A certain sound level increases by an additional 20 db. By how much does the intensity increase?

Answer:

A and C are different elements, while D is an isotope of C

Explanation:

To identify an atom, we simply use the atomic number. The atomic number is the number of protons in an atom.

For a neutral atom, the number of protons is the same as the number of electrons in such an atom.

Isotopes of an element have the same atomic number (protons) but different mass number(proton + neutron) as a result of their different number of neutrons.

From the given information about the atoms, only the second option is correct:

A and C are different elements because their atomic number differs:

          A : 3 electrons, 3 protons and 3 neutrons

          C : 4 electrons, 4 protons and 3 neutrons

D is an isotope of C:

          C : 4 electrons, 4 protons and 3 neutrons

          D : 4 electrons, 4 protons and 4 neutrons

They have the same atomic number but the mass number differs due to their different number of neutrons.

Answer:

The average separation is 0.002041 AU

Explanation:

Given data:

Mass of the neutron star, M₁ = 2.5[tex]M_{solar}[/tex]

Mass of the White dwarf, M₂ = 0.3[tex]M_{solar}[/tex]

Orbiting period (P)= 28 minutes

1 year = 365 × 24 × 60 minutes = 525600 minutes

or

1 minute = 1/525600 years

thus, 28 minutes = 28/525600 years = 5.327 × 10⁻⁵ years

now from the Kepler's third law we have,

MP² = a³

where, P is the period

M is the mass = M₁ + M₂

a is the size of the orbit

thus, by substituting the values in the equation we get

(2.5[tex]M_{solar}[/tex]+0.3[tex]M_{solar}[/tex])(5.327 × 10⁻⁵ years)² = a³

Also,

[tex]M_{solar}=1[/tex] when planets orbiting sun

thus,

2.8 ×(5.327 × 10⁻⁵ years)² = a³

or

a³ = 7.94 × 10⁻⁹

or

a = 1.99 × 10⁻³ AU

thus, the average separation is 0.001995 AU

Now

1 AU = 1.5 × 10⁸ km

thus,

0.001995 AU = 299281.61 km = 2.99 × 10⁵ km

in terms of sun's radius = (2.99 × 10⁵ km)/(7 ×10⁵) = 0.427

Thus, the this orbit system will fit inside the sun

Answer:

266.5 newton

Explanation:

Weight of first object = 51 kg * 9.81

                                   =  500 N

Weight of second object = 81 kg *9.81

                                         = 794.61 N

As length of ladder is 15 m, the center of mass will be at 7.5 m.

Level arm is always perpendicular to the force. So the level of arm is at

7.5 sin θ.

τ ( wall) = τ (first object ) + τ ( second object)

N cos 30 ×15 = 794.61 sin 30 × 4  + ( 500 sin 30 ×7.5 )

13 N =  1589.22 + 1875

13 N = 3464.22

   N= 266.5 newton  

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

Answer:

Option (B)

Explanation:

As the balloon is filled with helium, it rises up because the density of helium is less than the density of air. As it rises up the outside pressure means atmospheric pressure goes on decreasing and thus the inside pressure increases. It results teh volme of the balloon increases.

Explanation:

It is given that,

Speed of the electron in horizontal region, [tex]v=1.9\times 10^7\ m/s[/tex]

Vertical force, [tex]F_y=4.9\times 10^{-16}\ N[/tex]

Vertical acceleration, [tex]a_y=\dfrac{F_y}{m}[/tex]

[tex]a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}[/tex]  

[tex]a_y=5.37\times 10^{14}\ m/s^2[/tex]..........(1)

Let t is the time taken by the electron, such that,

[tex]t=\dfrac{x}{v_x}[/tex]

[tex]t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}[/tex]

[tex]t=1.26\times 10^{-9}\ s[/tex]...........(2)

Let [tex]d_y[/tex] is the vertical distance deflected during this time. It can be calculated using second equation of motion:

[tex]d_y=ut+\dfrac{1}{2}a_yt^2[/tex]

u = 0

[tex]d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2[/tex]

[tex]d_y=0.000426\ m[/tex]

[tex]d_y=0.426\ mm[/tex]

So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.

Answer:

[tex]y = 4.24 *10^{-4} m[/tex]

Explanation:

the vertical accerlaration acting on the electron

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

[tex]a = \frac{4.9*10^{-16}}{9.11*10^{-31}} = 5.37 *10^{14} m/s^{2}[/tex]

the time taken by the electron to cover the horizontal distance is

[tex]t =\frac{24*10^{-3}}{1.9*10^{7}}[/tex]

[tex]t = 1.26 *10^{-9} s[/tex]

[tex]v_i = o[/tex]

the vertical distance trvalled in time t is

[tex]y = v_i*t +\frac{1}{2} a_y*t^{2}[/tex]

[tex]y = 0 +\frac{1}{2}*(5.37 *10^{14})(1.26 *10^{-9})^{2}[/tex]

[tex]y = 4.24 *10^{-4} m[/tex]

Using the formula for velocity (v = u + at) with the given values, we find that the velocity of the ball 3 seconds after it's thrown upwards on a planet with a gravitational acceleration of 4.0 m/s2 is 12 m/s in the upward direction.

The question asks about the velocity of a ball 3 seconds after it's thrown upwards on a planet with a gravitational acceleration of 4.0 m/s2. We can use the formula for velocity which is the initial velocity plus acceleration times time (v = u + at). In this case, the initial velocity (u) is 24 m/s, the acceleration (a) is -4.0 m/s2 (negative because it's working against the upwards motion of the ball), and the time (t) is 3 seconds.

Substituting these values into the formula gives us v = 24 m/s + (-4.0 m/s2 * 3 s) = 24 m/s - 12 m/s = 12 m/s. So the velocity of the ball 3 seconds after it's thrown is 12 m/s in the upwards direction.

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Answer:

What is the fastest time trial for the first quarter checkpoint? 2.02 seconds

What is the slowest time trial for the first quarter checkpoint? 2.15 seconds

What is the range of times measured for this checkpoint? 0.13 seconds

Answer:

2.02

2.15

0.13

Explanation:

i did it on edge

Answer:

B,A,C

Explanation:

Hope this helps!!!

Answer:

B, A, C

Explanation:

Momentum:

[tex]P = m*v[/tex]

[tex]P = momentum[/tex]

[tex]m = mass[/tex]

[tex]v = velocity[/tex]

Momentum of car A

[tex]P_{A} = 1500kg*10m/s = 15,000kg*m/s[/tex]

Momentum of car B

[tex]P_{B} = 1500kg*25m/s = 37,500kg*m/s[/tex]

Momentum of car C

[tex]P_{C} = 1000kg*10m/s = 10,000kg*m/s[/tex]

In decreasing order:

B, A, C

Answer:

750 mm

Explanation:

Given:

d = 1.8 mm

R = 4.8 m

m = 5

y = 1

Using the equation

y = (mLR)/d ,

where,  

m gives a distance 'y' to that particular slit image.  

R = distance from the double slits to the screen

d = double slit separation distance.  

L = wavelength of the light.

substituting the values in the given equation

we get

L = [tex]\frac{1\times 1.8\times 10^{-3}}{5\times 4.8}[/tex]

or

L = 750 mm

Answer:

The wavelength of the monochromatic light is [tex]7.5\times10^{-7}\ m[/tex]

Explanation:

Given that,

Distance between the slits d = 1.8 mm

Distance of fringe from the slits D =4.8 m

Number of fringe m =1

Distance between the fringes = 1 cm

We need to calculate the wavelength of monochromatic light

Using formula of young's double slits

[tex]\lambda=\dfrac{Yd}{mD}[/tex]

Where, d = Distance between the slits

D = Distance of fringe from the slits

m = Number of fringe

y = Distance between the fringes

Put the value in to the formula

[tex]\lambda=\dfrac{1\times10^{-2}\times1.8\times10^{-3}}{5\times4.8}[/tex]

[tex]\lambda =7.5\times10^{-7}\ m[/tex]

Hence, The wavelength of the monochromatic light is [tex]7.5\times10^{-7}\ m[/tex]

Answer:

55738.539 Pa

Explanation:

Gvien:

The mass of the man  = 73 kg

Mass of the chair = 7.4 kg

radius of the leg of the chair = 1.5 cm = 0.015 m

Now,

the force of gravity due to both the masses, F = (73+7.4) kg x 9.8 = 787.92 N

Also,

Pressure = force / area

Area of a circle of leg = πr² = π x 0.015² = 7.068 x10⁻⁴ m2

Now,

there are 2 legs so the force will be  divided evenly in each leg

thus,

F' =  [tex]\frac{787.92}{2}=393.96N[/tex]

Hence, the pressure on each leg will be

Pressure = [tex]\frac{393.96N}{7.068\times 10^{-4}}=55738.539 Pa[/tex]

Answer:

7 mph

Explanation:

Let the speed of river is c mph.

The effective downstream speed = 14 + c

The effective upstream speed = 14 - c

Total time = 4 hrs

Total distance = 42 miles

Time for upstream + time for downstream = 4 hrs

21 / (14 + c) + 21 / (14 - c) = 4

21 (14 - c + 14 + c) = 4 (196 - c^2)

21 x 28 = 4 (196 - c^2)

c^2 = 196 -147 = 49

c = 7 mph

The speed of the river is 7 miles/h.

Speed of Bob in still water is 14 miles/h.

Speed of Bob downstream = (14 + x) miles/h

Speed of Bob upstream = (14 - x) miles/h

here x = speed of the river.

The total distance downstream and back is 42 miles. Hence,

Distance upstream = Distance downstream = 21 miles

Given that total time of journey upstream and downstream is 4 hours. Then, using the formula:

[tex]t = \frac{distance}{speed}[/tex], we get:

[tex]\frac{21 \hspace{0.5mm} miles} {(14 + x) \hspace{0.5mm} miles/h} + \frac{21 \hspace{0.5mm} miles}{(14 - x)\hspace{0.5mm}miles/h} =4 \hspace{0.5mm} hours[/tex]

or, [tex]\frac{2 \times 21 \times 14 \hspace{0.5 mm} h}{14^{2} -x^{2} } = 4 \hspace{0.5mm} hours[/tex]

or, 21 × 7 h = 14² - x²

or, x² = 49

or, x = 7 miles/h

Answer:

Farther

Explanation:

In acceleration, if you were taking distance measurements of a free falling object at 1 second intervals, you see the object moving farther at each measurement interval.

Answer:

New time constant is 7.27 ms

Explanation:

Time constant of RC circuit is given as

[tex]\tau = RC[/tex]

now we know that time constant is given as 10 ms for capacitor of 3 micro farad

then it is given as

[tex]10\times 10^{-3} s = (3\mu F)(R)[/tex]

[tex]R = 3333.33 ohm[/tex]

now the capacitor is connected with 8 micro farad capacitor in series then we have

[tex]C = \frac{(3\mu F)(8\mu F)}{3\mu F + 8\mu F}[/tex]

[tex]C = 2.18 \mu F[/tex]

now the new time constant is given as

[tex]\tau' = (2.18\mu F)(3333.3)[/tex]

[tex]\tau' = 7.27 ms[/tex]

The resulting time constant after adding an additional 8.0 µF capacitor in series to the original 3.0 µF capacitor is approximately 7.26 ms.

The time constant of an RC circuit is determined by the product of resistance (R) and capacitance (C), denoted by the equation T = RC. When capacitors are connected in series, the total capacitance (Ctotal) is decreased compared to individual capacitors. The formula for capacitors in series is 1/Ctotal = 1/C1 + 1/C2 + ... + 1/Cn. For the given initial time constant of 10.0 ms and an original capacitor of 3.0 µF, we can determine the original resistance (R). When an additional 8.0 µF capacitor is added in series, the new time constant can be calculated by finding the new total capacitance and then applying the RC formula.

Calculating the initial resistance:
R = T / Coriginal = 10.0 ms / 3.0 µF = 3.33 kΩ (approx.)
Calculating the total capacitance with the additional capacitor:
1/Ctotal = 1/3.0 µF + 1/8.0 µF = 0.333 µF-1 + 0.125 µF-1 = 0.458 µF-1
Ctotal = 1 / 0.458 µF-1 ≈ 2.18 µF
Now, calculate the new time constant:
Tnew = R * Ctotal = 3.33 kΩ * 2.18 µF ≈ 7.26 ms (approx.)

Energy and time since,

[tex]W=\dfrac{E}{t}[/tex]

Hope this helps.

r3t40

Answer:

When a balloon deflates air moves out of the balloon because the pressure inside the balloon is higher than the pressure outside the balloon.

Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.

This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.

Final answer:

Air leaves a deflating balloon as the cooler air inside has lower pressure than the outside, causing the higher pressure external air to push its way in, equalizing the pressure until it's the same inside and outside the balloon.

Explanation:

When a balloon is deflating, air leaves the balloon due to the difference in pressure between the air inside the balloon and the air outside. The air inside the balloon is initially at a higher pressure than the surrounding air. According to the principles of thermal expansion, the air in the balloon cools and its pressure decreases. As the pressure inside becomes lower than the pressure outside, air naturally flows out to equalize the pressure.

The buoyancy that lifts a hot-air balloon is based on the air's density inside being lower than outside. When a hot air balloon is inflating, the air inside becomes less dense due to heating, and it floats. However, as the air cools or if there is a leak in the balloon's material, the air inside loses its buoyancy and starts to deflate, releasing air until pressure equilibrium is reached or until the material cannot hold the air any longer.

Answer:

Velocity of the car decreases.

Explanation:

We can understand the situation if we apply the conservation of energy principle to the situation

Let the initial mass of the freight be [tex]m_{f}[/tex]

Initial velocity of the freight be [tex]v_{fi}[/tex]

Thus the initial Kinetic energy of the freight will be [tex]K.E=\frac{1}{2}m_{f}v_{if}^{2}[/tex]

When a Coal Block of mass M falls into the freight it's energy will become

[tex]K.E=\frac{1}{2}(m_{f}+M)v_{ff}^{2}[/tex]

Equating both the energies we get final velocity as[tex]v_{ff}[/tex]

[tex]\frac{1}{2}m_{f}v_{if}^{2}=\frac{1}{2}(M+m_{f})v_{ff}^{2}\\\\v_{ff}=\sqrt{\frac{m_f}{(M+m_{ff})}}\cdot v_{if}[/tex]

As we see that [tex]\sqrt{\frac{m_f}{(M+m_{ff})}}[/tex] is less than 1 we can infer that velocity decreases.

Final answer:

When coal is dumped into a moving freight car on a frictionless track, its velocity decreases due to the conservation of linear momentum, and the kinetic energy of the car decreases as a result.

Explanation:

If a large load of coal is suddenly dumped into an open freight car moving at a constant speed, the car's velocity will decrease due to the conservation of momentum. Since we are ignoring air resistance and assuming the track is frictionless, this scenario can be analyzed with the principle of conservation of linear momentum, which asserts that the total momentum of a closed system is constant if external forces are absent. In this case, the load of coal suddenly added to the freight car significantly increases the mass of the system without adding any horizontal momentum, as the coal is initially at rest relative to the horizontal motion of the car.

Considering the professional example provided, the initial momentum of the 30,000-kg freight car at a velocity of 0.850 m/s is m₁v₁. The final momentum, after 110,000 kg of scrap metal is added, must be the same as the initial momentum due to conservation of momentum. Thus, the final velocity v₂ can be found using the equation m₁v₁ = (m₁ + m₂)v₂, where m₂ is the mass of the scrap metal.

For the kinetic energy, initially, the car has a certain amount of kinetic energy due to its motion. After the additional scrap metal is dumped, the total mass of the car increases, and its velocity decreases, which corresponds to a decrease in kinetic energy. This lost kinetic energy can be computed by subtracting the final kinetic energy from the initial kinetic energy of the car before the coal was added.

Answer:

Voltage-gated K+ channels

The correct answer is that potassium ion channels mediate the falling phase of an action potential.

During the falling phase of an action potential, the membrane potential of the neuron must return to its resting state after being depolarized. This is primarily achieved through the efflux of potassium ions (K^+) out of the cell. The opening of voltage-gated potassium channels allows for this efflux, which repolarizes the membrane potential back towards the resting membrane potential.

 Here is the sequence of events during an action potential:

 1. Resting potential: The neuron is at its resting membrane potential ,typically around -70 mV, due to the concentration gradients of ions across the membrane and the selective permeability of the membrane to potassium ions via leak channels.

 2. Rising phase (Depolarization): When a stimulus reaches the threshold level, voltage-gated sodium channels open, allowing sodium ions (Na^+) to rush into the cell. This influx of positive charges depolarizes the membrane potential towards +30 mV.

3. Peak of the action potential: The membrane potential reaches its peak when the sodium channels become inactivated, stopping the influx of sodium ions.

4. Falling phase (Repolarization): Voltage-gated potassium channels open, and potassium ions move out of the cell, restoring the membrane potential towards the resting potential. The movement of potassium ions out of the cell is slower than the initial sodium influx, which is why the falling phase is slower than the rising phase.

5. Overshoot: Sometimes, the membrane potential temporarily overshoots the resting potential, becoming more negative than at rest, due to the continued efflux of potassium ion.

6. Return to resting potential: The potassium channels close, and the sodium-potassium pump restores the original ion distribution across the membrane, bringing the membrane potential back to the resting potential.

In summary, the falling phase of an action potential is mediated by the opening of voltage-gated potassium channels, which allows for the efflux of potassium ions, thereby repolarizing the neuron's membrane potential."

Answer:

7.4 cm

Explanation:

K = 2.17 x 10^3 N/m

m = 4.71 kg

v = 1.78 m/s (It is maximum velocity)

The angular velocity

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

[tex]\omega =\sqrt{\frac{2.71\times 10^{3}}{4.71}}[/tex]

ω = 24 rad/s

Maximum velocity, v = ω x A

Where, A be the maximum displacement

1.78 = 24 x A

A = 0.074 m = 7.4 cm

Final answer:

The maximum displacement of the mass-spring system is 0.035 m.

Explanation:

If we have a mass-spring system with a spring constant of 2.17e3 N/m and a mass of 4.71 kg attached to it, we can calculate the maximum displacement using the formula for potential energy in a spring: U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium. At maximum displacement, the potential energy is converted to kinetic energy, which is given by K = (1/2)mv^2, where m is the mass and v is the velocity. Equating the two equations and solving for x, we can find the maximum displacement.

First, let's find the equilibrium position by taking the weight of the mass as force due to gravity: F = mg, where m is the mass and g is the acceleration due to gravity. Then, we can calculate the maximum displacement using the formula x = (U/mg)^0.5.

Plugging in the given values, we have: x = ((1/2)(2.17e3 N/m)(x)^2) / (4.71 kg)(9.8 m/s^2), which gives us x = 0.035 m.

Answer: The car if it weighs more than the horse

Explanation: Kinetic energy is proportional to the object's mass and velocity, so higher the mass and velocity the more kinetic energy it will have. In this case as all the objects have the same velocity we have to consider their masses and thus, the car assumingly has the greatest kinetic energy

Answer:

A car traveling at 35 miles per hour

Explanation:

Kinetic energy is the energy possessed by an object having motion. The kinetic energy depends on the mass and speed of the object:

K.E. = 0.5 mv²

In the given options, all the things have same speed. So, the kinetic energy can be compared on the basis of the mass. The car is the heaviest among the given four options. Thus, the car traveling at the 35 miles per hour has the maximum kinetic energy.

Yes.  That's a true statement.  Ultrasound is indeed the name given to sounds with frequencies above the human range of hearing.  

Ultrasound is not different from "normal" (audible) sound in its physical properties, except that humans can't hear it.

Answer:

Work done on the loop to stop it  = 21,269.1496 J

Average power P = 924.7456 watt

Explanation:

Mass of the wheel M = 26.0 Kg

Radius r = 1.30 m

The wheel is rotating at a speed of 297 rev/min

1 minute = 60 seconds,

So,

The wheel is rotating at a speed of 297/60 rev/sec

Initial angular speed (ω₁) = 2π(297/60) = 31.1143 rad/s

Final angular speed (ω₂) = 0  rad/s

Time taken to stop , (t) = 23 s econds

Moment of inertia I of a circular hoop around its central axis = Mr²

Where m is the mass of the wheel and r is the radius of the wheel

Thus, I = 26.0×(1.30)² Kgm²  = 43.94 Kgm²

(a)

Work done to stop it is the difference in the kinetic energy of the initial and the final system. So,

Work done W = (1/2)I(ω₂² - ω₁²) = 0.5*43.94(0 - (31.1143)²)  = -21,269.1496 J

Thus, work done on the loop to stop it  = - 21,269.1496 J

The answer has to be answered in absolute value so, Work  = |-21,269.1496 J| = 21,269.1496 J

(b)

Average power P = |W|/t = 21,269.1496 J/23.0 s = 924.7456 watt

Answer:

[tex]7.6\cdot 10^9 N/C[/tex]

Explanation:

The relationship between force exerted on a charge and strength of the electric field is given by

[tex]F=qE[/tex]

where

F is the strength of the electric force

q is the charge of the particle

E is the magnitude of the electric field

For the particle in the problem, we have

[tex]q=3.3\cdot 10^{-18} C[/tex]

[tex]F=2.5\cdot 10^{-8} N[/tex]

So the magnitude of the electric field at the location of the particle is

[tex]E=\frac{F}{q}=\frac{2.5\cdot 10^{-8}}{3.3\cdot 10^{-18}}=7.6\cdot 10^9 N/C[/tex]

Final answer:

The magnitude of the electric field at the given location is [tex]7.6 × 10^9 N/C.[/tex]

To calculate the magnitude of the electric field at a location, we can use the formula E = F/q, where E represents the electric field, F represents the force, and q represents the charge. In this case, we are given the charge of one particle as  [tex]+3.3 × 10^-18 C[/tex] between the particles as  [tex]2.5 × 10^-8 N[/tex]hese values into the formula, we get:

[tex]E = (2.5 × 10^-8 N) / (3.3 × 10^-18 C)[/tex]

Evaluating this expression, we find that the magnitude of the electric field at this location is approximately  [tex]7.6 × 10^9 N/C.[/tex]

Answer:

Infinite Distance

Explanation:

The electric potential due to a point charge can be expressed by the following equation:

[tex]V=\frac{kQ}{r}[/tex]

Here,

V is the electric potential due to the point charge

k is the proportionality constant

Q is the magnitude of the point charge

r is the distance from the charge

As the value of r increases, the value of V decreases since there is an inverse relation between the two. The value of V can be absolutely 0 when the distance from the charge is infinite i.e. r is infinite. Mathematically, dividing a number by infinity results in zero. Also theoretically speaking, at infinite distance the electric field lines won't approach and hence the electric potential would be zero.

The electric potential due to a point charge is 0 V at a distance r from the charge when the charge is infinite.

The electric potential due to a point charge is 0 V at a distance r from the charge when the charge is infinite. In other words, if the point charge is far away from the observation point, the electric potential becomes zero. This is because the electric potential decreases with distance.

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Answer:

autonomic specificity

Explanation:

The hypothesis which states that the different emotions in our body involve different and unique physiological profiles is called autonomic specificity hypothesis.

Viewed from this hypothesis perspective, emotions in the human body can be seen as time-tested solutions to timeless problems and challenges. With emotions, evolution in the human body has provided the human with  at least one generalized response to tackle these problems.

Answer:

The sled slides 0.68m before rest.

Explanation:

Vi= 2 m/s

m= 60kg

μ= 0.3

N= m * g = 60kg * 9.8 m/s²= 588 N

Fr= μ * N

Fr= 176.4 N

a= Fr/m

a= -2.94m/s²

t= Vi/a

t= 0.68 seg

d= Vi*t - a*t²/2

d= 0.68m

Answer:

0.149 m, [tex]321.8^{\circ}[/tex]

Explanation:

Let's start by resolving each vector into its components along the x- and y- direction:

[tex]E_x = E cos \theta = (0.111) cos 90^{\circ}=0\\E_y = E sin \theta = (0.111) sin 90^{\circ}=0.111 m[/tex]

And

[tex]F_x = F cos \theta = (0.234) cos 300^{\circ} = 0.117 m\\F_y = F sin \theta = (0.234) sin 300^{\circ} = -0.203 m[/tex]

So the components of the vector sum are

[tex]R_x = E_x + F_x = 0+0.117 = 0.117 m\\R_y = E_y + F_y = 0.111 -0.203 = -0.092 m[/tex]

The magnitude of the vector sum is

[tex]R=\sqrt{R_x^2 +R_y^2 }=\sqrt{(0.117)^2+(-0.092)^2}=0.149 m[/tex]

And the direction is

[tex]\theta=tan^{-1} (\frac{|R_y|}{R_x})=-tan^{-1} (\frac{0.092}{0.117})=-38.2^{\circ}=321.8^{\circ}[/tex]

Answer:

Option (a) is the correct answer

Explanation:

Option (a) smaller than P is the correct answer for the given question. This is because the pressure (P) at any depth 'd' from the top surface of the water with density (ρ)  is given as:

P = ρ × g × d

where,

g is the acceleration due to the gravity

thus from the above equation it can be concluded that the pressure at any depth 'd' is directly proportional to the density of the water.

thus,

in the given case the density of the water is lower than that in the first case.

Hence, the option (a) is the correct answer

Pressure at the bottom of a container filled with liquid depends on the density of the liquid, among other factors. When water is replaced with ethyl alcohol, which has a lower density, the resulting pressure at the bottom decreases. Therefore, the correct answer is (a) smaller than P.

The pressure at the bottom of a container filled with liquid is determined by the formula P = hρg, where P is the pressure, h is the height of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity. When the water (density 1000 kg/m³) is replaced with ethyl alcohol (density 806 kg/m³), the height of the liquid and gravity remain constant while the density decreases.

Since ρ in the term hρg has decreased, the overall product, i.e, the pressure at the bottom (P), also decreases. Therefore, when the glass is filled with ethyl alcohol, the pressure at the bottom of the glass is smaller than before. So, the correct answer is (a) smaller than P.

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Answer:

Neutrophils

Explanation:

Answer:

Assume that [tex]\rm g= 9.81\; N\cdot kg^{-1}[/tex]; [tex]\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}[/tex].

Density of the disk: approximately [tex]\rm 2.19\times 10^{3}\; kg\cdot m^{-3}[/tex].

Weight of the disk: approximately [tex]\rm 245\;N[/tex].

Buoyant force on the disk if it is submerged under water: approximately [tex]\rm 112\; N[/tex].

The disk will sink when placed in water.

Explanation:

Convert the dimensions of this disk to SI units:

The radius of a circle is 1/2 its diameter:

[tex]\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m[/tex].

Volume of this disk:

[tex]V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}[/tex].

Density of this disk:

[tex]\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}[/tex].

[tex]\rho(\text{disk}) >\rho(\text{water})[/tex] indicates that the disk will sink when placed in water.

Weight of the object:

[tex]W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N[/tex].

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately [tex]\rm 0.0114009\; m^{3}[/tex] of water. The buoyant force on the disk will be:

[tex]\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}[/tex].

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

In a Northern Hemisphere extratropical cyclone, the surface winds blow counterclockwise and inward due to the Coriolis effect, a result of the Earth's rotation.

Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow counterclockwise and inward about the center. This is due to the Coriolis effect, a phenomenon caused by the Earth's rotation which influences the direction that wind travels across the globe. It results in winds in the Northern Hemisphere curving to the right, thus causing cyclone winds to move counterclockwise.

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In a Northern Hemisphere extratropical cyclone, the surface winds blow in a counterclockwise and inward direction. This pattern is influenced by the Coriolis force, which causes winds to be deflected to the right, leading to this counterclockwise rotation. The phenomenon also aligns with the fact that air is attracted towards low-pressure centers like cyclones.

In the Northern Hemisphere, when viewed from above, surface winds around an extratropical cyclone flow in a counterclockwise and inward direction. This phenomenon is influenced by the Coriolis force, which causes winds to be deflected to the right in the Northern Hemisphere. This deflection results in a counterclockwise rotation. The Coriolis force similarly influences tropical cyclones, causing them to display the same pattern of rotation.

A key point to note is that this rotation is associated with areas of low pressure, such as the center of these cyclone systems. This low-pressure center attracts air, causing winds to flow inward. As the air rises within these low-pressure zones, it cools and forms clouds, making such cyclonic weather patterns visible from space.

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