A ball is dropped and begins bouncing. On the first bounce, the ball travels 3 feet. Each consecutive bounce is 1/8 the distance of the previous bounce. What is the total distance that the ball travels? Round to the nearest hundredth.

Answer:

   V₂  =  225 m^{3}

so no exact option is match with the calculated answer.

Explanation:

According to Boyle's Law, we have

                                     P_1 V_1  =  P_2 V_2    ----------- (1)

Data Given;

                  P_1  =  1.0 atm

                  V_1  =  180 m³

                  P_2  =  0.80 atm

                  V_2  =  ?

Solving equation 1 for V₂,

                  [tex]V_2= \frac{P_1 V_1}{P_1}[/tex]

Putting values,

                   [tex]V_2 = \frac{1.0*180}{0.80}[/tex]

                  [tex]V_2 = 225 m^{3}[/tex]

so no exact option is match with the calculated answer.

Answer:

E

Explanation:

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:

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:

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:

a) Increase by a factor of more than 2

Explanation:

The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.

Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:

[tex]P=\frac{RT}{v}[/tex]

Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.

If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.

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

Applying the work-energy theorem, which states that the work done by nonconservative forces equates to the change in kinetic plus potential energy, the skateboarder's change in potential energy results in -18.4J, and the absolute change in height becomes 0.032 m.

The student is asking about the concept of conservation of energy in physics, particularly in the context of the work-energy theorem. This theorem states that work done by nonconservative forces (like friction) is equal to the change in the kinetic and potential energy of the system. According to the given problem, the initial kinetic energy of a 55.6-kg skateboarder is KE0 = 0.5 * m * v², and after performing 80.4J of work on himself, while friction does -244J of work on him, the final kinetic energy becomes KEf = m * v² / 2 where 'v' is the final speed of 7.24 m/s.

(a) Considering that there are no conservative forces, the work-energy theorem states that Wnc = ΔKE + ΔPE. From here, we can calculate that ΔPE or PEf - PE0 results in -18.4J.

(b)The change in vertical height can be obtained by Δh = ΔPE / (m*g), where 'g' is the gravitational constant. Thus, Δh =-18.4J divided by (55.6 * 9.8), which results in -0.032 m, but since we are asked for the absolute value, the height change is 0.032 m.

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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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Answer:

Option-(D): 10¹¹ waves per second.

Explanation:

The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).

Answer:

answer is A

Explanation:

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:

50°

Explanation:

It is given that the both scales are the same at 18˚

Now,

at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above

the 18˚ mark.

also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,

mathematically, we can write it as (by unitary method)

3˚ increase in the temperature on the Cantor =  8˚ increase on the Frobenius scale

or

1˚ increase in the temperature on the Cantor =  (8/3)˚ increase on the Frobenius scale

thus, for x˚ increase in the temperature on the Cantor =  ((8/3)˚ × x) increase on the Frobenius scale

hence, for 12° increase we have

12˚ increase in the temperature on the Cantor =  ((8/3)˚ × 12) increase on the Frobenius scale

or

12˚ increase in the temperature on the Cantor =  32° increase on the Frobenius scale

hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.

Answer:

Rate of change of area is [tex]75.398ft^{2}/sec[/tex]

Explanation:

[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]

Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]

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:

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

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.

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]

where,

[tex]\bar x[/tex] is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]

or

[tex]\bar x = \frac{691.83}{669}[/tex]

or

[tex]\bar x =1.034m[/tex]

Hence, the center of gravity of the entire body lies 1.034 m above the floor

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.

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]

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: 768000km

Explanation:

Velocity is given by the relation between the distance [tex]d[/tex] and the time it takes to travel that distance [tex]t[/tex]:

[tex]V=\frac{d}{t}[/tex]   (1)

In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":

[tex]t=2.56s[/tex]  (2)

In addition, we know radio waves are electromagnetic waves (light), and its velocity is:

[tex]V=3(10)^{8}m/s[/tex]   (3)

Substituting (2) and (3) in (1):

[tex]3(10)^{8}m/s=\frac{d}{2.56s}[/tex]   (4)

And finding [tex]d[/tex]:

[tex]d=(3(10)^{8}m/s)(2.56s)[/tex]   (5)

Finally we can obtain the distance:

[tex]d=768000000m=768000km[/tex]  

Answer:

384,000 km

Explanation:

Given

Velocity of radio wave [tex]v = 3.00 \times 10^{8}m/s[/tex]

Duration of echo T = 2.56 s

Solution

Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s

Since any time delays in the electronic equipment can be ignored

time taken for the radio wave to reach moon

[tex]t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s[/tex]

[tex]v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km[/tex]

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:

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:

i = 7.777 × 10⁻⁴ A = 0.77 mA

Explanation:

Given:

loop radius, r = 3.0 cm = 0.03 m

Area, A = π x r² = π x 0.03² = 0.0028 m²

Magnetic Field, B = 0.75 T

Loop resistance, R = 18 Ω

time, t = 0.15 seconds

Now,

the induced emf is given as:

EMF = [tex]-\frac{BA}{t}[/tex]

also

EMF = i x R

Where, i is the current flowing

equating both the formulas for EMF, we get

[tex]{i}{R}=-\frac{BA}{t}[/tex]

or

[tex]{i}=-\frac{BA}{tR}[/tex]

substituting the values in the above equation we get

[tex]{i}=-\frac{0.75\times 0.0028}{0.15\times 18}[/tex]

or

the magnitude of the current, i = 7.777 × 10⁻⁴ A = 0.77 mA

The electric current produced in a loop due to a changing magnetic field can be calculated using Faraday's law of electromagnetic induction. Based on the given data, the magnitude of the induced current in the loop is approximately 1.18 mA.

The scenario described in the question depicts a situation where a magnetic field decreases uniformly in magnitude, thus inducing an electric current within a circular loop according to Faraday's law of electromagnetic induction.

The electric current produced in the loop as a result of the change in the magnetic field can be calculated using the expression, I = ΔΦ/ Δt * R, where ΔΦ is the change in magnetic flux, Δt is the time interval, and R is the resistance of the loop.

In this case, the initial magnetic flux, Φ1 = B1 * A, where B1 is the initial magnetic field and A is the area of the loop. Since the magnetic field decreases to zero, the final magnetic flux, Φ2, is zero. So, ΔΦ = Φ2 - Φ1 = - B1 * A. Here, A = πr², where r is the radius of the loop.

So, I = - [ (B1* πr²) / ( Δt * R ) ]. Substituting the given values into the equation, we get I = (0.75 T * π * (0.03 m)²) / ( 0.15 s * 18 Ω ) = 0.00118 A or 1.18 mA.

So, the magnitude of the induced current is approximately 1.18 mA.

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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.

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:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle  = 12.5 m/s

Now, from the Newton's second law we have  

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

or

F = 78.875 N  

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

or

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

A torque of 102.5375 Nm must be exerted by the fireman

The torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.

Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

[tex]F=ma[/tex]

Here, (m) is the mass of the body and (a) is the acceleration.

For the flow of water, the second law of motion can be given as,

[tex]F=Q\times v[/tex]

As he rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. Thus the force of this can be given as,

[tex]F=6.31\times12.5\\F=78.875\rm N[/tex]

The torque for the fireman exert on the hose is equal to the product of force applied and the distance traveled. Therefore the value of torque is,

[tex]\tau=78.875\times1.30\\\tau=102.5375\rm Nm[/tex]

Thus, the torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.

Learn more about the Newtons second law of motion here;

https://brainly.com/question/25545050

Energy and time since,

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

Hope this helps.

r3t40

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:

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.)