A bicyclist in the Tour de France has a speed of 27.0 miles per hour (mi/h) on a flat section of the road. What is this speed in (a) kilometers per hour (km/h), and (b) meters per second (m/s)?

Answer:

Explanation:

Given

Race length [tex]L=100\ m[/tex]

Runner reaches maximum velocity in [tex]t_1=2.4\ s[/tex]

[tex]v_{max}=11.3\ m/s[/tex]

distance traveled during this time

[tex]v=u+at[/tex]

[tex]11.3=0+a\times 2.4[/tex]

[tex]a=\frac{11.3}{2.4}=4.708\ m/s^2[/tex]

distance traveled in this time

[tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]s=0+0.5\times 4.708\times (2.44)^2[/tex]

[tex]s=14.014\ m[/tex]

Remaining distance will be traveled with [tex]v_{max}[/tex]

remaining distance [tex]d=100-14.014=85.98\ m[/tex]

time taken [tex]t_2=\frac{d}{v_{max}}=\frac{85.98}{11.3}=7.609\ s[/tex]

total time [tex]t=t_1+t_2[/tex]

[tex]t=2.44+7.609=10.049\ s[/tex]        

The planet will take more time to orbit the Sun.

Explanation:

According to the Kepler's law of orbital motion, a planet which is far away from the Sun experiences a lower  gravitational pull towards the Sun, thus it will move with a lower speed in its orbit. Thus the farther planet takes more time to orbit the Sun compared to the planets closer to the Sun.

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Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

Answer:

Explanation:

Given

For single guitar noise level [tex]SL=70\ dB[/tex]

Intensity of sound

Sound level[tex]=10\log (\frac{I}{I_0})[/tex]

where I=Intensity of sound Produced

[tex]I_0=[/tex]Human threshold frequency [tex](10^{-12}\ W/m^2)[/tex]

[tex]70=10\log (\frac{I}{10^{-12}})[/tex]

[tex]I=10^{-12}\times 10^7[/tex]

[tex]I=10^{-5}\ W/m^2[/tex]

For 2 guitars

[tex]I'=2I=2\times 10^{-5}\ W/m^2[/tex]

[tex]SL=10\log (\frac{I}{I_0})[/tex]

[tex]SL=10\log (\frac{2\times 10^{-5}}{10^{-12}})[/tex]

[tex]SL=10(7+\log (2))[/tex]

[tex]SL=73.01\ dB[/tex]

Answer:

Average recoil force experienced by machine will be 200 N

Explanation:

We have give mass of each bullet m = 50 gram = 0.05 kg

There are 4 bullets

So mass of 4 bullets = 4×0.05 = 0.2 kg

Initial speed of the bullet u = 0 m/sec

And final speed of the bullet v = 1000 m/sec

So change in momentum [tex]P=m(v-u)=0.2\times (1000-0)=200kgm/sec[/tex]

Time is given per second so t = 1 sec

We know that force is equal to rate of change of momentum

So force will be equal to [tex]F=\frac{200}{1}=200N[/tex]

So average recoil force experienced by machine will be 200 N

The average recoil force experienced by the machine gun is 100N.

The impulse-momentum theorem states that the impulse applied to an object will be equal to the change in its momentum.

Force (F) * change in time (Δt) = change in momentum = mass (m) * velocity (v)

FΔt = mv

m = 50 g = 0.05 kg

F = mv / Δt

F = (0.05kg * 1000 m/s * 4 bullets)/ 1 second

F = 100 N

The average recoil force experienced by the machine gun is 100N.

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Answer: 0.44831m

Explanation:

Unwanted horizontal spring oscillator=2.65kg

Spring constant =38.5N/M

Speed=2.92m/s

Amplitude of oscillation=?

Potential energy=m*v2/2

=2.65*2.92/2

=3.8695J

Potential energy=kinetic energy

Potential energy=1/2kx^2

3.869=1/2*38.5*x^2

3.869=19.25x^2

Dividing both sides by 19.25

3.869/19.25=x^2

So therefore, x^2=√0.200987

x=0.44831m

Answer:

The electric flux through the surface is equal to 3.878 x 10³ Nm²/C

The field distance r is equal to half the length of each side of the cube.

From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C

The total flux area available to this electric field is 6x32cm²

Explanation:

The full solution can be found in the attachment below.

Thank you for reading this post and I hope it is helpful to you.

The force of attraction between sodium and chloride ions in a salt crystal can be calculated using Coulomb's law, considering isotropic forces and applying the Madelung constant.

The force of attraction between the sodium and chloride ions in a salt crystal can be calculated using Coulomb's law, which involves the product of the charges divided by the distance between them squared. The force is proportional to the ionic charges and inversely proportional to the square of the distance between the ions. In a sodium chloride (NaCl) crystal, the resulting ions (Na+ and Cl¯) produce isotropic forces, which means the force is the same in all directions.

Given the distance mentioned in your question (2.82 Angstroms), you can use this along with the known charges of sodium and chloride ions to calculate the force. However, please note that this calculation needs to take into account the Madelung constant, which is a factor for considering the interaction of a sodium ion with all the nearby chloride and sodium ions in a tightly arranged three-dimensional lattice structure. For a NaCl crystal, the Madelung constant is approximately 1.75.

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Answer:

44.21053 m

279.69925 seconds

Explanation:

Time taken by the squirrel to reach the finish line

[tex]\dfrac{1200}{3.5}=342.85714\ seconds[/tex]

Time taken by the horse to cover 0.96 km

[tex]\dfrac{960}{19}=50.52631\ seconds[/tex]

Time taken by the horse to cover 1.2 km

[tex]\dfrac{1200}{19}=63.15789\ seconds[/tex]

Let the distance that the squirrel is from the finish line when the horse resumes the race be x.

During this time the horse also reaches the finish line.

We deduce that

The time taken by the horse from one stop = Time taken by squirrel before x distance from finish line.

[tex]3.5\times (63.15789-50.52631)=x\\\Rightarrow x=44.21053\ m[/tex]

The squirrel is 44.21053 m from the finish line when the horse resumes the race.

Duration is given by

[tex](342.85714-63.15789)=279.69925\ s[/tex]

The duration is 279.69925 seconds

When the horse resumed the race, the squirrel was 240m from the finish line. The horse was stationary for approximately 68.57 seconds.

The subject of this question is Physics, specifically it deals with the concept of speed and distance. The grade level is high school as it involves basic kinematics.

(a) The horse has traveled 0.96 km, which is 960 m. It leaves 240 m for the rest of the journey. When the horse stopped, the squirrel kept moving. By the time the horse started again, the squirrel must have traveled more than 240 m. We know the horse and the squirrel finished the race at the same time, hence, they must have started the remaining journey at the same time. So, the distance from the finish line where the squirrel was when the horse began to run again equals to the rest of the horse's journey, which is 240m.

(b) To figure out how long the horse was stationary we need to calculate how long it took the squirrel to cover the distance. This is done by dividing the distance the squirrel traveled (240m) by its speed (3.5 m/s). Hence, time = distance / speed = 240m / 3.5 m/s = 68.57 seconds.

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Answer: K =24 psi

Explanation:

Given: Standard deviation =3psi

Internal pressure strength =157psi

Number of random bottle =n=64

K= 3 × square root of 64

K= 3×8=24 psi

If mean internal pressure K fall below K,

157-1.3=155.7psi

At 2%:

0.16×64 = 10.24

Final answer:

The value of K is 150.835 psi.

Explanation:

To find the value of K, we need to calculate the critical value of the sample mean that separates the lower 2% of the distribution from the upper 98%. First, we need to determine the z-score corresponding to the desired probability of 2%. We can find this value using a standard normal distribution table or a calculator. The z-score for a 2% probability is approximately -2.055. Next, we can calculate the value of K by multiplying the z-score by the standard deviation and adding it to the mean:

K = 157 + (-2.055 * 3) = 150.835 psi

Therefore, if the sample mean falls below 150.835 psi, the bottler will conclude the vendor's claim about the mean internal pressure strength to be false.

Answer:

The stitches and dimples around a baseball  and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.

Explanation:

The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.  

A smooth ball with no stitches or dimples has more air drag that opposes the motion.

A golf ball is smooth ball with dimples to create a thin turbulent boundary layer of air that clings to the ball's surface. This allows the smoothly flowing air to follow the ball's surface a little farther around the back side of the ball, thereby decreasing the size of the wake, and allowing the ball to travel more easily.

Positively charged particles always follow the path set by electric field lines. This is because electric field lines are defined by the path that the positive test charges would naturally travel.

The options that best complete the statement are 'always' and 'electric field lines are defined by the path positive test charges travel.' This is because the trajectory of positively charged particle trajectories always follow the direction of electric field lines. The electric field lines are essentially mapped out pathways that positive test charges would naturally follow due to the electric force acting upon them. Hence, a positively charged particle will always move along these field lines, from a region of higher potential to a region of lower potential.

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Answer:

The acceleration of the sprinter is 1.4 m/s²

Explanation:

Hi there!

The equation of position of the sprinter is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the sprinter at a time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:

x = 1/2 · a · t²

At t = 9.9 s, x = 71 m

71 m = 1/2 · a · (9.9 s)²

2 · 71 m / (9.9 s)² = a

a = 1.4 m/s²

The acceleration of the sprinter is 1.4 m/s²

The sprinter's acceleration can calculated by solving two equations from physics: the first formula of motion (final velocity equals initial velocity plus acceleration times time) and the formula for final velocity (distance divided by time). The result is an approximate acceleration of 0.72 m/s^2.

The question requires the use of kinematics, a topic in physics. The formula to calculate uniform acceleration is given by a = 2*(final velocity - initial velocity) / time. Here, final velocity is the velocity achieved after 71m, initial velocity is 0 (standing start), and time is 9.9s. However, we are not given final velocity. We will have to use the first formula of motion, v = u + at, in which v is the final velocity, u is the initial velocity, a is acceleration and t is time. Simultaneously, we have v = d/t where d is distance and t is time. Solving these two equations will give the acceleration (a).

From the first formula of motion and the equation v = d/t, it's clear the sprinter's final velocity (when he/she stops accelerating) is 71m/9.9s which equals approximately 7.17 m/s. Substituting into the first formula of motion gives 7.17 m/s = 0 + a*9.9s which simplifies to a = 7.17 m/s / 9.9s. This gives us an acceleration of approximately 0.72 m/s2.

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Answer:

Part A  B.  Part B C.

Explanation:

A) The elementary charge e (without sign) is equal to the charge of one electron (with negative sign) or to the charge of one proton (with positive sign), so the proton must have a total charge of +e.

if u = +2/3 e and d= -1/3 e, we need a combination which sum gives +3/3 e.

Combination A adds to 6/3e, so it is not possible. C adds to zero, and D gives a negative result.

The only remaining choice is udd:

uud ⇒ +2/3 e + 2/3 e -1/3 e = +3/3 e = +e

So, the statement B is true.

B) As the neutron has no net charge, we need to find a combination which sum adds to zero.

So, A and D are not possible, as they are combinations of the same type of quarks, so the sum is either positive or negative, but not zero.

uud gives +e (we chose it to make a proton in part A), so the only remaining choice is udd:

udd⇒ +2/3 e -1/3 e - 1/3 e = 0

So the statement C is true.

To solve this problem we will use the concepts of the moment of rotational inertia, angular acceleration and the expression of angular velocity.

The rotational inertia is expressed as follows:

[tex]I = \sum mr^2[/tex]

Here,

m = Mass of the object

r = Distance from the rotational axis

The rotational acceleration in terms of translational acceleration is

[tex]\alpha = \frac{a}{R}[/tex]

Here,

a = Acceleration

R = Radius of the circular path of the object

The expression for the rotational speed of the object is

[tex]\omega = \frac{\Delta \theta}{\Delta t}[/tex]

Here,

[tex]\Delta \theta[/tex] is the angular displacement of the object

The explanation by which when climbing a mountain uphill is changed to a larger pinion, is because it produces a greater torque but it is necessary to make more pedaling to be able to travel the same distance. Basically every turn results in less rotations of the rear wheel. Said energy that was previously used to move the rotation of the wheel is now distributed in more turns of the pedal. Therefore option a and c are correct.

This would indicate that the correct option is D.

Shifting to a larger-diameter gear in a 10-speed bicycle allows for an increased force exerted by the chain on the gear and a greater torque on the wheel, making it easier to ride uphill.

When riding a 10-speed bicycle up a hill, shifting to a larger-diameter gear attached to the back wheel is preferred compared to a smaller gear because it increases the force exerted by the chain on the gear and allows the cyclist to exert a greater torque on the wheel.

By shifting to a larger gear, the chain wraps around a larger portion of the gear's circumference, resulting in a greater force being applied to rotate the wheel. This increased force allows the cyclist to overcome gravity more efficiently and climb the hill with less effort.

The larger gear also allows the cyclist to apply a greater torque to the wheel. Torque is the rotational equivalent of force and represents the ability to turn the wheel. With a larger gear, the cyclist can pedal with more force and generate a larger torque, which is necessary to propel the bike up the hill.

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Answer:

0.71 cm

Explanation:

[tex]q_1=75.0nC=75\times 10^{-9}C[/tex]

[tex]1nC=10^{-9}C[/tex]

[tex]q_2=75.0nC=75\times 10^{-9}C[/tex]

Force between two charges=1 N

Coulomb's law of force

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Where k=[tex]9\times 10^9Nm^2/C^2[/tex]

Using the formula

[tex]1=\frac{9\times 10^9\times 75\times 10^{-9}\times 75\times 10^{-9}}{r^2}[/tex]

[tex]r^2=0.50625\times 10^{-4}[/tex]m

[tex]r=\sqrt{0.50625\times 10^{-4}}=0.71\times 10^{-2} m[/tex]

[tex]r=0.71\times 10^{-2}\times 10^{2}=0.71\times 10^{-2+2}=0.71\times 10^0=0.71 cm[/tex]

Using formula

[tex]1 m=10^2cm, a^x\cdot a^y=a^{x+y}, a^0=1[/tex]

Hence, the distance between two charges =r=0.71 cm

Answer:

At least 6 N, assuming that the rope has a length of more than 2 meters.

Explanation:

In general, if it takes a force of [tex]F(h)[/tex] newtons to lift an object at height [tex]h[/tex], the work done lifting the object from [tex]h = a[/tex] to [tex]h = b[/tex] can be found using the definite integral about [tex]h[/tex]:

[tex]\displaystyle W = \int \limits_{a}^{b} F(h)\, dh[/tex].

If the value of [tex]F(h)[/tex] is a constant [tex]m \cdot g[/tex] regardless of height [tex]h[/tex], then the result of the integral would be

[tex]\displaystyle \int \limits_{a}^{b} (m \cdot g)\, dh = \left[m \, g\, h \right]_a^b = m\cdot g \, (b - a)[/tex].

However, in this case the value of [tex]F(h)[/tex] does depend on the value of [tex]h[/tex].

In other words, [tex]F(h) = 3\; h[/tex] where [tex]F[/tex] is in Newtons and [tex]h[/tex] is in meters.

Evaluate the integral:

[tex]\begin{aligned} W &= \int \limits_{a}^{b} F(h)\, dh \cr &= \int \limits_{0}^{2}3\, h \, dh && \text{Apply the power rule.}\cr &= \left[\frac{3}{2}\,h^2\right]_{0}^{2}\cr &= \frac{3}{2} \times 2^2 \cr &= 6\end{aligned}[/tex].

The work required to raise one end of the rope to a height of 2 meters with an applied force of 3 N is 6 joules.

We can calculate the work required to raise one end of the rope to a height of 2 meters as follows:

[tex] W = F*d*cos(\theta) [/tex]

Where:

F: is the force exerted = 3 N (newton: unit of force)

d: is the displacement = 2 m

θ: is the angle between the applied force and the displacement

Since the force and the displacement are in the same direction, θ = 0, so:

[tex] W = F*d*cos(\theta) = Fdcos(0) = F*d [/tex]

Hence, the work done is:

[tex] W = 3 N*2m = 6 J [/tex]

Therefore, it is required 6 J of work.

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Answer

given,

mass of block 1, m = 0.3 Kg

speed of block 1, v = 4 m/s

mass of second block,M = 2 Kg

initial speed of block = 0 m/s

spring constant, k = 100 N/m

a) for block 1

  linear momentum before collision

   P₁ = m v = 0.3 x 4 = 1.2 Kg.m/s

  Kinetic energy

   [tex]KE_1 = \dfrac{1}{2}mv^2[/tex]

   [tex]KE_1 = \dfrac{1}{2}\times 0.3\times 4^2[/tex]

   [tex]KE_1 =2.4\ J[/tex]

b) After impact

  final velocity calculation

 using conservation of momentum

  m v  = (m + M )v_f

   0.3 x 4 = 2.3 x v_f

    v_f = 0.522 m/s

   Linear momentum

   P₂ = (m+M) v_f

   P₂ = 1.5 x 0.522

   P₂ = 0.783 kg.m/s

   Kinetic energy

   [tex]KE_2= \dfrac{1}{2}(M+m)v^2[/tex]

   [tex]KE_2= \dfrac{1}{2}\times 2.3\times 0.522^2[/tex]

   [tex]KE_2=0.313\ J[/tex]

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

Final answer:

The question requires applying Coulomb's law to calculate the net electric force on particle C due to particles A and B, and summing the vector forces to find the resultant.

Explanation:

The question involves finding the net electric force on particle C, which is at the position (0, 3.00 m) in the presence of particles A and B with given charges and positions. The situation can be analyzed using Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

We would sum the vector forces from both charges on C to find the resultant force. This is a classical physics problem typically addressed in high school or introductory college physics courses.

Answer:

B (z) = 2,467 10⁻¹¹ cos (1,486 10⁸ x - 5.571 10¹⁵  t )

Explanation:

An electromagnetic wave is a wave that is sustained in the perpendicular fluctuations of the electric and magnetic fields, the equation of the wave is

             E (y) = Eo cos (kx –wt)

             B (z) = Bo cos (kx-wt)

Let's look for the terms to build these equations. The speed of the wave is given by

            c = λ f

The frequency and period are related

           f = 1 / T

Let's start by applying this equation our case

             f = c /λ

             f = 3 10⁸/225 10⁻⁹

             f = 1.33 10¹⁵ Hz

The angular velocity and the wave number are

              w = 2π f

              k = 2π /λ

              w = 2π 1.33 10¹⁵ = 8.38 10¹⁵ rad / s

              k = 2π / 225 10⁻⁹ = 2.79 10⁷ m⁻¹

It indicates that the period increases by a factor of 1.5, let's look for the new frequency

                  T = 1.5 T₀

                  f = 1 / T

                  f = 1 / 1.5T₀

                  f = 1 / 1.5 f₀

                  f = 1 / 1.5 1.33 10¹⁵ = 8.87 10¹⁴ Hz

                  c = λ f

                 λ = c / f

                 λ = 3 10⁸ / 8.87 10¹⁴ = 4,229 10⁻⁸ m

Let's find the new w and k

                w = 2π f

                w = 2π 8.87 10¹⁴ = 5.571 10¹⁵ rad/s

                 k = 2π / λ

                 k = 2π / 4,229 10⁻⁸ = 1,486 10⁸ cm⁻¹

We use the relationship that the fields are in phase

                 c = E₀ / B₀

                 B₀ = E₀ / c

                 B₀ = 7.4 10⁻³ / 3 10 ⁸ = 2.467 10⁻¹¹ T

With these values ​​we can build the equation of the magnetic field

               B (z) = 2,467 10⁻¹¹ cos (1,486 10⁸ x - 5.571 10¹⁵  t )

Answer:

The wavelength of the electron is [tex]\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]

Explanation:

Given that,

Temperature = 300 K

We know that,

The energy of free electron is

[tex]E=\dfrac{(\hbar)^2k^2}{2m}[/tex]

[tex]k=\dfrac{\sqrt{2mE}}{\hbar}[/tex]

Where, k = wave number

The momentum of the electron is

[tex]p=\hbar k[/tex]

Th effective mass is

[tex]m=am_{0}[/tex]

We need to calculate the wavelength of the electron

Using formula of wave number

[tex]k=\dfrac{2\pi}{\lambda}[/tex]

[tex]\lambda=\dfrac{2\pi}{k}[/tex]

Put the value of k

[tex]\lambda=\dfrac{2\pi}{\dfrac{\sqrt{2mE}}{\hbar}}[/tex]

[tex]\lambda=\dfrac{h}{\sqrt{2am_{0}E}}[/tex]

We know that,

Thermal energy of electron

[tex]E=3kT[/tex]

The de Broglie wavelength of the electron is

[tex]\lambda=\dfrac{h}{\sqrt{2am_{0}\times3kT}}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times3\times300\times1.380\times10^{-23}\times a}}[/tex]

[tex]\lambda=\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]

Hence, The wavelength of the electron is [tex]\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]

The de Broglie wavelength of an electron can be calculated using de Broglie's equation, λ = h/p. To find the wavelength for an electron in a semiconductor at 300 K, one needs to know the electron's velocity, which is related to its kinetic energy at that temperature. Precise calculation would require details about the effective electron mass in the semiconductor.

The de Broglie wavelength λ of a particle can be determined using de Broglie's equation, λ = h/p, where 'h' is Planck's constant (6.626 × 10-34 m2 kg/s) and 'p' is the momentum of the particle. The momentum of an electron moving at thermal velocities in a semiconductor at room temperature (T = 300 K) can be approximated using the formula p = mv, where 'm' is the mass of an electron (9.11 × 10-31 kg) and 'v' is the velocity. The average thermal velocity can be obtained from the kinetic theory of gases, which states that the average kinetic energy (KE) of a particle is ⅓kT, where 'k' is the Boltzmann constant (1.38 × 10-23 J/K) and 'T' is the temperature in kelvins. Therefore, v can be calculated as √(3kT/m). Substituting this velocity into the momentum formula and then into de Broglie's equation yields the de Broglie wavelength of the electron.

However, a precise value for the de Broglie wavelength of an electron in a semiconductor at T = 300 K would require additional information not provided in the question, such as the effective mass of the electron in the semiconductor, which can differ from the free electron mass due to the crystal structure's influence.

Answer:

D. waning quarter

Explanation:

The Moon is a natural satellite of Earth and reflects the light of Sun to become visible from the Earth. It shows various phases during its revolution around the Earth depending on how much part of the lit up portion of the Moon is towards Earth. Just like Moon shows phases, if we go on Moon the Earth will show various phases.

When the Moon will be in waxing quarter phase i.e. as seen from Earth, we will see a semicircular Moon with its right side lit up. At the same time if someone from the Moon sees Earth, the Earth will show a waning quarter phase. It will be seen as semicircular but left side will be lit up.

If you are an observer standing on the Moon looking back at Earth during the waxing quarter phase of the Moon, you would see a waning quarter phase of the Earth. Thus, the correct option is D.

If you are an observer standing on the Moon looking back at Earth, the phase of the Earth that you would see depends on the relative positions of the Moon, Earth, and Sun. In the waxing quarter phase of the Moon, you would see a waning quarter phase of the Earth. This is because when the Moon is in the waxing quarter phase, the illuminated portion of the Moon is on the right side, meaning the Earth would be on the left side of the Moon and in the waning quarter phase.

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Answer:

Part A:

[tex]E=127500N/C\\E=1.275*10^{5} N/C[/tex]

Part B:

Option B (Towards the South)

Explanation:

Part A:

Magnitude if electric field E:

E=Force/charge

Force=2.04×10−14 N

Charge=1.6×10−19 C

[tex]E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C[/tex]

Part B:

Option B (Towards the South)

As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.

Final answer:

The correct option is (b) toward the south. The magnitude of the electric field is 1.275 N/C, and the direction of the electric field is toward the south, as it attracts the positively charged proton in that direction.

Explanation:

To determine the magnitude of the electric field experienced by a proton, we use the formula for electric force: F = Eq, where F is the electric force, E is the electric field strength, and q is the charge of the particle.

Given that the force experienced by the proton is 2.04×10⁻¹⁴ N and the charge of a proton (q) is approximately 1.6×10⁻¹⁹ C, we can rearrange the formula to solve for E: E = F/q. Substituting the values, we obtain E = (2.04×10⁻¹⁴ N) / (1.6×10⁻¹⁹ C) = 1.275 N/C. Therefore, the magnitude of the electric field is 1.275 N/C.

Regarding the direction of the electric field, it is crucial to understand that electric fields exert forces on positive charges away from the source of the field and toward negative charges or areas of lower potential.

Since the proton is positively charged and experiences a force toward the south, the electric field's direction must be toward the south to attract the proton in that direction. So, the correct choice for the direction of the electric field is (b) toward the south.

The unit vector that has the same direction as the given vector [tex]\(-3i + 2j - k\) is \(\frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\).[/tex]

To find a unit vector in the same direction as the given vector [tex]\(-3i + 2j - k\)[/tex], we first need to calculate the magnitude of the given vector and then divide each component by that magnitude.

The magnitude of the given vector is given by the formula:

[tex]\[|v| = \sqrt{(-3)^2 + (2)^2 + (-1)^2} = \sqrt{14}\][/tex]

Now, we can find the unit vector [tex]\(\hat{u}\)[/tex] by dividing each component of the given vector by its magnitude:

[tex]\[\hat{u} = \frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\]\[\hat{u} = \frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\][/tex]

So, the unit vector that has the same direction as the given vector [tex]\(-3i + 2j - k\) is \(\frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\).[/tex]

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To find a unit vector with the same direction as the given vector -3i + 2j - k, divide the given vector by its magnitude.

To find a unit vector with the same direction as the given vector -3i + 2j - k, we need to divide the given vector by its magnitude. The magnitude of the given vector is the square root of the sum of the squares of its components, which is sqrt((-3)^2 + 2^2 + (-1)^2) = sqrt(14).

So, the unit vector with the same direction as the given vector is (-3i + 2j - k) / sqrt(14).

Final answer:

The volume of the bag at the new pressure and temperature is 0.991 L.

Explanation:

To solve this problem, we can use the combined gas law:

P1V1/T1 = P2V2/T2

Where:

P1 = 795 torr (initial pressure)

V1 = 976 mL (initial volume, convert to L: 976 mL / 1000 = 0.976 L)

T1 = 25°C (initial temperature, convert to Kelvin: 25°C + 273 = 298 K)

P2 = 553 torr (final pressure)

T2 = 10°C (final temperature, convert to Kelvin: 10°C + 273 = 283 K)

Now we can plug in these values into the equation:

P1V1/T1 = P2V2/T2

(795 torr)(0.976 L)/(298 K) = (553 torr)(V2)/(283 K)

Solving for V2, we get:

V2 = (795 torr)(0.976 L)(283 K) / (553 torr)(298 K) = 0.991 L

Therefore, the volume of the bag at the new pressure and temperature is approximately 0.991 L.

Answer:

0.0005217 m/s or 5.217×10⁻⁴ m/s

Explanation:

Speed: This can be defined as the ratio of the distance covered by a body to the time taken to cover that distance. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude only.

Mathematically, speed is expressed as

S = d/t ......................................................... Equation 1

Where S = speed, d = total distance, t = time taken to cover the distance

Given: d = 120 mm = (120/1000) m = 0.12 m, t = 230 s.

Substituting into equation 1

S = 0.12/230

S = 0.0005217 m/s

Hence the speed of the security guard = 0.0005217 m/s or 5.217×10⁻⁴ m/s

The speed of the security guard is 0.52 mm/ss.

To find the speed of the security guard, we need to divide the distance traveled by the time taken. In this case, the distance is 120 mm and the time is 230 ss. The formula to calculate speed is:

Speed = Distance / Time

Substituting the given values:

Speed = 120 mm / 230 ss

Simplifying the expression, we get:

Speed = 0.52 mm/ss

Therefore, the speed of the security guard is 0.52 mm/ss.

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Answer:

 λ = 381 nm

Explanation:

given,

width of slit,a = 0.580 x 10⁻³ m

distance of the screen from the slit, D = 1.56 m

width of central maximum,y = 2.05 x 10⁻³ m

Distance of the edge of central maximum from center of the central maximum y ‘ = y/ 2  

                      Y ‘ = 1.025 x  10⁻³ m

for first maximum

d sin θ = λ

sin θ = tan θ = y'/D

[tex]\lambda = \dfrac{y'a}{D}[/tex]

[tex]\lambda = \dfrac{1.025\times 10^{-3}\times 0.580\times 10^{-3}}{1.56}[/tex]

 λ = 381 x 10⁻⁹ m

 λ = 381 nm

wavelength of light is equal to 381 nm.

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. [tex]x = 6 * 3.2^2 = 61.44 m[/tex]

b) at t = 3.2 + Δt. [tex]x = 6*(3.2 + \Delta t)^2[/tex]

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

[tex]v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}[/tex]

[tex]v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}[/tex]

[tex]v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}[/tex]

[tex]v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}[/tex]

[tex]v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}[/tex]

[tex] v = 38.4 + \Delta t[/tex]

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

a. The position of box at t = 3.20s is 61.44 meters

b.  The position at t = 3.20+ Δt is,  [tex]x(3.20+ \Delta t)=6*(3.20+ \Delta t)^{2}[/tex]

c. The velocity is 38.4 meter per second.

The position of box is given by as a function shown below,

                     [tex]x(t)=6t^{2}[/tex]

where x is in meters and t is in seconds.

a. The position of box at t = 3.20s is,

              [tex]x(3.2)=6*(3.2)^{2}\\ \\x(3.2)=61.44m[/tex]

b. The position at t = 3.20+ Δt,

             [tex]x(3.20+ \Delta t)=6*(3.20+ \Delta t)^{2}[/tex]

c.  We have to find

                          [tex]\frac{\Delta x}{\Delta t}=\frac{x(3.20+\Delta t)-x(3.2)}{\Delta t}\\\\\frac{\Delta x}{\Delta t}=\frac{6[(3.2)^{2}+(\Delta t)^{2}+6.4\Delta t - (3.2)^{2} ]}{\Delta t} \\\\\frac{\Delta x}{\Delta t}=6 \Delta t +38.4[/tex]

When [tex]\Delta t[/tex] approaches to zero.

 Velocity =    [tex]\frac{\Delta x}{\Delta t}=38.4 m/s[/tex]

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To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,

[tex]m_1 = 1300kg[/tex]

[tex]m_2 = 2400kg[/tex]

[tex]u_1 = 12m/s i[/tex]

[tex]u_2 = 18m/s j[/tex]

Using conservation of momentum,

[tex]m_1u_1+m_2u_2 = (m_1+m_2)v_f[/tex]

[tex]1300*12i-2400*18j = (1300+2400)v_f[/tex]

Solving for [tex]v_f[/tex]

[tex]v_f = 4.2162i-11.6756j[/tex]

Using the properties of vectors to find the magnitude we have,

[tex]|v| = \sqrt{(4.2162^2)+(-11.6756)^2}[/tex]

[tex]|v| = 12.4135m/s[/tex]

Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s

For the two cars the final velocity of the wreckage of the two cars immediately after the collision is 12.4135 m/s.

Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

The mass of the car one is 1300 kg and the mass of the second car is 2400  kg.

The initial velocity of the first car is 12 m/s in the positive x direction, and that of the second car is 18 m/s in the positive y direction.

Final velocity of the two cars after the collision is equal. Thus, the final velocity of the wreckage of the two cars immediately after the collision by the law of conservation can be given as,

[tex](1300)12\hat i-(2400)18\hat j=(1300+2400)v\\v=4.2162\hat i-11.6756\hat j[/tex]

Solve the above equation further using the property of vectors as,

[tex]v=\sqrt{(4.2162)^2+(-11.6756)^2}\\v=12.4135\rm m/s[/tex]

Hence, For the two cars the final velocity of the wreckage of the two cars immediately after the collision is 12.4135 m/s.

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Answer:

The woman's distance from the right end is 1.6m = (8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.

Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.

Theses are the conditions for static equilibrium.

Explanation:

The step by step solution can be found in the attachment below.

Thank you for reading this solution and I hope it is helpful to you.

The woman is standing 6.4 meters from the left end of the board. Since the board is 8 meters long, she is standing [tex]\( 8 \, \text{m} - 6.4 \, \text{m} = 1.6 \, \text{m} \)[/tex] from the right end of the board.

To solve this problem, we need to apply the principles of static equilibrium to the board. The board is in equilibrium because it is not moving, which means the sum of the forces acting on it must be zero, and the sum of the torques (or moments) about any point must also be zero.

The total weight of the board and the woman is [tex]\( 500 \, \text{N} + 100 \, \text{N} = 600 \, \text{N} \)[/tex]. This total weight is balanced by the support forces at the ends of the board. Therefore, the sum of the support forces is equal to the total weight:

[tex]\[ F_L + F_R = 600 \, \text{N} \][/tex]

Substituting [tex]\( F_R = 3F_L \)[/tex] into the equation, we get:

[tex]\[ F_L + 3F_L = 600 \, \text{N} \] \[ 4F_L = 600 \, \text{N} \] \[ F_L = \frac{600 \, \text{N}}{4} \] \[ F_L = 150 \, \text{N} \][/tex]

Now we can find [tex]\( F_R \)[/tex]:

[tex]\[ F_R = 3F_L \] \[ F_R = 3 \times 150 \, \text{N} \] \[ F_R = 450 \, \text{N} \][/tex]

Next, we need to consider the torques about one of the support points. Let's choose the left end as our pivot point. The torque due to the woman's weight is the product of her weight and her distance from the left end, which we will call [tex]x[/tex]. The torque due to the board's weight acts at the center of the board (since the board is uniform), which is 4 meters from either end. The torque due to the support force [tex]\( F_R \)[/tex] acts at the right end.

Setting the sum of the torques equal to zero, we have:

[tex]\[ -F_R \times 8 \, \text{m} + 500 \, \text{N} \times x + 100 \, \text{N} \times 4 \, \text{m} = 0 \][/tex]

Substituting [tex]\( F_R = 450 \, \text{N} \)[/tex] and rearranging terms, we get:

[tex]\[ -450 \, \text{N} \times 8 \, \text{m} + 500 \, \text{N} \times x + 100 \, \text{N} \times 4 \, \text{m} = 0 \] \[ -3600 \, \text{N} \cdot \text{m} + 500 \, \text{N} \times x + 400 \, \text{N} \cdot \text{m} = 0 \] \[ 500 \, \text{N} \times x = 3600 \, \text{N} \cdot \text{m} - 400 \, \text{N} \cdot \text{m} \] \[ 500 \, \text{N} \times x = 3200 \, \text{N} \cdot \text{m} \] \[ x = \frac{3200 \, \text{N} \cdot \text{m}}{500 \, \text{N}} \] \[ x = 6.4 \, \text{m} \][/tex]

Therefore, the woman is standing 6.4 meters from the left end of the board. Since the board is 8 meters long, she is standing [tex]\( 8 \, \text{m} - 6.4 \, \text{m} = 1.6 \, \text{m} \)[/tex] from the right end of the board.