A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.

PART A)
Which of the following describes the electricfield inside this conductor?

a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.The charge density inside theconductor is:

a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinitePART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.

Answer:

DUNNO

Explanation:

BIT BAD FOR 65% WOULD GET -100%

Answer:

v = 31.5 m/s

Explanation:

given,

Speed of air balloon, u = 11.9 m/s

height of balloon above the ground, h = 43.4 m

acceleration due to gravity = 9.8 m/s²

Speed of the bag when it hit the ground = ?

using equation of motion

[tex]v^2 = u^2 + 2 g h[/tex]

[tex]v^2 = 11.9^2 + 2\times 9.8\times 43.4[/tex]

[tex]v= \sqrt{992.25}[/tex]

    v = 31.5 m/s

Speed of the bag before it hit the ground is equal to v = 31.5 m/s

Applying the physics of kinematics, specifically the equations of motion, the speed of the sandbag when it hits the ground can be calculated to be 29 m/s.

The speed of the sandbag when it hits the ground can be calculated with the concepts of kinematics in physics, in particular, the equations of motion. Considering that the positive direction is upwards and the balloon is moving upward at a velocity of 11.9 m/s, the initial velocity (vi) of the sandbag is +11.9 m/s. The sandbag is then acted upon by gravity, which imparts a downward acceleration of -9.8 m/s2. The final speed (vf), when the sandbag hits the ground, can be calculated using the equation vf2 = vi2 + 2ad, where a is acceleration and d is the distance. Substituting the given values: vf2 = (11.9 m/s)2 + 2*(-9.8 m/s2)*(-43.4 m), we get vf ≈ 29 m/s, which is the speed of the sandbag when it hits the ground.

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The primary current for the given transformer is 3 A which is determined by considering the principle of a transformer, the turns ratio and the law of conservation of power, and calculating secondary current using Ohm's law.

To find the primary current of a transformer, we need to first understand the principle of a transformer. A transformer operates on the electromagnetic induction principle, and the ratio of turns in the primary and secondary coil determines how much voltage is changed. In your transformer, the primary voltage is 240 V and the secondary voltage is 60 V. Therefore, the turns ratio is 240/60 which gives a value of 4.

Now, considering the law of conservation of power in an ideal transformer, the power output is equal to the power input. Power is given by the product of voltage and current.

So, primary voltage (V1) times primary current (I1) is equal to secondary voltage (V2) times secondary current (I2). The secondary current (I2) can be calculated using Ohm's law as V2/R, where R is the resistance of the load connected. Plugging in the given values we get

I2 = 60/5 = 12 A.

Now, substituting the known variables in the power conservation equation we get

240*I1 = 60*12.

Solving for I1 gives a value of 3A.

Therefore, the primary current for the given transformer is 3 A.

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Answer:

1.19×10²² m/s

Explanation:

Kinetic Energy: This can be defined as the the energy of a body due to motion.

The formula for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass of proton. v = velocity of proton.

Making v the subject of the equation,

v = √(2Ek/m).................. Equation 2.

Given: Ek = 0.995×10⁻⁵ J, m = 1.673×10⁻²⁷ kg.

Substitute into equation 2

v = √(2×0.995×10⁻⁵/1.673×10⁻²⁷ )

v = 1.19×10²² m/s.

Therefore, the velocity of proton = 1.19×10²² m/s

Answer:

a.-6.75 kgm/s

b.[tex]3375 N[/tex]

Explanation:

We are given that

Mass of baseball=0.15 kg

Initial speed=[tex]u=20m/s[/tex]

Final speed=[tex]v=-25m/s[/tex]

a.We know that

Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]

Momentum=[tex]mass\times velocity[/tex]

Using the formula

Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]

b.Time=[tex]2\times 10^{-3} s[/tex]

Force=[tex]\frac{Impulse}{time}[/tex]

Using the formula

Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N

Average force exerted by the bat on the ball=[tex]3375N[/tex]

Answer:

Approximately 500kJ

Explanation:

According to the law of conservation of energy which states that energy can neither be created nor destroyed but can be converted from one form to another.

A parachutist jumping out of the airplane covering a particular height and under the influence of gravity will possess potential energy during fall. Since PE = mass × acceleration due to gravity × height

PE = 39.4×1340×9.8

PE = 501,562Joules

If the body lands on the ground with a speed of 5.78m/s, this means the body possesses kinetic energy at the point of landing. The kinetic energy on landing is 1/2mv²

KE = 1/2×39.4×5.78²

KE = 658.14Joules

The amount of energy lost due to friction will be PE-KE

= 501,562-658.14

= 500,903Joules approximately 500kJ

Final answer:

The solution involves calculating the parachutist's initial potential energy and final kinetic energy, then finding the difference to determine the energy lost to air friction during the jump.

Explanation:

The question asks how much energy was lost to air friction during the jump of a parachutist who has a mass of 39.4 kg, jumps from a height of 1340 m, and lands with a speed of 5.78 m/s. To find the energy lost to air friction, we first calculate the potential energy at the start and the kinetic energy at the end of the jump, then find the difference.

Initial potential energy (PE_initial) is given by mgh, where m is mass, g is acceleration due to gravity (9.8 m/s²), and h is height. Therefore, PE_initial = 39.4 kg * 9.8 m/s² * 1340 m.

Final kinetic energy (KE_final) is given by 1/2 mv², where m is mass and v is velocity at landing. Thus, KE_final = 1/2 * 39.4 kg * (5.78 m/s)².

The energy lost to air friction equals the initial potential energy minus the final kinetic energy. Subtracting KE_final from PE_initial gives the amount of energy lost to air resistance.

Answer:

2.57 m/s

Explanation:

mass of Erica (ME) = 39 kg

velocity of Erica (VE) = 0 m/s (since she is at her high point of her jump, her velocity will be 0)

mass of Danny (MD) = 45 kg

velocity of Danny (VD) = 4.8 m/s

from the conservation of momentum total initial momentum is equal to the total final momentum

total initial momentum = final initial momentum

(ME)(VE) + (MD)(VD) = (ME + MD) V

where "V" is the velocity of Erica and Danny after Danny grabs her.

(39 x 0) + (45 x 4.8) = (39 + 45) x V

0 + 216 = 84 V

V = 216 / 84 = 2.57 m/s

The speed just after Danny grabs Erica is equal to 2.57 m/s.

Given the following data:

To determine their speed just after Danny grabs Erica, we would apply the law of conservation of momentum:

[tex]M_EV_E + M_DV_D = (M_E + M_D)V_{f}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]39 \times 0 + 45 \times 4.8 = (39 +45)V_f\\\\0+216=84V_f\\\\V_f = \frac{216}{84} \\\\V_f = 2.57 \;m/s[/tex]

Final speed = 2.57 m/s

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Answer:

The concepts of basic chemistry and physics teach us that the more stable a compound would be, the more difficult it will be to break that compound. On the other hand, compounds which are not stable have weak binds and can be  broken down easily.

For example, CFC's are stable compounds. It is difficult to break down these compounds and hence they exist in the environment without being broken down. They cause harmful effects on the ozone layer of the Earth.

Answer: THE SAME

Explanation: visible light is an electromagnetic wave, which has the properties of both magnet and electrical. The speed of light in air has Generally been estimated to be

299,792 kilometers per second. For for a distant start moving towards the Earth at a speed one-fourth (1/4th) of 299,792kilometers per second compared to a flashing light from the Earth,the speed of the light from the star is expected to be the same.

Answer:

The angle of the incident ray at the air-glass interface is 62.86°

Explanation:

Given that,

Refractive index of glass =1.60

Angle = 42°

We need to calculate the angle of the incident ray at glass-water interface

Using Snell's law

[tex]n_{2}\sin\theta_{2}=n_{3}\sin\theta_{3}[/tex]

Where, [tex]n_{2}[/tex] = refractive index of glass

[tex]n_{3}[/tex] = refractive index of water

[tex]\theta_{3}[/tex] = angle of refraction

Put the value into the formula

[tex]1.6\sin\theta_{2}=1.33\sin42[/tex]

[tex]\sin\theta_{2}=\dfrac{1.33\sin42}{1.6}[/tex]

[tex]\theta_{2}=\sin^{-1}(\dfrac{1.33\sin42}{1.6})[/tex]

[tex]\theta_{2}=33.8^{\circ}[/tex]

We need to calculate the angle of the incident ray at the air-glass interface

Using Snell's law

[tex]n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}[/tex]

Where, [tex]n_{1}[/tex] = refractive index of air

[tex]n_{2}[/tex] = refractive index of glass

[tex]\theta_{1}[/tex] = angle of incident

Put the value into the formula

[tex]1\sin\theta_{1}=1.6\sin33.8[/tex]

[tex]\theta=\sin^{-1}(1.6\sin33.8)[/tex]

[tex]\theta=62.86^{\circ}[/tex]

Hence, The angle of the incident ray at the air-glass interface is 62.86°

Answer:

62.8 degree

Explanation:

Let the incident ray incident at an angle [tex]\theta_1[/tex] at air glass surface.

[tex]\theta_3=42^{\circ}[/tex]=Angle of refraction when ray travel from glass to water

[tex]\theta_2=[/tex]Angle of refraction when the ray travel from air to glass

Refractive index of glass,[tex]n_2=1.6[/tex]

We know that

Refractive index of water=[tex]n_3=1.33[/tex]

Snell's law

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

Where [tex]\theta_1[/tex]=Angle of incidence

[tex]\theta_2=[/tex]Angle of refraction

[tex]n_1=[/tex]Refractive index of medium 1

[tex]n_2=[/tex]Refractive index of medium 2

When the ray travel from glass to water

[tex]n_2sin\theta_2=n_3sin\theta_3[/tex]

Where [tex]n_2=[/tex]Refractive index of medium 1(Glass)

[tex]n_3[/tex]=Refractive index of medium 2 (Water)

[tex]\theta_2=[/tex]Angle of incidence

[tex]\theta_3[/tex]=Angle of refraction

Substitute the values

[tex]1.6sin\theta_2=1.33sin42[/tex]

[tex]sin\theta_2=\frac{1.33sin42}{1.6}[/tex]

[tex]sin\theta_2=0.556[/tex]

[tex]\theta_2=sin^{-1}(0.556)=33.8^{\circ}[/tex]

Angle of refraction when ray travel from air to glass= Angle of incidence of when ray travel from glass to water

Angle of refraction when the ray travel from air to glass=33.8 degree

Refractive index of air=[tex]n_1=1[/tex]

Again apply Snell's law

[tex]n_1sin\thet_1=n_2sin\theta_2[/tex]

[tex]1\times sin\theta_1=1.6sin(33.8)[/tex]

[tex]sin\theta_1=1.6\times 0.556=0.8896[/tex]

[tex]\theta_1=sin^{-1}(0.8896)=62.8^{\circ}[/tex]

Hence, the angle of the incident ray at the air-glass interface=62.8 degree

Juanita did more work on the square box

Explanation:

According to the law of conservation of energy, the work done in lifting an object is equal to the increase in gravitational potential energy of the object. This is due to the fact that the work done on the object is converted into potential energy.

The potential energy of an object (GPE) is given by

[tex]GPE=mgh[/tex]

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object

In this problem, we have two objects:

- The roud box is lifted and its GPE increases by 50 J --> this means that the work done by Juanita on the box is 50 J

- Thr square box is lifted and its GPE increases by 100 J --> this means that the work done by Juanita on the box is 100 J

Therefore, Juanita did more work on the square box.

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The square box received more work by Juanita than the round box based on the increase in gravitational potential energy.

The square box received more work done by Juanita compared to the round box. When the gravitational potential energy increase for the square box is 100 J, which is greater than the 50 J increase in the round box, it indicates that Juanita did more work on the square box.

Answer:985.71 Hz

Explanation:

Given

Speed of ambulance (source) [tex]v_s=108\ kmph\approx 30\ m/s[/tex]

Original Frequency of sound wave [tex]f=900\ Hz[/tex]

speed of sound waves [tex]v=345\ m/s[/tex]

According to Doppler apparent frequency will be different to original frequency when speed of source is moving w.r.t to observer.

[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]

where f'=apparent frequency

f=original frequency

[tex]v_o[/tex]=observer speed

[tex]f'=900\times (\frac{345+0}{345-30})[/tex]

[tex]f'=900\times (\frac{345}{315})[/tex]

[tex]f'=985.71\ Hz[/tex]        

Final answer:

The tension in the pendulum at the lowest point is 12.1 N. Without additional information, we cannot determine the exact angle at which the pendulum reaches its highest point. The tension in the pendulum at the highest point is 4.9 N.

Explanation:

Tension in the Pendulum Cable

To determine the tension in the pendulum cable at the lowest point, we can use the formula for circular motion.

T = mg + m(v^2/r)

Where:

g is the acceleration due to gravity (9.8 m/s²)

r is the length of the pendulum

Substituting the values given, we have:

T = 0.500 kg * 9.8 m/s² + 0.500 kg * (3.40 m/s)^2 / 0.80 m

Now, we can calculate the tension:

T = 4.9 N + 7.225 N

T ≈ 12.1 N

The Angle at the Highest Point

To find the angle at the highest point, we use the principle of conservation of energy. The kinetic energy at the lowest point equals potential energy at the highest point.

mgh = 1/2 mv^2

Solving for h gives us h = v^2/(2g), and then use trigonometry to find the angle θ.

h = r(1 - cosθ)

Without sufficient information to calculate the speed of the pendulum, we must leave the answer as an open calculation.

Tension at the Highest Point

At the highest point, the tension is equal to the weight of the pendulum alone since it is momentarily at rest, and there is no centripetal force required.

T = mg

Thus, the tension at the highest point will be:

T = 0.500 kg * 9.8 m/s² = 4.9 N.

Answer:

the new volume =93,750 Liters

Explanation:

From Boyles law

P1V1 = P2V2

1.2 X 125000 = 1.6 X V2

V2 = 93,750 Liters

Answer:

after the sun sets or just as it is setting

Explanation:

a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.

The best time to find the thinnest crescent moon just after the new moon is in the early evening, just after sunset, when the moon starts to reflect sunlight towards the earth, showing the crescent shape. The appearance of the Moon's surface can vary significantly with its phase and this can be better viewed through binoculars.

To find the thinnest crescent moon just after new moon, the best time to look is usually in the early evening just after sunset. The Moon, moving eastward each day in its 30 days cycle around the Earth, moves roughly 12° in the sky each day. A day or two after the new phase, the thin crescent first appears, as we begin to see a small part of the Moon's illuminated hemisphere reflecting a little sunlight toward us.

Because the Moon is moving eastward away from the Sun, it rises later and later each day. Therefore, after a new moon, the thin crescent will be seen in the west just after sunset.

Keep in mind that the bright crescent increases in size on successive days as the Moon moves farther and farther around the sky away from the direction of the Sun.

Bear in mind that the brighter the Moon is in the night sky, the harder it is to see the faint flashes of meteors. Furthermore, as seen through a good pair of binoculars, the appearance of the Moon's surface changes dramatically with its phase, revealing more topographic details when sunlight streams in from the side, causing topographic features to cast sharp shadows.

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Answer : The energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J

Explanation :

As we are given that:

1 watt = 1 J/s

So,

100 watt = 100 J/s

Now we have to calculate the energy consumed by bulb in 24 hours.

As we know that:

1 hr = 60 min

1 min= 60 sec

So,

24 hr = 24 × 60 × 60 sec = 86400 sec

As, the energy consumed by bulb in 1 second = 100 J

So, the energy consumed by bulb in 86400 second = 86400 × 100 J

                                                                                       = 8.64 × 10⁶ J

Thus, the energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J

To calculate the energy consumed by a 100-watt electric incandescent light bulb in 24 hours, we use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s). In this case, the energy consumed is 8640000 J.

To calculate the electrical energy used by a 100-watt electric incandescent light bulb in 24 hours, we can use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s).

First, we need to convert the time from hours to seconds. There are 3600 seconds in an hour. So, 24 hours is equal to 24 x 3600 = 86400 seconds.

Now, we can substitute the values into the formula. E = 100W x 86400s = 8640000 J.

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Answer:

Distance = 26.0m Displacement = 4.0m

Explanation:

Distance specifies only how far an object has traveled while displacement is the distance traveled in a specified direction.

Total distance traveled by the object will be distance travelled through north + distance travelled through south i.e 15.0m + 11.0m = 26.0m

Displacement is gotten by using the Pythagoras theorem. Since the object traveled in the same vertical direction (15.0m through north which is upward i.e positive y direction and 11.0m through south i.e in the negative y direction), the displacement will be 15.0m - 11.0m = 4.0m

The distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m

First, we will define the terms distance and displacement

Distance is the total movement of an object without any regard to direction.

Displacement is the difference between the original and final position of a path taken by an object.

Since, the object moves 15.0 m north and then 11.0 m south,

Then,

Distance traveled = 15.0 m + 11.0 m

Distance traveled = 26.0 m

For the magnitude of the displacement,

The object moves 15.0 m north and then 11.0 m south, which is in the opposite (negative) direction

Then,

Magnitude of displacement = 15.0 m - 11.0 m

Magnitude of displacement = 4.0 m

Hence, the distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m

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Answer:

What is the coefficient of kinetic friction = 0.432

Explanation:

The detailed steps and derivation with appropriate substitution is as shown in the attached file.

This question involves the concepts of the law of conservation of energy and frictional energy.

a) The speed of the child when he reaches the bottom of the slide is "6 m/s".

b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".

a)

According to the law of conservation of energy in this situation:

Loss in Potential Energy = Gain in Kinetic Energy

[tex]mgh = \frac{1}{2}mv^2\\\\2gh=v^2\\v=\sqrt{2gh}\\\\[/tex]

where,

v = velocity = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = 6 ft = 1.83 m

Therefore,

[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]

v = 6 m/s

b)

Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:

Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy

[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]

where,

l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]

[tex]\mu_k[/tex] = coefficient of kinetic friction = ?

Therefore,

[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]

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The attached picture explains the law of conservation of energy.

Answer:

A & B are correct

Explanation:

Certain automotive sensors like crank shaft position sensor uses a.c. while most automotive components require DC to work properly through the battery which is incorporated with an alternator to keep it charged through a diode while continual supply of DC is sustained.

Technician A is incorrect as DC circuits are commonly used in automobiles, and Technician B is partially correct since some automotive sensors can generate an AC signal.

In the scenario presented, Technician A says that dc circuits are not normally used in automotive circuits, which is incorrect. DC circuits are indeed commonly used in automobiles for various functions such as starting the engine, powering lights, and running the dashboard instruments. Technician B says that certain automotive computer sensors use ac current. While most sensors in a car operate on DC current, there are some sensors, like variable reluctance sensors, that can generate an AC signal as they operate. Therefore, both statements are not entirely accurate; however, Technician B's statement has some validity regarding the existence of sensors that produce an AC signal within automotive applications.

Answer: The answer is attached

Explanation:

Final answer:

In quantum mechanics, the probability of a spin-flip and the potential for achieving a spin flip with certainty are explored through the application of magnetic fields in different directions.

Explanation:

P(a): The probability that the spin-flip would occur can be calculated using quantum mechanics. When applying magnetic fields B along x and y, the probability of finding the spin of the electron oriented in the negative z-direction after is over is determined by the transition probabilities.

P(b): The time to ensure a spin flip with certainty (P = 1) can be calculated by adjusting the duration of the magnetic field application. By manipulating the duration, the spin can be controlled to achieve a certain outcome at a specific time.

Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

[tex]294.3sin\theta = 20[/tex]

[tex]sin\theta = 20/294.3 = 0.068[/tex]

[tex]\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0[/tex]

So the ramp cannot be larger than 3.9 degrees

Answer:

d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.

Explanation:

In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.

On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.

Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.

The statement is true that an astronaut would float away during a spacewalk if not tethered. The astronaut and ISS, although influencing each other gravitationally to a small extent, they would continue along their separate paths under the influence of Earth's gravity and microgravity.

This statement is true. When an astronaut steps out for a spacewalk, she and the International Space Station (ISS) are both in orbit around Earth but the astronaut will not stay in place relative to the ISS without a tether holding her to the station.

The reason for this is due to microgravity and the properties of motion in space. The astronaut and the ISS are both in free-fall around the Earth, so in the absence of other forces, they would continue along their separate paths. Without a tether, even a small force (like the push off the astronaut does to get away from the airlock) can cause her to drift away from the station.

While the ISS and the astronaut do influence each other gravitationally, this effect is extremely small compared to the force of Earth's gravity. So, without a tether, the astronaut can float away from the ISS.

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Answer:

-387883.3 m/s²

0.000286168546055 seconds

Explanation:

t = Time taken

u = Initial velocity = 370 m/s

v = Final velocity = 259 m/s

s = Displacement = 9 cm

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2[/tex]

The acceleration of the bullet is -387883.3 m/s²

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s[/tex]

The time taken is 0.000286168546055 seconds

Final answer:

The acceleration of the bullet is approximately -111 m/s^2 and the total time the bullet is in contact with the board is approximately 0.00081 seconds.

Explanation:

The acceleration of the bullet can be calculated using the equation of motion:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. In this case, the initial velocity of the bullet is 370 m/s (positive because it is in the direction of motion) and the final velocity is 259 m/s (positive because it is also in the direction of motion). The time it takes for the bullet to decelerate can be calculated using the equation:

x = ut + (1/2)at^2

where x is the displacement, u is the initial velocity, t is the time, and a is the acceleration. In this case, the displacement is the thickness of the wooden board, which is 9.0 cm (or 0.09 m). Rearranging the equation, we get:

t = (v - u) / a

Substituting the values, we have:

t = (259 - 370) / a

Combining the equations for time and acceleration, we get:

t = (0.09) / (370 - 259) = (0.09) / (111) = 0.00081 seconds.

So, the acceleration of the bullet is approximately -111 m/s^2 (negative because it is in the opposite direction of the bullet's motion) and the total time the bullet is in contact with the board is approximately 0.00081 seconds.

Answer: did you get the answer(s) by chance?

Explanation:

Answer:

The last one

Explanation:

Answer:

about 10 trillion kilometers

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

t = Seconds in one year = [tex]365.25\times 24\times 60\times 60[/tex]

1 Light year

[tex]1\ ly=ct\\\Rightarrow 1\ ly=3\times 10^8\times 365.25\times 24\times 60\times 60=9.46728\times 10^{15}\ m\\ =9.46728\times 10^{15}\times 10^{-3}\\ =9.46728\times 10^{12}\ km\approx 10\ trillion\ km[/tex]

The answer is a. about 10 trillion kilometers

Answer:

Explanation:

The electric flux through a Gaussian surface can be calculated using Gauss's law. (a) Calculate electric flux for q1 and q2, (b) Calculate electric flux for q2 and q3, (c) Calculate electric flux for all three charges.

The electric flux through a Gaussian surface can be calculated using Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

(a) To find the electric flux through the Gaussian surface enclosing only charges q1 and q2, we need to calculate the net charge enclosed by the surface, which is the sum of the two charges. Then, we divide this sum by the permittivity of free space to obtain the electric flux.

(b) Following the same procedure, we can find the electric flux through the Gaussian surface enclosing only charges q2 and q3.

(c) To find the electric flux through the Gaussian surface enclosing all three charges, we calculate the net charge enclosed by the surface, which is the sum of all three charges. Again, we divide this sum by the permittivity of free space to obtain the electric flux.

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Answer:

A) An interference pattern

Explanation:

the two slit experiment is key to understand the microscopic world. The wave-like properties of light were demonstrated by the famous experiment first performed by Thomas Young in the early nineteenth century. In original experiment, a point source of light illuminates two narrow adjacent slits in a screen, and the image of the light that passes through the slits is observed on a second screen.

Key Points

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching  the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

[tex]F\propto\frac{d}{dt} (p)[/tex]

[tex]\Rightarrow F=\frac{d}{dt} (m.v)[/tex]

since mass is constant

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

when [tex]dt=10t[/tex]

then,

[tex]F'=m.\frac{v}{10\times t}[/tex]

[tex]F'=\frac{1}{10} \times \frac{m.v}{t}[/tex]

[tex]F'=\frac{F}{10}[/tex] the body will experience the tenth part of the maximum force.

where:

[tex]\frac{d}{dt} =[/tex] represents the rate of change in dependent quantity with respect to time

[tex]p=[/tex] momentum

[tex]m=[/tex] mass of the person jumping

[tex]v=[/tex] velocity of the body while hitting the ground.

Final answer:

The detailed response covers the maximum height reached by the egg, the time it takes to hit the ground, and its velocity just before landing.a)60.0m,b)2.45 seconds,c)-6.0m/s.

Explanation:

a) Maximum Height: The maximum height above the ground reached by the egg can be calculated using the kinematic equation. It will be twice the initial height. In this case, it would be 60.0m.

b) Time to Hit Ground: You can determine the time it takes for the egg to hit the ground by using the kinematic equation for vertical motion. The time would be approximately 2.45 seconds after its release.

c) Velocity Before Landing: The velocity of the egg immediately before hitting the ground would be the same as the initial velocity when it was thrown, but in the opposite direction, which is -6.0m/s.

Answer:

Explanation:

Given

Speed of ball [tex]u=3.2\ m/s[/tex]

Plane is inclined at an angle [tex]20^{\circ}[/tex]

To win the Game we need to hit the target at [tex]x=2.4\ m[/tex] away

Launch angle of ball [tex]\theta [/tex]

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is [tex]g\sin 20[/tex]

Range of Projectile is given by

[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]

for [tex]R=2.4\ m[/tex]

[tex]2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}[/tex]

[tex]\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}[/tex]

[tex]\sin 2\theta =0.7855[/tex]

[tex]2\theta =51.77[/tex]

[tex]\theta =25.88^{\circ}[/tex]

so ball must be launched at an angle of [tex]25.88^{\circ}[/tex]

To answer this, we use projectile motion principles with an equation specifically structured for the problem. Inserting the given values into the equation derived from the horizontal and vertical equations of motion, we can find the required launching angle to hit the target.

To solve this problem, we can apply the concepts found in projectile motion physics. Given the initial speed or velocity (3.2 m/s) and the horizontal distance of the target (2.4 m) we can find the necessary angle (theta) to hit the target. The angle θ can be given by the equation of motion for a projectile which can be derived from the horizontal and vertical equations of motion, θ = atan[(vf² ± sqrt(vf⁴ - g*(g*x² + 2*y*vf²)) / (g*x)] where g = 9.8 m/s² is the acceleration due to gravity, x = 2.4 m is the horizontal distance, y = 0 (height difference), and vf = 3.2 m/s is the final velocity, the speed at which the ball is launched.

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