately 0.68 MeV.

The electric potential energy of a system depends upon the charge of the components and the distance between them. In this case, we can calculate the electric potential energy using Coulomb's law which states that the electric potential energy 'V' between two charges is given by the equation V = k*q1*q2/r where 'k' is Coulomb's constant (8.99 × 10^9 N m^2/C^2), 'q1' and 'q2' are the two charges, and 'r' is the distance between the charges. Given that the charges are two protons, they both have the same charge (1.6 × 10^-19 Coulombs). The distance 'r' is given as 2.1 × 10^-15 m. Substituting these values in, we get: V = (8.99 × 10^9 N m^2/C^2)* (1.6 × 10^-19 C) * (1.6 × 10^-19 C) / (2.1 × 10^-15 m) which results in an electric potential energy of approximately 0.68 MeV (mega electron volts).

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To solve this concept we will apply the mathematical equations concerning the calculation of Volume in a sphere and the relation of density as a function of mass and volume, that is

The volume of the neutron star is

[tex]V = \frac{4\pi }{3}R^3[/tex]

[tex]V = \frac{4 \pi}{3} (\frac{17*10^{5}cm}{2})^3[/tex]

[tex]V = 25.72^{17}cm^3[/tex]

Now the density of the neutron star is

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

[tex]\rho = \frac{1.989*10^{30}kg(\frac{10^3g}{1kg})}{25.72*10^{17}cm^3}[/tex]

[tex]\rho = 7.733*10^{14}g/cm^3[/tex]

Therefore the density of the neutron star is [tex]\rho = 7.733*10^{14}g/cm^3[/tex]

Final answer:

Determining if a friction force is required for a car rounding a banked curve depends on the car's speed relative to the curve's ideal speed. At 95 km/h, a frictional force may be needed if this speed is not the ideal speed for the 68 m radius curve banked at 16 degrees.

Explanation:

When a 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees, we need to determine if a friction force is required when the car is traveling at 95 km/h. If the car is traveling at the correct banked curve speed, it could complete the turn without any frictional force. However, if the car travels at a speed higher or lower than this optimal speed, a frictional force will be necessary either to prevent the car from slipping outward or to prevent it from falling inward towards the center of the curve.

To find out whether a friction force is needed, we first need to calculate the ideal speed for this banked turn. This involves calculating the speed at which the components of the normal force provide enough centripetal force for the turn. The ideal speed is reached when no friction force is needed to keep the car on the path, meaning the force of gravity, the normal force, and the centripetal force are in perfect balance.

However, if the car is indeed traveling at 95 km/h, faster or slower than this ideal speed, then either a static frictional force acting upwards along the bank or a static frictional force opposite to the car's direction would be required to maintain its circular path without slipping.

Answer:

Constructive Interference.

Explanation:

Constructive Interference.

Definition:

Two waves meet in such a way their highs(Crests) combine to form a new waves whose magnitude is the sum of magnitude of combining waves.

Since two waves have same wavelength and are in phase so when they combine they well form a way which has the magnitude equal to the sum of the magnitude of both waves.

Reasons why it is Constructive Interference:Waves

To solve this problem we will apply the concepts related to linear velocity and angular velocity to perform the respective conversion with the given values. To find the velocity in the upper part of the tire we will use the mathematical relation that expresses that it is twice the linear velocity. Let's start

PART A)  

[tex]\omega = \frac{v}{r}[/tex]

[tex]\omega = \frac{29}{0.3}[/tex]

[tex]\omega = 96.66 rad/s[/tex]

Now we now that [tex]2\pi rad = 1 rev[/tex], then

[tex]\omega = 96.66rad/s (\frac{1 rev}{2\pi rad})[/tex]

[tex]\omega = 15.38rev/s[/tex]

PART B)

[tex]v = 2v_0[/tex]

[tex]v = 2(29)[/tex]

[tex]v = 58m/s[/tex]

To solve this problem we will apply the laws of proportion between the resistance / weight ratio that may occur depending on the cases mentioned. The strength-to-weight ratio for each object is,

For the nano tube,

[tex]\frac{S}{W} = \frac{1}{10*10^{-9}} =1*10^{8}[/tex]

For the leg of a flea,

[tex]\frac{S}{W}= \frac{1}{100*10^{-6}}=10000[/tex]

For the elephant,

[tex]\frac{S}{W} = \frac{1}{2}= 0.5[/tex]

From this we can conclude that the resistance / weight ratio of the nano tube is 10 thousand times better than the flea's leg and [tex]10 ^ 8[/tex] times better than the elephant's leg.

Answer:

the difference is due to resistance tolerance

Explanation:

In mathematical calculations, either done by hand or in a computer program, the heat taken from the resistors is the nominal value, which is the writing in its color code, so all calculations give a result, but the Resistors have a tolerance, indicated by the last band that is generally 5%, 10%, 20% and in the expensive precision resistance can reach 1%.

   This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values ​​and the values ​​measured with the instruments, multimeters.

   Another source of error also occurs due to temperature changes in the circuit that affect the nominal resistance value, there is a very high resistance group that indicates the variation with the temperature, they are only used in critical circuits, due to their high cost

In summary, the difference is due to resistance tolerance.

Answer with Explanation:

We are given that

Mass , m=372 g=[tex]\frac{372}{1000}=0.372 Kg[/tex]

1 kg=1000g

Maximum acceleration, a=[tex]17.6 m/s^2[/tex]

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a=[tex]A\omega^2[/tex]

Maximum speed, v=[tex]\omega A[/tex]

[tex]17.6=A\omega^2[/tex]

[tex]1.75=A\omega[/tex]

[tex]\frac{17.6}{1.75}=\frac{A\omega^2}{A\omega}=\omega[/tex]

Angular frequency,[tex]\omega=10.06 rad/s[/tex]

b.Substitute the value of angular frequency

[tex]1.75=A(10.06)[/tex]

[tex]A=\frac{1.75}{10.06}=0.17 m[/tex]

Hence, the amplitude=0.17 m

c.Spring constant,k=[tex]m\omega^2[/tex]

Using the formula

[tex]k=0.372\times (10.06)^2[/tex]

Hence, the spring constant,k=37.6 N/m

Final answer:

The angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.

Explanation:

To determine the angular frequency, we can use the formula:

ω = √(k/m)

where ω is the angular frequency in radians per second, k is the spring constant in Newtons per meter, and m is the mass in kilograms.

Given that the maximum acceleration is 17.6 m/s^2 and the mass is 372 g (or 0.372 kg), we can calculate the spring constant:

k = m * a

k = 0.372 kg * 17.6 m/s^2 = 6.5472 N/m

Now we can find the angular frequency:

ω = √(6.5472 N/m / 0.372 kg) ≈ 7.686 rad/s

To determine the amplitude, we can use the formula:

A = vmax / ω

where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.

Given that the maximum speed is 1.75 m/s and the angular frequency is 7.686 rad/s, we can calculate the amplitude:

A = 1.75 m/s / 7.686 rad/s ≈ 0.227 m

Therefore, the angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.

To solve this problem we apply the kinematic equations of linear motion. For which the speed is described as the distance traveled in a time interval. This would be,

[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]

Our values are given as,

[tex]\text{The speed of the boat} = v_b = 3.8m/s[/tex]

[tex]\text{The speed of the ocean} = V = 6.8m/s[/tex]

[tex]\text{The wave length of the wave is the same distance traveled by boat} = d = \lambda = 30m[/tex]

[tex]\text{The relative speed of the boat} = v_r = -3.8 +6.8 = 3m/s[/tex]

Replacing we have,

[tex]t = \frac{d}{v_r}[/tex]

[tex]t = \frac{30m}{3m/s}[/tex]

[tex]t = 10s[/tex]

Therefore will take until the boat is next on a crest around to 10s

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

[tex]F_1 = F \frac{{343}}{(343-V)}[/tex]

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

[tex]F_2 = F\frac{343}{(343+V)}[/tex]

We are also so told the perceived frequency decreases by 11.9%

[tex]F_2 = F_1 - 9.27\% \text{ of } F_1[/tex]

[tex]F_2 = F_1-0.0927F_1[/tex]

[tex]F_2 = 0.9073F_1[/tex]

Equating,

[tex]F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})[/tex]

[tex]\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}[/tex]

[tex]0.9073(343+V) = 343-V[/tex]

[tex](0.9073)(343)+(0.9073)V = 343-V[/tex]

[tex]V+0.9073V = 343-(0.9073)(343)[/tex]

Solving for V,

[tex]V = 16.67 m/s[/tex]

Final answer:

The distance from the charge is 30 meters.

Explanation:

The distance from the point charge can be calculated using the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance. In this case, the electric field strength is the same at any point 5.00 mm away from the charge. Therefore, we can set up the equation as follows:

E = kQ/r^2

5.00 x 10^-3 m = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/r^2

By rearranging the equation and solving for r, we can find the distance:

r^2 = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/(5.00 x 10^-3 m)

r^2 = 900 m^2

r = sqrt(900 m^2) = 30 m

Therefore, the charge is located 30 meters away.

Answer:

F1 = 1210.65 N

Q = 65.7081 degrees

Explanation:

Sum of Forces in x - direction:

F1 * cos (Q) + F2*sin(30) - F3*(3/5) = Fres*cos (theta)

F1 * cos (Q) + (500)*(0.5) - 450*(3/5) = 570*cos(33)

F1*cos (Q) = 498.0422237 N    .... Eq1

Sum of Forces in y - direction:

F1 * sin (Q) - F2*sin (60) - F3 * (4/5) =  Fres*sin (theta)

F1 * sin (Q) - 500*sin(60) - 450 * (4/5) =  570*sin (33)

F1 * sin (Q) = 1103.45692 N  .... Eq 2

Divide Eq 2 by Eq 1

tan (Q) = 2.21558916

Q = arctan (2.21558916) = 65.7081 degrees

F1 = 1210.65 N

Final answer:

To calculate the centripetal acceleration and average speed of the Sun in its galactic orbit, we utilize relevant formulas and astronomical data. The concept of centripetal acceleration is fundamental in understanding circular motion in celestial bodies like the Sun as it orbits the Milky Way galaxy.

Explanation:

Centripetal acceleration is the acceleration directed toward the center of a circular path. To calculate the centripetal acceleration of the Sun in its galactic orbit, we use the formula: a = v^2 / r, where v is the speed of the Sun and r is the radius of its orbit. Given that the Sun's orbit radius is 3.00 x 10^4 light years and it takes 2.60 x 10^8 years to orbit the Milky Way galaxy, we can also calculate the average speed of the Sun in its galactic orbit. To determine if a nearly inertial frame of reference can be located at the Sun, we need to consider the motion relative to the galactic center.

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by [tex]2\pi r=3.5[/tex]

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to [tex]=\frac{3.5}{2}=1.75km[/tex]

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

As the gas (air) in the balloon warms, it expands due to thermal expansion, causing its density to decrease. This lower density compared to the surrounding air leads to buoyancy, making the balloon rise.

When the sun warms the gas (air) inside a balloon, the air expands due to an increase in temperature. This process, known as thermal expansion, causes the molecules in the air to move faster and spread out more, occupying a larger volume. As a result, the density of the air inside the balloon decreases because density is defined as mass per unit volume, and while the mass of the air remains constant, its volume increases. In the context of a hot air balloon, or any closed system where air is heated, this decrease in density compared to the cooler surrounding air leads to buoyancy. Buoyancy is the force that makes things float, which in the case of the balloon, causes it to rise since the hot air inside it is less dense than the cooler external environment.

Answer:

Integrated circuit (IC)

Explanation:

An integrated circuit ( IC ) is a semiconductor which contains multiple electronic components

interconnected to form a complete electronic function. Integrated circuits are the most essential part of all electronic products.

Modern integrated circuits contain as much as billions of circuit components such as transistors , diodes , resistors , and capacitors

onto a single monolithic die .

The correct answer is option b. An Integrated Circuit (IC), or microchip, is a small electronic component consisting of transistors and miniaturized parts etched onto a piece of silicon, pivotal in the development and miniaturization of electronic devices

The small electronic component made up of transistors (tiny switches) and other miniaturized parts is called an Integrated Circuit (IC). An Integrated Circuit, sometimes referred to as a microchip, is an electronic circuit of transistors etched onto a small piece of silicon. This technology allows for complex circuitry to be compacted into a tiny chip, which is crucial for the functionality of modern electronic devices like computers and cell phones. The invention of the IC was pivotal in launching the modern computer revolution because it significantly reduced the size and complexity of electronic devices, replacing bulky vacuum tubes and complicated wiring with a compact, efficient solution.

Integrated Circuits are designed to handle both analog and digital signals, but in the realm of digital electronics, they are essential for managing binary code, the series of ones and zeroes that computers use to process data. This is achieved through the hundreds, thousands, or even millions of transistors that can act as on-off switches within a single IC. The miniaturization and efficiency of ICs have been fundamental in advancing the technological capabilities of electronic devices, making them smaller, faster, and more accessible to the general public.

Answer:

Isogonic line is a line joining the points of equal magnetic declination.

Explanation:

Isogonal line is a line that joins the places of equal declination. Also isogonal line is known as the line which connects the point having the same magnetic declination.

An isogonic line joins points of equal magnetic declination on a map. This is important for navigation, as it reflects the angle difference between true north and magnetic north. Isotherms, isohyets, and isobars are other types of isolines used in geography. Thus option 5. isogonic line is correct.

A isogonic line is a line joining points of equal magnetic declination. Magnetic declination is the angle between magnetic north (the direction the north end of a compass needle points) and true north. These lines are important for navigational purposes and are often shown on special maps known as isogonic charts.

Other types of isolines include:

Understanding these different lines helps in various geographical and meteorological analyses, making it easier to interpret maps and forecasts.

Complete question.

A(n) _________ is a line joining the points of equal magnetic declination.

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of [tex] 62 lb / ft ^ 3 [/tex] (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

[tex]W = \gamma A * \int_0^15 dy[/tex]

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

[tex]W = (62)(14*7)\int^{15}_0 (15-y)dy[/tex]

[tex]W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0[/tex]

[tex]W = (14*7*62)[15(15)-\frac{(15)^2}{2}][/tex]

[tex]W = 683550ft-lbs[/tex]

Therefore the total work in the system is [tex]683550ft-lbs[/tex]