Metal surfaces on spacecraft in bright sunlight develop a net electric charge.

Do they develop a negative or a positive charge?

Answer: d.

Explanation:

The sandstone beds become thicker to the west.

Based on the information provided, there is no explicit trend mentioned in the sandstone beds from east to west. To find a trend, one would need to look for changes in grain size, sediment type, or sedimentary structures, none of which are detailed in the given data.

The trend visible in the sandstone beds as they are traced from east to west is not explicitly listed in the provided information. However, looking at the descriptions, it seems that no direct trend such as grading into conglomerate, becoming thinner, or becoming thicker is mentioned.

To determine such a trend, geologists might look for changes in the grain size, sediment type, or sedimentary structures within the sandstone layers. For example, if the beds were observed to contain increasingly larger grain sizes, or a transition to a different rock type such as conglomerate, that might indicate a trend in energy conditions and depositional environments.

Since no such trend is described in the information given, the most appropriate answer based on the provided options would be There is no information on how the sandstone beds change from east to west, which corresponds to option a.

Answer:

Velocity of the first car after the collision, [tex]v_1=8.93\ m/s[/tex]

Explanation:

It is given that,

Mass of the car, [tex]m_1 = 480\ kg[/tex]

Initial speed of the car, [tex]u_1 = 14.4\ m/s[/tex]    

Mass of another car, [tex]m_2 = 570\ kg[/tex]

Initial speed of the second car, [tex]u_2 = 13.3\ m/s[/tex]  

New speed of the second car, [tex]v_2 = 17.9\ m/s[/tex]  

Let [tex]v_1[/tex] is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]m_1u_1+m_2u_2-m_2v_2=m_1v_1[/tex]

[tex]480\times 14.4+570\times 13.3-570\times 17.9=480v_1[/tex]

[tex]v_1=8.93\ m/s[/tex]

So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.

Answer:

Explanation:

Given

Radius of Track [tex]r_1=75 m[/tex]

coefficient of Static Friction [tex]\mu _s=0.78[/tex]

Here centripetal Force is Balanced by Friction Force      

thus

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

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

[tex]v=\sqrt{\mu _srg}[/tex]

[tex]v=\sqrt{0.78\times 75\times 9.8}[/tex]

[tex]v=23.94 m/s[/tex]

(b)For [tex]r_2=25 m[/tex]

[tex]\mu _s=0.12[/tex]

[tex]v=\sqrt{\mu _sr_2g}[/tex]

[tex]v=\sqrt{0.12\times 25\times 9.8}[/tex]

[tex]v=5.42 m/s[/tex]

Final answer:

The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. This can be calculated using the formula v = sqrt(μs * g * r). For the given scenarios, the maximum speeds are 30.5 m/s and 7.36 m/s respectively.

Explanation:

The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. The maximum speed can be calculated using the formula:

v = sqrt(μs * g * r)

where v is the maximum speed, μs is the coefficient of static friction, g is the acceleration due to gravity (9.8 m/s²), and r is the radius of the turn.

For the first scenario with r = 75.0 m and μs = 0.780, the maximum speed is:

v = sqrt(0.780 * 9.8 * 75.0) = 30.5 m/s

For the second scenario with r = 25.0 m and μs = 0.120, the maximum speed is:

v = sqrt(0.120 * 9.8 * 25.0) = 7.36 m/s

Answer:

Angular acceleration will be [tex]18.84rad/sec^2[/tex]

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity [tex]\omega _i=0rev/sec[/tex]

And final angular velocity [tex]\omega _f=24rev/sec[/tex]

Time is given as t = 8 sec

From equation of motion

We know that [tex]\omega _f=\omega _i+\alpha t[/tex]

[tex]24=0+\alpha \times 8[/tex]

[tex]\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2[/tex]

So angular acceleration will be [tex]18.84rad/sec^2[/tex]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

[tex]\tau = \vec{F} \times \vec{r}[/tex]

[tex]\tau = (8i+6j)\times(-3i+4j)[/tex]

[tex]\tau = (8*4)(i\times j)+(6*-3)(j\times i)[/tex]

[tex]\tau = 32k +18k[/tex]

[tex]\tau = 50 k[/tex]

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

[tex]cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}[/tex]

[tex]cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}[/tex]

[tex]cos\theta = -0.24[/tex]

[tex]\theta = cos^{-1} (-0.24)[/tex]

[tex]\theta = 103.88\°[/tex]

Therefore the angle between the ratio and the force is 103.88°

Answer:

D) The spring will contract, raising the weight.

Explanation:

According to the statement there is current that will enter the current through the metal ions that it has in its stratum. The passage of the current will generate within the spring a magnetic field that travels in a loop. That is, while the upper part of the spring which is also that of the spring acts as a north pole, the lower part of the spring and the magnetic field will act as the south pole. The position of the poles will generate an opposition effect that will generate an attraction to each other which will generate a contraction in the spring and an increase in weight on it.

Answer:

a) parallel to the ground True

c) parallel to the ground towards man True

Explanation:

To examine the possibilities, we propose the solution of the problem.

Let's use Newton's second law

      F = m a

The force is exerted by the arm and the centripetal acceleration of the golf club, which in this case varies with height.

In our case, the stick is horizontal in the middle of the swing, for this point the centripetal acceleration is directed to the center of the circle or is parallel to the arm that is also parallel to the ground;

Ask the acceleration vector

a) parallel to the ground True

b) down. False

c) parallel to the ground towards True men

d) False feet

e) the head. False

Answer:

d=13.81 mm

Explanation:

Given that

P = 15 KN  ,L = 50 m

E= 200 GPa

ΔL = 25 mm

σ  = 150 MPa

Lets take d=Diameter

There are we have two criteria to find out the diameter of the wire

Case I :

According to Stress ,σ  = 150 MPa

P = σ  A

[tex]A=\dfrac{P}{\sigma}[/tex]

[tex]d=\sqrt{\dfrac{4P}{\pi \sigma}}[/tex]

By putting the values

[tex]d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}[/tex]

d= 11.28 mm

Case II:

According to elongation ,ΔL = 25 mm

[tex]\Delta L=\dfrac{PL}{AE}[/tex]

[tex]A=\dfrac{PL}{E\Delta L}[/tex]

[tex]A=\dfrac{4PL}{\pi E\Delta L}[/tex]

[tex]d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}[/tex]

d=13.81 mm

Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.

The diameter is 13.81 mm

A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa.

As per the Stress ,σ  = 150 MPa

P = σ  A

[tex]d = \sqrt{\frac{4\times 15000}{\pi \times 150} }[/tex]

= 11.28mm

Now

According to elongation ,ΔL = 25 mm

[tex]= \sqrt{\frac{4\times 15000 \times 50000}{\pi \times 200\times 1000 \times 25} }[/tex]

= 13.81 mm

For determining the wire of the smallest diameter, the above formulas should be used.

Therefore, we can conclude that the diameter is 13.81 mm.

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Answer:

P₂ = 392720.38 Pa = 392.72 kPa

Explanation:

Given

D₁ = 5 cm = 0.05 m

D₂ = 10 cm = 0.10 m

v₁ = 8 m/s

P₁ = 380 kPa = 380000 Pa

α = 1.06

ρ = 1000 kg/m³

g = 9.8 m/s²

We can use the following formula

(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL

knowing that z₁ = z₂ we have

(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL   (I)

Where

V₂ can be obtained as follows

V₁*A₁ = V₂*A₂      ⇒     V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)

⇒    V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)  

⇒    V₂ = 2 m/s

and

hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)

⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)

⇒ hL = 1.9469 m

Finally we get P₂ using the equation (I)

⇒  P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)

⇒  P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)

⇒  P₂ = 392720.38 Pa = 392.72 kPa

The velocity of the water in the larger section of the pipe is 3.2 m/s and the pressure is 412 kPa. The downstream pressure P2 is 356.7 kPa and the error if Bernoulli's equation had been used is approximately 55.3 kPa.

First, we can calculate the velocity of the water in the larger section of the pipe using the continuity equation, which states that the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region. Using this equation, we can find that the velocity in the larger section is 3.2 m/s.

Next, we can use Bernoulli's equation to solve for the pressure in the larger section of the pipe. Bernoulli's equation states that the pressure in a fluid decreases as the velocity increases. Plugging in the given values and the calculated velocity of 3.2 m/s, we can find that the pressure in the larger section is 412 kPa.

Finally, we can use the pressure drop equation, P2 - P1 = RQ, to solve for the downstream pressure P2. The resistance R can be calculated using the kinetic energy correction factor and the given values. Plugging in all the values, we can find that the downstream pressure P2 is 356.7 kPa. To estimate the error that would have occurred if Bernoulli's equation had been used, we can calculate the difference between the actual downstream pressure and the pressure calculated using Bernoulli's equation. This error is approximately 55.3 kPa.

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Answer:

[tex]3.81972\times 10^{-7}\ A[/tex]

Explanation:

B = Magnetic field = [tex]3\times 10^{-8}\ G[/tex]

d = Diameter of loop = 16 cm

r = Radius = [tex]\frac{d}{2}=\frac{16}{2}=8\ cm[/tex]

i = Current

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]

The magnetic field of a loop is given by

[tex]B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A[/tex]

The current needed to produce such a field at the center of the loop is [tex]3.81972\times 10^{-7}\ A[/tex]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = [tex]\sqrt{6gL}[/tex]

Hence, v0 = 6.86 m/s

To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. The equation (3/2)gh = 4(v0)^2 can then be used to solve for v0. The value of v0 is equal to the square root of (3/8)gh.

To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. When the rod is released, its potential energy is converted to kinetic energy and rotational kinetic energy. At the moment the rod comes to a halt, all of its initial kinetic energy will be converted back into potential energy. Since the rod is initially hanging vertically downward, we can equate the initial potential energy to the potential energy at the moment of momentary halt:

mgh = (1/2)Iω2

Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, I is the moment of inertia of the rod about the pivot point, and ω is the angular velocity of the rod at the moment of momentary halt. The moment of inertia of a rod rotating about one end is given by I = (1/3)mL2, where L is the length of the rod. Therefore, we can rewrite the equation as:

mgh = (1/2)(1/3)mL2ω2

Simplifying the equation, we have:

(3/2)gh = ω2L2

To find ω, we need to relate it to the linear speed v0. Since the rod is rotating about a pivot point, the linear speed of a point on the rod is related to the angular velocity by the equation v = ωr, where r is the distance of the point from the pivot. In this case, the distance r is equal to half the length of the rod, so r = L/2. Substituting this into the equation, we have:

v0 = ω(L/2)

From this equation, we can solve for ω:

ω = (2v0)/L

Substituting this into the previous equation, we get:

(3/2)gh = ((2v0)/L)2L2

Simplifying further, we have:

(3/2)gh = 4(v0)2

Finally, solving for v0, we get:

v0 = √((3/8)gh)

Answer:

a. [tex]d=99x10^{8}m[/tex]

b. [tex]r=14.22 R_s[/tex]

Explanation:

The eccentricity of an asteroid's is 0.0442 so

a.

to find the focus distance between both focus is

[tex]d=2*e*a[/tex]

[tex]e=1.12x10^{11}m\\a=0.0442[/tex]

So replacing numeric

[tex]d=2*1.12x10^{11}m*0.0442=9900800000[/tex]

[tex]d=99x10^{8}m[/tex]

b.

Now to find the ratio of that distance between the solar radius and the distance

[tex]r=\frac{d_1}{d_s}[/tex]

[tex]r=R_s*\frac{99x10^8m}{6.96x10^8m}[/tex]

[tex]r=14.22 R_s[/tex]

Answer: 1.39 m3

Explanation:

If we consider the helium to be an ideal gas, at any condition, we can apply the Ideal Gas Equation as follows:

P V = n R T

Taking the initial state as (1), we can write:

P1V1 = n R T1  (1)

In the initial state, we have P= 1.3 atm, V=1.6 m3, and T= 273 K + 16 K =289 K.

Let’s call (2) to the final state, so we can write as follows:

P2V2 = n R T2 (2)

In the final state, our givens are P= 1.2 atm, and T= 273 K -41 K = 232 K

So, dividing both sides in (1) and (2), we can solve for V2, as follows:

V2 = (1.3 atm. 1.6 m3.232 K) / 1.2 atm. 289 K = 1.39 m3

Answer:

[tex]F=19.8kN=4442lb[/tex]

Explanation:

On Earth the atmospheric presure is [tex]P_E=101325 N/m^2[/tex]. This will be the pressure inside the lander. Outside, the pressure on Mars will be [tex]P_M=650 N/m^2[/tex]. This means that the net force will be outward (since inside the pressure is higher) and, since the area of the hatch is [tex]A=\pi r^2[/tex], of value:

[tex]F=(P_E-P_M)\pi r^2=(101325N/m^2-650N/m^2)\pi (\frac{0.5m}{2})^2=19768N=19.8kN[/tex]

Since 1lb in weight  is equal to 4.45N, we can write:

[tex]F=19768N=19768N\frac{1lb}{4.45N}=4442lb[/tex]

Answer:

300 nm

Explanation:

R = Gas constant = 8.314 J/molK

r = Atomic radii = [tex]1\times 10^{-10}\ m[/tex]

d = Atomic diameter = [tex]2r=2\times 10^{-10}\ m[/tex]

At STP

T = Temperature = 273.15 K

P = Pressure = 100 kPa

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]

The mean free path is given by

[tex]\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm[/tex]

The answer that best represents the mean free path for gas molecules is 300 nm

Answer:

b. end on the n = 2 shell.

Explanation:

When hydrogen atoms move from higher energy level to lower energy level then it shows spectral lines and these lines are known as Balmer series. The only four lines are visible and other liens are not in the visible range.

The Balmer series formed by hydrogen electron and it ends when n = 2.

Therefore the answer is b.

b. end on the n = 2 shell.

To solve this problem it is necessary to apply the concepts related to the energy released through the mass defect.

Mass defect can be understood as the difference between the mass of an isotope and its mass number, representing binding energy.

According to the information given we have that the reaction presented is as follows:

[tex]^{232}U_{92} \Rightarrow ^{228}Th_{90}+^4He_2[/tex]

The values of the atomic masses would then be:

Th = 232.037146 u

Ra = 228.028731 u

He = 4.0026

The mass difference of the reaction would then be represented as

[tex]\Delta m = 232.037146 u - (228.028731 u + 4.002603 u )[/tex]

[tex]\Delta m = 0.005812 u[/tex]

From the international measurement system we know that 1 atomic mass unit is equivalent to 931.5 MeV,

[tex]\Delta m = 0.005812 u (\frac{931.5MeV}{1u})[/tex]

[tex]\Delta m = 5.414MeV[/tex]

Therefore the energy is 5.414MeV

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer ,  so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. =  λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

[tex]T^2 = \frac{4\pi^2}{GM}a^3[/tex]

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

[tex]T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s[/tex]

PART A) Replacing the values to find a, we have

[tex]a^3= \frac{T^2 GM}{4\pi^2}[/tex]

[tex]a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}[/tex]

[tex]a^3 = 6.46632*10^{39}[/tex]

[tex]a = 1.86303*10^{13}m[/tex]

Therefore the semimajor axis is [tex]1.86303*10^{13}m[/tex]

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

[tex]R = a(1-e)[/tex]

[tex]R = 1.86303*10^{13}(1-0.997)[/tex]

[tex]R= 5.58*10^{10}m[/tex]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

[tex]\tau = I \alpha[/tex]

Where,

I=Inertial Moment

[tex]\alpha =[/tex] Angular acceleration

Also Torque with linear equation is defined as,

[tex]\tau = F*d[/tex]

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

[tex]d*F = I\alpha[/tex]

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

[tex]d*F= \frac{mR^2a}{R}[/tex]

[tex]d*F= mRa[/tex]

[tex]a = \frac{rF}{ mR}[/tex]

[tex]a = \frac{0.04*20}{1.5*0.3}[/tex]

[tex]a=1.77 m/s^2[/tex]

Then the velocity of the wheel is

[tex]V = a *t \\V=1.77*4 \\V=7.11 m/s[/tex]

Therefore the correct answer is D.

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (in this case the -z direction) and the curling of the fingers point in the direction of the magnetic field  B→ (in this case the +x direction), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

Final answer:

To find the velocities of pieces A and B as measured by an observer in the laboratory, use the relativistic velocity addition formula.

Explanation:

To find the velocities of pieces A and B as measured by an observer in the laboratory, we need to use the relativistic velocity addition formula. Let's call the initial velocity of the uranium nucleus as v. Piece A is moving forward with a speed of 0.43c relative to the original nucleus and piece B is moving backward at 0.35c relative to the original nucleus.

The velocity of piece A as measured by an observer in the laboratory is given by vA = (v + vA') / (1 + v*vA'/c^2), where vA' is the velocity of piece A relative to the original nucleus. Plugging in the values, we get vA = (v + 0.43c) / (1 + v*0.43c/c^2).

The velocity of piece B as measured by an observer in the laboratory is given by vB = (v - vB') / (1 - v*vB'/c^2), where vB' is the velocity of piece B relative to the original nucleus. Plugging in the values, we get vB = (v - 0.35c) / (1 - v*0.35c/c^2).

Answer:

x = 1.26 sin 3.16 t

Explanation:

Assume that the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

[tex]V=\dfrac{dx}{dt}[/tex]

V= A ω cosωt

Maximum velocity

V(max)= Aω

Given that F= 32 N

F = K Δ

K=Spring constant

Δ = 0.4 m

32 =0.4 K

K = 80 N/m

We know that  ω²m = K

8 ω² = 80

ω = 3.16 s⁻¹

Given that V(max)= Aω = 4 m/s

3.16 A = 4

A= 1.26 m

Therefore the general equation of displacement

x = 1.26 sin 3.16 t

Answer:

Explanation:

mass, m = 8 kg

extension, y = 0.4 m

force, F = 32 N

maximum velocity, v = 4 m/s

maximum velocity , v = ω A

where, ω be the angular velocity and A be the amplitude

4 = ω x 0.4

ω = 10 rad/s

position

x = A Sin ωt

x = 0.4 Sin 10 t

Answer:

Green laser

Explanation:

In the double slit experiment, the distance between two consecutive bright spots or two consecutive dark spots is defined as the fringe width

We are assuming that the two slits are on a vertical line

The formula for fringe width is (D× wavelength of light) ÷ d

where D is the distance from the slit to the screen where the interference pattern is observed

d is the distance between the two slits

As it is given that both lasers have same configurations which means they both have same value of D and d

∴ Fringe width directly depends on the wavelength of the light

As green laser has more wavelength when compared to blue laser

∴ Green laser would produce an interference pattern that has the largest spacing between bright spots when compared to blue laser

Answer: [tex]2.89(10)^{-3} m[/tex]

Explanation:

The Heisenberg uncertainty principle postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.  

In other words:  

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

[tex]\Delta x \geq \frac{h}{4 \pi m \Delta V}[/tex] (1)

Where:

[tex]\Delta x[/tex] is the uncertainty in the position of the electron

[tex]h=6.626(10)^{-34}J.s[/tex] is the Planck constant

[tex]m=9.11(10)^{-31}kg[/tex] is the mass of the electron

[tex]\Delta V[/tex] is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is [tex]0.001\%[/tex] of the velocity of the electron [tex]V=2 km/s=2000 m/s[/tex], then [tex]\Delta V[/tex] is:

[tex]\Delta V=2000 m/s(0.001\%)[/tex]

[tex]\Delta V=2000 m/s(\frac{0.001}{100})[/tex]

[tex]\Delta V=2(10)^{-2} m/s[/tex] (2)

Now, the least possible uncertainty in position [tex]\Delta x_{min}[/tex] is:

[tex]\Delta x_{min}=\frac{h}{4 \pi m \Delta V}[/tex] (3)

[tex]\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)}[/tex] (4)

Finally:

[tex]\Delta x_{min}=2.89(10)^{-3} m[/tex]

The least possible uncertainty within which we can determine the position of this electron is 0.29 micrometers. This value is derived using the Heisenberg's Uncertainty Principle, which states that the more precisely one measures the position of a particle, the less precisely one can measure its speed, and vice versa.

The subject question is dealing with the principle of uncertainty in quantum mechanics. According to Heisenberg's Uncertainty Principle, the position and momentum of a particle cannot both be accurately measured at the same time. The more precisely one measures the position of a particle, the less precisely one can measure its speed, and vice versa.

The accuracy is given as 1 part in 105, that is 0.001% or 1e-5. The speed of the electron is known to be 2.0 km/s so the uncertainty in velocity (Δv) would be 2.0 km/s * 1e-5 = 2e-8 m/s. The mass of the electron (melectron) is given to be 9.11 × 10-31 kg.

Momentum is the product of mass and velocity, so Δp = melectron x Δv = (9.11 × 10-31 kg) * (2e-8 m/s) = 1.822e-38 kg m/s. According to Heisenberg's Uncertainty Principle (ΔxΔp ≥ h/4π) the least uncertainty in position (Δx) = h / (4πΔp), where h is Planck's constant 6.626 × 10-34 J.s. Substituting the values, Δx = 6.626 × 10-34 J.s / (4π * 1.822e-38 kg m/s) = 2.9e-7 m or 0.29 μm.

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To solve the problem, it is necessary to apply the related concepts to Newton's second law as well as the Normal and Centripetal Force experienced by passengers.

By Newton's second law we understand that

[tex]F = mg[/tex]

Where,

m= mass

g = Gravitational Acceleration

Also we have that Frictional Force is given by

[tex]F_r = \mu N[/tex]

In this particular case the Normal Force N is equivalent to the centripetal Force then,

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

Applying this to the information given, and understanding that the Weight Force is statically equivalent to the Friction Force we have to

[tex]F = F_r[/tex]

[tex]mg = \mu N[/tex]

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

Re-arrange to find v,

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

[tex]v = \sqrt{\frac{(9.8)(2)}{0.25}}[/tex]

[tex]v = 8.9m/s[/tex]

From the last expression we can realize that it does not depend on the mass of the passengers.

The minimum speed required for passengers to remain stuck to the wall of a cylindrical carnival ride with a radius of 2.0 meters and a coefficient of static friction of 0.25 is d) 8.9 m/s.

To determine the minimum speed required to prevent passengers from sliding down the wall of a cylindrical carnival ride with a 2.0-m radius, we can use the principles of centripetal force and friction.

The centripetal force needed to keep a rider in circular motion is provided by the normal force (N) which acts horizontally. This force is balanced by the frictional force (f) acting vertically upwards to counteract the gravitational force (mg) pulling the rider down.

The frictional force is given by:

f = μN

Where μ is the coefficient of static friction (0.25). The normal force is equivalent to the centripetal force needed for circular motion:

N = mv² / r

Thus, the frictional force equation becomes:

μ(mv² / r) = mg

Solving for v:

μv² / r = g

v² = rg / μ

[tex]v = \sqrt{\frac{rg}{\mu}}[/tex]

Substituting the given values (r = 2.0 m, g = 9.8 m/s², μ = 0.25):

[tex]v = \sqrt{\frac{2.0 \cdot 9.8}{0.25}}[/tex]

v = √(78.4)

v = 8.9 m/s

Therefore, the correct option is d) as the minimum speed required is 8.9 m/s.

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

[tex]I=\dfrac{CB^2}{2\mu_0}[/tex]

[tex]I=\dfrac{CE^2}{2\mu_0 C^2}[/tex]

[tex]E=\sqrt{{2I\mu_0 C}}[/tex]

[tex]E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}[/tex]

E=930.84 N/C

Therefore answer is 930.84 N/C

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

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Answer:

a) [tex]P_{L}=199.075W[/tex]

b) [tex]P_{L}=1.991x10^{-4}W[/tex]

Explanation:

1) Notation

Power on the refrigerator: [tex]P=IV=3Ax110V=330W[/tex]

Voltage [tex]V=110V[/tex]

[tex]D=8.252mm[/tex], so then the radius would be [tex]r=\frac{8.252}{2}=4.126mm[/tex]

[tex]L=2x10km=20km=20000m[/tex], representing the length of the two wires.

[tex]\rho=2.65x10^{-8}\Omega m[/tex], that represent the resistivity for the aluminum founded on a book

[tex]P_L[/tex] power lost in the transmission.

2) Part a

We can find the total power adding all the individual values for power:

[tex]P_{tot}=(330+100+60+3)W=493W[/tex]

From the formula of electric power:

[tex]P=IV[/tex]

We can solve for the current like this:

[tex]I=\frac{P}{V}[/tex]

Since we know [tex]P_{tot}[/tex] and the voltage 110 V, we have:

[tex]I=\frac{493W}{110V}=4.482A[/tex]

The next step would be find the cross sectional are for the aluminum cables with the following formula:

[tex]A=\pi r^2 =\pi(0.004126m)^2=5.348x10^{-5}m^2[/tex]

Then with this area we can find the resistance for the material given by:

[tex]R=\rho \frac{L}{A}=2.65x10^{-8}\Omega m\frac{20000m}{5.348x10^{-5}m^2}=9.910\Omega[/tex]

With this resistance then we can find the power dissipated with the following formula:

[tex]P_{L}=I^2 R=(4.482A)^2 9.910\Omega=199.075W[/tex]

And if we want to find the percentage of power loss we can use this formula

[tex]\% P_{L}=\frac{P_L}{P}x100[/tex]

3) Part b

Similar to part a we just need to change the value for V on this case to 110KV.

We can solve for the current like this:

[tex]I=\frac{P}{V}[/tex]

Since we know [tex]P_{tot}[/tex] and the voltage 110 KV=110000V, we have:

[tex]I=\frac{493W}{110000V}=4.482x10^{-3}A[/tex]

The cross sectional area is the same

The resistance for the material not changes.

With this resistance then we can find the power dissipated with the following formula:

[tex]P_{L}=I^2 R=(4.482x10^{-3}A)^2 9.910\Omega=1.991x10^{-4}W[/tex]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

[tex]PE=\frac{GMm}{r}[/tex]

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

[tex]m = \rho V[/tex]

[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]

[tex]m = 7.210*10^{11}Kg[/tex]

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]

PART A) Potential energy when the ocean is at its furthest point to the moon,

[tex]PE_1 = \frac{GMm}{r_1}[/tex]

[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]

[tex]PE_1 = 9.05*10^{15}J[/tex]

PART B) Potential energy when the ocean is at its closest point to the moon

[tex]PE_2 = \frac{GMm}{r_2}[/tex]

[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]

[tex]PE_2 = 9.361*10^{15}J[/tex]

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

[tex]\Delta KE = \Delta PE[/tex]

[tex]\frac{1}{2}mv^2 = PE_2-PE_1[/tex]

[tex]v=\sqrt{2(PE_2-PE_1)/m}[/tex]

[tex]v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}[/tex]

[tex]v = 29.4m/s[/tex]

Answer:

θ =607.33°

Explanation:

Given that

Angular acceleration α = 3.45 rad/s²

Initial angular speed ,ω = 1.85 rad/s

The angle rotates by wheel in time t

[tex]\theta=\omega t +\dfrac{1}{2}\alpha t^2[/tex]

Now by putting the values

[tex]\theta=\omega t +\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta=1.85\times 2 +\dfrac{1}{2}\times 3.45\times 2^2[/tex]

θ = 10.6 rad

[tex]\theta=\dfrac{180}{\pi}\times 10.6\ degree[/tex]

θ =607.33°

Therefore angle turn by wheel in 2 s is θ =607.33°