The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x) = (2.0 {\rm J/m}^{3})x^{3}- (15 {\rm J/m}^{2})x^{2}+ (36 {\rm J/m})x - 23 {\rm J}. Find the location(s) where the force on the mass is zero.

Answer:

Speed of larger piece is [tex]\dfrac{V\sqrt{2}}{3}[/tex]

Explanation:

We apply the principle of conservation of momentum.

The watermelon is initially at rest. The initial momentum = 0 kg m/s in all directions.

After the collision,

Vertical momentum = momentum of piece in y-direction + y-component of momentum of larger piece = [tex]mV + 3mv_{ly}[/tex]

Here, [tex]v_{ly}[/tex] is the y-component of velocity of larger piece.

This is equal to 0, since the initial momentum is 0.

[tex]v_{ly}=\dfrac{V}{3}[/tex]

Horizontal momentum = momentum of piece in x-direction + x-component of momentum of larger piece = [tex]mV + 3mv_{lx}[/tex]

Here, [tex]v_{lx}[/tex] is the x-component of velocity of larger piece.

This is also equal to 0, since the initial momentum is 0.

[tex]v_{lx}=\dfrac{V}{3}[/tex]

The velocity of the larger piece, [tex]v_l[/tex], is the resultant of [tex]v_{lx}[/tex] and [tex]v_{ly}[/tex]. Since they are mutually perpendicular,

[tex]v_l = \sqrt{v_{ly}^2+v_{lx}^2}= \sqrt{\left(\dfrac{V}{3}\right)^2+\left(\dfrac{V}{3}\right)^2}[/tex]

[tex]v_l = \dfrac{V\sqrt{2}}{3}[/tex]

Answer:

[tex]v(1.5)=0.7648\ m/s[/tex]

Explanation:

Dynamics

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

[tex]\displaystyle a=\frac{F}{m}[/tex]

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

[tex]F=2cos1.1t[/tex]

The variable acceleration is calculated by:

[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]

[tex]a=0.8439cos1.1t[/tex]

The instant velocity is the integral of the acceleration:

[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]

[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]

Integrating

[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]

[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]

[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]

Final answer:

To find the x-component of velocity at t= 1.5 seconds for a particle under a sinusoidally varying force, we derive the equation of motion, integrate the acceleration, and calculate the velocity, resulting in [tex]v_x[/tex] = 0.567 m/s.

Explanation:

The question pertains to finding the x-component of velocity (vx) of a particle at t= 1.5 seconds, given the mass and the sinusoidally varying force. Since the force applied on the particle varies as Fx = F0cos(ωt), where F0 = 2 N and ω = 1.1 rad/s, we can find the acceleration and then integrate it with respect to time to find velocity. The acceleration ax is given by Fx/m = (2cos(1.1t))/2.37. To find the change in velocity, we integrate ax with respect to time, giving us vx = ∫ ax dt = ∫ (2cos(1.1t))/2.37 dt. Evaluating this integral from 0 to 1.5 s, we use the definite integral which simplifies to vx(t) = (2/2.37)(sin(1.1(1.5))-sin(1.1(0)))/1.1. The calculation yields vx = 0.567 m/s, indicating the particle's velocity in the x-direction at 1.5 seconds.

The speed of the particle to pass through the Selector is 2875 m/s

Explanation:

Given -

Electric field,(We can represent as E) = 460 N/C

Magnetic field, (We can represent as B) = 0.16 T

Speed,(We can represent as  v) = ?

We know that the formula for finding the velocity ,

[tex]v = \frac{E}{B} \\\\v = \frac{460}{0.16} \\\\v = 2875m/s[/tex]

Therefore, speed of the particle to pass through the Selector is 2875 m/s

Answer:

10044 N

Explanation:

The acceleration of the cab is calculated using the equation of motion:

[tex]v^2 = u^2+2as[/tex]

v is the final velocity = 0 m/s in this question, since it is brought to rest

u is the initial velocity = 10 m/s

a is the acceleration

s is the distance = 35 m

[tex]a = \dfrac{v^2-u^2}{2s} = \dfrac{(0 \text{ m/s})^2-(10 \text{ m/s})^2}{2\times (35\text{ m})} = -1.43\text{ m/s}^2[/tex]

Since it accelerates downwards, its resultant acceleration is

[tex]a_R = g + a[/tex]

g is the acceleration of gravity.

[tex]a_R = (9.8-1.43)\text{ m/s}^2 = 8.37\text{ m/s}^2[/tex]

The tension in the cable is

[tex]T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}[/tex]

Answer: 0.321 seconds

Explanation:

Let assume that air has a temperature of 20 °C. Sound speed at given temperature is [tex]343 \frac{m}{s}[/tex]. As sound spreads at constant speed, time can be easily found by using this formula:

[tex]\Delta t = \Delta t_{far} - \Delta t_{near}[/tex]

[tex]\Delta t = \frac{x_{far}-x_{near}}{v_{sound,air}}[/tex]

[tex]\Delta{t} = \frac{120 m - 10 m}{343 \frac{m}{s} }\\\\\Delta {t} = 0.321 sec[/tex]

Answer:

he phase difference is π the destructive interference and the lujar is a still or silent place

Explanation:

This is a sound interference exercise where the amino difference is equal to the phase difference of the sound.

      Δr / λ = ΔФ / 2π

Let's find the path difference

     r₁ = 4m

     r₂ = √ (4² + 3²) = 5 m

     Δr = r₂ - r₁

     Δr = 5-4 = 1m

Let's find the wavelength of the sound

        v =  λ f

         λ = v / f

         λ = 340/170

         λ = 2m

Let's find the phase difference between the two waves

      ΔФ = Δr 2π / λ

      ΔФ = 1 2π / 2

      ΔФ = π

Since the phase difference is π the destructive interference and the lujar is a still or silent place

Final answer:

To determine if Sam is standing on a loud or quiet spot, the path difference of the sound waves from the speakers was calculated, which is half the wavelength, indicating that Sam stands at a loud spot due to constructive interference.

Explanation:

The question relates to interference patterns created by the sound waves from two speakers and whether the spot where Sam stands is a loud or quiet spot. To determine this, we need to calculate the path difference of the sound waves reaching Sam from both speakers. The speed of sound in air is given as 340 m/s, and the frequency of the tone is 170 Hz.

The wavelength (λ) of the sound can be found using the formula speed = frequency × wavelength, which results in λ = 340 m/s / 170 Hz = 2 m. The path difference is the difference in distance from each speaker to Sam. For one speaker, the distance is 4.0 m. For the other speaker, we use the Pythagorean theorem since Sam is standing in front of the first speaker: √(4.0^2 + 3.0^2) = 5.0 m. The path difference is therefore 5.0 m - 4.0 m = 1.0 m, which is exactly half the wavelength.

Since the path difference corresponds to half a wavelength, this results in constructive interference, and Sam is indeed standing at a loud spot.

Answer:

The distance between the ships changing at 6PM is 21.29Km/h

Explanation:

Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h

Given

dx/dt= 35

dy/dt= 25

dv/dt= ???? at t= 6PM - 2PM= 4

Therefore t=4

We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction

So, we use:

[tex] D^2 = (150 - x)^2 + y^2 [/tex];

[tex] D^2 = (150 - 140)^2 + y^2 [/tex]

But ship B travels at t=4, at 4.25 =100 in the y-direction

so, let's use the equation:

[tex] D^2 = 10^2 + 100^2 [/tex]

[tex] = D= sqrt*(10 + 100) [/tex]

Lets use 2DD' = 2xx' + 2yy'

Differentiating with respect to t we have:

D•d(D)/dt = -(10)•dx/dt + 100•dy/dt

=100.5 d(D)/dt = (-10)•35 + (100)•25

When t=4, we have x=(140-150) =10 and y=100

[tex]= D = sqrt*(10^2 + 100^2) [/tex]

=100.5

= 100.5 dD/dt = 10.35 +100.25

= dD/dt = 21.29km/h

The distance between Ship A and Ship B is changing at approximately 21.4 km/h at 6 P.M.

Here, we use related rates. Let's define the positions of the ships -

Ship A's position at 6 P.M. :

Ship B's position at 6 P.M. :

Let’s denote the distance between the two ships at time t as D, x as the east-west distance (positive east) between Ship A and Ship B, and y as the north-south distance (positive north) from the initial position of Ship B.

Given:

We use the Pythagorean theorem to relate x, y, and D:

Differentiating both sides with respect to t:

Simplifying and solving for [tex]\frac{dD}{dt}[/tex] gives:

Plugging in the values:

Therefore, the distance between the ships is changing at approximately 21.4 km/h at 6 P.M.

Answer:

The overall thermal efficiency is 30%.

Explanation:

Given;

Output power = 500 kWh

input energy  per kg of coal = 6 MJ = 6 x 10⁶ J = 1.66667 kWh

1000 kg of coal will produce 1000 x 1.66667 kWh = 1666.67 kWh

Thus, total input power = 1666.67 kWh

Overall thermal efficiency = Total output power/Total input Power

Overall thermal efficiency = (500/1666.67) *100

Overall thermal efficiency = 0.29999 *100

Overall thermal efficiency = 30%

Therefore, the overall thermal efficiency is 30%.

Question is not complete and the missing part is;

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

Answer:

0.828 m/s

Explanation:

Resolving vertically, we have;

Fn and Fg act vertically. Thus,

Fn - Fg = 0 - - - - eq(1)

Resolving horizontally, we have;

Ff = ma - - - - eq(2)

Now, Fn and Fg are both mg and both will cancel out in eq 1.

Leaving us with eq 2.

So, Ff = ma

Now, Frictional force: Ff = μmg where μ is coefficient of friction.

Also, a = v²/r

Where v is linear speed or velocity

Thus,

μmg = mv²/r

m will cancel out,

Thus, μg = v²/r

Making v the subject;

rμg = v²

v = √rμg

Plugging in the relevant values,

v = √0.14 x 0.5 x 9.8

v = √0.686

v = 0.828 m/s

Final answer:

To determine the linear speed of the coin when it just begins to slip, we can use the equation frictional force = centripetal force for circular motion. By equating these two forces and solving for the linear speed, we can find the answer.

Explanation:

To determine the linear speed of the coin when it just begins to slip, we can use the equation:
frictional force = centripetal force for circular motion

The frictional force can be calculated using the equation:
frictional force = coefficient of static friction x normal force

And the centripetal force can be calculated using the equation:
centripetal force = mass of coin x acceleration towards the center of the disk

By equating these two forces and solving for the linear speed, we can find the answer.

In this case, we are given the coefficient of static friction as 0.50. We can also assume that the normal force is equal to the weight of the coin, which is the mass of the coin multiplied by the acceleration due to gravity. By plugging in these values, we can find the linear speed of the coin.

Answer:

Entropy Change = 0.559 Times

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

Answer:

r = 0.1 Ω

Explanation:

We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.

When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.

[tex]0.15 = (1.5)r\\r = 0.1~\Omega[/tex]

Answer:

(1)

The velocity of the center of mas of the system is= 0.801 m/s

(2)

Initial velocity of car 1 in the center of mass reference frame is =4.099 m/s

(3)

The final velocity of car 1 in the center of mass reference frame is

 - 4.099 m/s

(4)

The final velocity of car 1 in the ground (original ) reference frame is = -3.298 m/s

(5)

The final velocity of car 2 in the ground (original) reference frame is = 7.166  m/s

Explanation:

Given

m₁ = 109 kg

v₁= 4.9 m/s

m₂= 83 kg

v₂= -3.6 m/s

The two cars have an elastic collision.

(1)

The velocity of the center of mas of the system is

[tex]V_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]

       [tex]= \frac{109.4.9+83(-3.4)}{109+83}[/tex] m/s

      = 0.801 m/s

(2)

Initial velocity of car 1 in the center of mass reference frame is

[tex]V_{1,i}[/tex] = initial velocity - [tex]V_{cm}[/tex]

    = (4.9 - 0.801) m/s

    =4.099 m/s

(3)

Since the collision is elastic, the car 1 will bounce of opposite direction.

The final velocity of car 1 in the center of mass reference frame is

 [tex]V_{1,f}[/tex] = - 4.099 m/s

(4)

The final velocity of car 1 in the ground (original ) reference frame

[tex]V'_{1,f}[/tex]  =  [tex]V_{cm}+V_{1,f}[/tex]

      =(0.801- 4.099) m/s

      = - 3.298 m/s

(5)

The momentum is conserved,

[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]

[tex]\Rightarrow v'_2=\frac{m_1}{m_2}(v_1-v'_1) +v_2[/tex]  

Here [tex]v'_1= V'_{1,f}[/tex] =  - 3.298 m/s

[tex]\Rightarrow v'_2=\frac{109}{83}[4.9-(-3.298)]+(-3.6)[/tex]

      =7.166 m/s

The final velocity of car 2 in the ground (original) reference frame is = 7.166  m/s

Final answer:

The velocity of the center of mass of the bumper car system is 1.225 m/s. Car 1 has an initial velocity of 3.675 m/s in the center-of-mass reference frame. After the elastic collision, car 1's final velocity is -2.45 m/s, and car 2's final velocity is 4.825 m/s in the ground reference frame.

Explanation:

Velocity of the Center of Mass, Velocities in Reference Frames, and Final Velocities After an Elastic Collision

The velocity of the center of mass (V cm) of a system in one dimension is given by the formula:

V cm = (m1 * v1 + m2 * v2) / (m1 + m2)

In this case, we have m1 = 109 kg moving at v1 = 4.9 m/s and m2 = 83 kg moving at v2 = -3.6 m/s. The velocity of the center of mass is therefore:

V cm = (109 kg * 4.9 m/s + 83 kg * (-3.6 m/s)) / (109 kg + 83 kg)

Calculating this we get:

V cm = (534.1 kg*m/s - 298.8 kg*m/s) / 192 kg = 235.3 kg*m/s / 192 kg = 1.225 m/s (rounded to three significant figures)

The initial velocity of car 1 in the center-of-mass reference frame is given by:

u1 = v1 - V cm

Substituting the known values:

u1 = 4.9 m/s - 1.225 m/s = 3.675 m/s

In an elastic collision, velocities in the center-of-mass reference frame are mirrored. Thus, the final velocity of car 1 in the center-of-mass reference frame remains the same but in the opposite direction:

u1' = -u1 = -3.675 m/s

To find the final velocity of car 1 in the ground reference frame, we add the velocity of the center of mass:

v1' = u1' + V cm = -3.675 m/s + 1.225 m/s = -2.45 m/s

For car 2, the same principle applies. The final velocity in the center-of-mass reference frame will be the opposite of the initial, and thus:

v2' = -v2 + V cm = 3.6 m/s + 1.225 m/s = 4.825 m/s

Answer:

See the answers and the explanation below.

Explanation:

To solve this problem we must make a free body diagram with the forces applied as well as the direction. In the attached images we can see the nomenclature of the direction of the forces and the free body diagram.

a)

Sum of forces in y-axis

Fy = 140 - (220*sin(10))

Fy = 101.8 [N]

Sum of forces in x-axis

Fx = - (220*cos(10))

Fx = - 216.65 [N]

b)

For the above result, for there to be balance we realize that we need one equal to the resulting y-axis but in the opposite direction and another opposite force in direction but equal in magnitude on the x-axis.

Fy = - 101.8 [N]

Fx =   216.65 [N]

c )

Now we need to use the Pythagorean theorem to find the result of these forces.

[tex]F = \sqrt{(101.8)^{2} +(216.65)^{2} } \\F=239.4[N][/tex]

And the direction will be as follows.

α = tan^(-1) (101.8 / 216.65)

α = 25.16° (south to the east)

F = 216.65 i - 101.8 j [N]

Answer:

Incomplete question

Check attachment for the diagram of the problem.

Explanation:

The acceleration of the car A is given as

a=3ft/s²

Car B is rounding a curve of radius

r=440ft

Car B is moving at constant speed of Vb=30mi/hr.

Car A reach a speed of 45mi/hr

Note, 1 mile = 5280ft

And 1 hour= 3600s

Then

Va=45mi/hr=45×5280/3600

Va=66ft/s

Also,

Vb=30mi/hour=30×5280/3600

Vb=44ft/s

Now,

a. Let write the relative velocity of car B, relative to car A

Vb = Va + Vb/a

Then,

Using triangle rule, because vectors cannot be added automatically

Vb/a²= Vb²+Va²-2Va•VbCosθ

From the given graphical question the angle between Va and Vb is 60°.

Vb/a²=44²+66² - 2•44•66Cos60

Vb/a²=1936+ 4356 - 5808Cos60

Vb/a² = 3388

Vb/a = √3388

Vb/a = 58.21 ft/s

The direction is given as

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Va/SinA = Vb/SinB = (Vb/a)/SinC

66/SinA = 44/SinB = 58.21/Sin60

Then, to get B

44/SinB = 58.21/Sin60

44Sin60/58.21  = SinB

0.6546 = SinB

B=arcsin(0.6546)

B=40.89°

b. The acceleration of Car B due to Car A.

Let write the relative acceleration  of car B, relative to car A.

Let Aa be acceleration of car A

Ab be the acceleration of car B.

Ab = Aa + Ab/a

Given the acceleration of car A

Aa=3ft/s²

Then to get the acceleration of car B, using the tangential acceleration formular

a = v²/r

Ab = Vb²/r

Ab = 44²/440

Ab = 4.4ft/s²

Using cosine rule again as above

Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ

Ab/a²= 3²+4.4²- 2•3•4.4•Cos30

Ab/a²= 9+19.36 - 22.863

Ab/a² = 5.497

Ab/a = √5.497

Ab/a = 2.34ft/s²

To get the direction using Sine rule again, as done above

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Aa/SinA = Ab/SinB = (Ab/a)/SinC

3/SinA = 4.4/SinB = 2.34/Sin30

Then, to get B

4.4/SinB = 2.34/Sin30

4.4Sin30/2.34 = SinB

0.9402 = SinB

B=arcsin(0.9402)

B=70.1°

Since B is obtuse, the other solution for Sine is given as

B= nπ - θ.   , when n=1

B=180-70.1

B=109.92°

To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. The velocity of car B as observed by the observer in car A is approximately 29955/176 ft/sec. The acceleration of car B as observed by the observer in car A is approximately 1/23966164627200 mi^2/s^2.

To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. Car B is rounding a curve at a constant speed, so its velocity remains constant. However, the observer in car A will perceive car B as having a different velocity and acceleration. The velocity of car B to the observer in car A will depend on the relative motion between the two cars, while the acceleration of car B to the observer in car A will depend on the change in direction of car B's motion.

Let's calculate the velocity and acceleration of car B as observed by an observer in car A:

Velocity: Since car B is rounding a curve with a radius of 440 ft and a constant speed of 30 mi/hr, we can use the formula v = rω to find the angular velocity ω. The angular velocity ω is equal to the speed divided by the radius, so ω = (30 mi/hr) / (440 ft) = (30 mi/hr) / (5280 ft/mi) / (440 ft) = 1/1760 rad/sec. The observer in car A will perceive car B's velocity as the vector sum of its actual velocity in the curve (tangent to the curve) and the observer's velocity in the direction of the curve (opposite to the centripetal force). Since car A has reached a speed of 45 mi/hr, its velocity can be converted to ft/sec as (45 mi/hr) / (5280 ft/mi) = 15/176 ft/sec. Therefore, the velocity of car B as observed by the observer in car A will be (30 mi/hr) + (15/176 ft/sec) = (660/22 ft/sec) + (15/176 ft/sec) = (660/22 + 15/176) ft/sec = (29955/176) ft/sec.

Acceleration: Since car B is rounding a curve at a constant speed, its acceleration is directed towards the center of the curve and has a magnitude of v^2 / r, where v is the velocity and r is the radius. Substituting the values, we get the acceleration as (30 mi/hr)^2 / (440 ft) = ((30 mi/hr)^2) / ((5280 ft/mi) / (440 ft)) = (900 mi^2/hr^2) / (5280 ft/mi) * (440 ft) = (900 mi^2 * ft^2) / (5280 hr^2) * (440) ft = (900 * 5280 * 440) ft^2 / hr^2 = (2119680000/5280) ft^2 / hr^2 = (400800 ft^2/hr^2) = (400800 ft^2/hr^2) * (1/3600 hr^2/s^2) * (1 mi^2 / (5280 ft)^2) = (400800 / 3600) * (1/5280)^2 mi^2/s^2 = (111/9900) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (1/266611200) mi^2/s^2 = (1/266611200) * (5280 ft/mi)^2 = (1/266611200) * 5280^2 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/19) ft^2/s^2 = (1/19) * (1/5280)^2 mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/26268580800) mi^2/s^2 = (1/26268580800) * (5280 ft/mi)^2 = (1/26268580800) * 5280^2 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/237896) ft^2/s^2 = (1/237896) * (1/5280)^2 mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/23966164627200) mi^2/s^2.

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To calculate the distance L between the screen and the slit, use the single-slit diffraction minimum condition combined with the known distance of the red laser's third minimum and its wavelength.

The student is working on a Physics problem related to single-slit diffraction. To find the distance L between the screen and the slit, we can use the formula for the position of a diffraction minimum, y = L × tan(θ), where θ is the angle of the diffraction minimum. However, for small angles (as in most diffraction problems), tan(θ) ≈ sin(θ), so the formula simplifies to y = L × sin(θ). The condition for the minima in a single-slit diffraction pattern is given by a × sin(θ) = m × λ, where a is the width of the slit, m is the order number of the minimum, and λ is the wavelength of the light. Given that the third diffraction minimum (m = 3) for the red laser is at a distance y3,red = 4.05 cm from the pattern's center, and the wavelength of the red light is λ = 633.0 nm, we can write 0.150 mm × sin(θ) = 3 × 633.0 nm. Solving for sin(θ) and then using y = L × sin(θ) where y = 4.05 cm, we can calculate the distance L.

Answer:

[tex]9.198\times 10^6 m/s[/tex]

Explanation:

We are given that

Magnetic field, B=1.2 T

Radius of circular path, r=0.080 m

[tex]q_p=1.6\times 10^{-19} C[/tex]

[tex]m_p=1.67\times 10^{-27} kg[/tex]

[tex]\theta=90^{\circ}[/tex]

We have to find the speed of proton.

We know that

Magnetic force, F=[tex]qvBsin\theta[/tex]

According to question

Magnetic force=Centripetal force

[tex]q_pvBsin90^{\circ}=\frac{m_pv^2}{r}[/tex]

[tex]1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}[/tex]

[tex]v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}[/tex]

[tex]v=9.198\times 10^6 m/s[/tex]

To solve this problem we will apply the concepts related to the Friction force and work. The friction force can be defined as the product between the Normal Force (Mass by gravity) and the dynamic friction constant. In the case of Work this is defined as the product of the distance traveled by the applied force. Then we will solve the points sequentially to find the answer to each point,

PART A) The friction force with the given data  is,

[tex]F_f = \mu_k mg[/tex]

Here,

[tex]\mu_k[/tex] = Kinetic coefficient

m = Mass

g = Gravitational acceleration

[tex]F_f = (0.25)(30kg)(9.8m/s^2)[/tex]

[tex]F_f = 73.5N[/tex]

PART B) The work done by the worker is the distance traveled for the previously force found, then

[tex]W_w = rF_f[/tex]

[tex]W_w = (4.5m)(73.5N)[/tex]

[tex]W_w = 330.75J[/tex]

PART C) The work done by the friction force would be the distance traveled with the previously calculated force, therefore

[tex]W_f = -rF_f[/tex]

[tex]W_f = -(4.5m)(73.5N)[/tex]

[tex]W_f = -330.75J[/tex]

[tex]W_f = -331J[/tex]

PART D) The work done by the normal force is,

[tex]W_N = N (r) Cos(90)[/tex]

[tex]W_N = 0J[/tex]

The work done by gravitational force is

[tex]W_g = rF_gcos(90)[/tex]

[tex]W_g = 0J[/tex]

PART E) The expression for the total work done is,

[tex]W_{net} = W_f +W_w +W_N+W_g[/tex]

[tex]W_{net} = 330.75-330.75+0+0[/tex]

[tex]W_{net} = 0J[/tex]

Therefore the net work done by the system is 0J

Complete Question:

A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. b) How much kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\

b) what is the speed?

Answer:

a) the kinetic energy gained =  6.42 * 10⁻¹⁶ J

b) the speed of the particle, v = 619328.3 m/s

Explanation:

q = 1.602 *10⁻¹⁹C

V = 4.01 kV  = 4.01 * 10³ V

Work done by the deuteron = qV

Work done by the deuteron = 1.602 * 10⁻¹⁹ *  4.01 *10³

Work done = 6.42 * 10⁻¹⁶ J

Kinetic Energy gained = work done

Kinetic Energy gained by the deuteron =  6.42 * 10⁻¹⁶ J

B) The formula for Kinetic Energy is given by:

KE = 1/2 Mv²

Let the mass of the proton be m₁ = 1.67262 × 10⁻²⁷kg

Let the mass of the neutron be m₂ =   1.67493 × 10−27 kg

M = m₁ + m₂

KE = 1/2 ( m₁ + m₂)v²

Let v = speed of the deuteron

From part (a)

KE = 6.42 * 10⁻¹⁶ J

 1/2 ( m₁ + m₂)v²=  6.42 * 10⁻¹⁶

0.5 * (1.67262 + 1.6749) *10⁻²⁷ * v² =  6.42 * 10⁻¹⁶

v = 619328.3 m/s

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

[tex]\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }[/tex] where [tex]V_{molar} = \frac{A}{\rho }[/tex] an and A is the atomic mass of iron.

Therefore:

[tex]\frac{2}{a^{3} } = \frac{N_{A} * p }{A}[/tex]

This implies that:

[tex]A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }[/tex]

  = [tex]\frac{4}{\sqrt{3} }r[/tex]

Assuming that there is no phase change gives:

[tex]\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }[/tex]

  = 8.60 g/m³

Answer:

[tex]v=12.65\ m.s^{-1}[/tex]

Explanation:

Given:

During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:

[tex]m.\frac{v^2}{r} =\mu.N[/tex]

where:

[tex]\mu=[/tex] coefficient of friction

[tex]N=[/tex] normal reaction force due to weight of the car

[tex]v=[/tex] velocity of the car

[tex]1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)[/tex]

[tex]v=12.65\ m.s^{-1}[/tex] is the maximum velocity at which the vehicle can turn without skidding.

Final answer:

The maximum speed at which a car can safely make a turn on a flat road, given the mass of the car, the turn's radius of curvature, and the coefficient of friction, is approximately 25 m/s. This calculation demonstrates that the car's load does not influence its ability to negotiate the turn safely on a flat surface.

Explanation:

The student is asking how to calculate the maximum speed at which a car can safely make a turn on a flat road, given the car's mass, the turn's radius of curvature, and the coefficient of friction between the tires and the road. First, let's denote the mass of the car as m, gravitational acceleration as g, the radius of curvature as R, and the coefficient of friction as μ. To find the maximum speed v, we use the fact that the centripetal force needed to make the turn must be less than or equal to the maximum static friction force, which is μmg. This gives the condition mv²/R ≤ μmg. Solving for v, we find v = √(μgR).

By plugging in the numbers: m = 1350 kg, R = 71 m, g = 9.8 m/s², and μ = 0.23, we get v = √(0.23 * 9.8 * 71) which calculates to be approximately 25 m/s. Note, because coefficients of friction are approximate, the answer is given to only two digits.

This result is quite significant as it shows that the maximum safe speed is independent of the car's mass due to the proportional relationship between friction and normal force, which in turn is proportional to mass. This implies that how heavily loaded the car is does not affect its ability to negotiate the turn, assuming a flat surface.

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have [tex]m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}[/tex] , where:

[tex]l[/tex] is the length

[tex]m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3[/tex]

We divide the first equation by the second equation to get:

[tex]\frac{m}{R} = \frac{d A^2}{\rho}[/tex]

[tex]A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2[/tex]

Using this Area, we find the diameter of the wire:

[tex]D = \sqrt{\frac{4A}{\pi}}[/tex]

[tex]D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}[/tex]

[tex]D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm[/tex]

To find the length, we multiply the two equations stated initially:

[tex]mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}[/tex]

[tex]l^2 = 8.534\\l = 2.92 \ m[/tex]

Answer:

a) Current in wire 1 = 0.132 A

Current in wire 2 = 0.101 A

b) Resistance of wire 1 = R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ

Resistance of wire 2 = R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ

Explanation:

Current density, J = (current) × (cross sectional area)

Current density for both wires = J = 2950 A/m²

For wire 1,

Cross sectional Area = πr² = π(0.00378²)

A₁ = 0.00004491 m²

For wire 2,

With the assumption that the strands are well banded together with no spaces in btw.

Cross sectional Area = 19 × πr² = π(0.000756)²

A₂ = 0.00003413 m²

Current in wire 1 = I₁ = J × A₁ = 2950 × 0.00004491 = 0.132 A

Current in wire 2 = I₂ = J × A₂ = 2950 × 0.00003413 = 0.101 A

b) Resistance = ρL/A

ρ = resistivity for both wires = (1.69 x 10⁻⁸) Ω.m

L = length of wire = 1.00 m for each of the two wires

A₁ = 0.00004491 m²

A₂ = 0.00003413 m²

R₁ = ρL/A₁ = (1.69 x 10⁻⁸ × 1)/0.00004491

R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ

R₂ = ρL/A₂ = (1.69 x 10⁻⁸ × 1)/0.00003413

R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ

Hope this helps!!

Answer:

0.0399 m

Explanation:

We are given that

Mass of coffee=375g=[tex]\frac{375}{1000}=0.375 kg[/tex]

1kg=1000g

Depth=h=7.5 cm=[tex]7.5\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

Density of coffee=[tex]\rho=1000kg/m^3[/tex]

We have to find the inside radius  of coffee mug.

We know that

[tex]\rho=\frac{m}{V}[/tex]

Substitute the values

[tex]1000=\frac{0.375}{\pi r^2h}[/tex]

[tex]r^2=\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}[/tex]

By using [tex]\pi=3.14[/tex]

[tex]r=\sqrt{\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}}[/tex]

[tex]r=0.0399 m[/tex]

Hence, the inside radius=0.0399 m

Answer:

Analyte⇒ one of analgesics

stationery phase⇒ silica

mobile phase⇒ solvent

Explanation:

during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.

Answer:

The velocity in the smaller arteries will be reduced by a factor of 0.139

Explanation:

The flow rate of blood is going to stay the same when it is transferred from the major artery to the smaller ones.

flow rate = Velocity * Area

Since the flow rate remains constant, we have:

Flow rate in major artery = combined flow rate in smaller arteries

Velocity in Major artery * 1.00 = Velocity in smaller artery * (0.4 * 18)

[tex]V_M * 1 = V_S * (18*0.4)[/tex]

[tex]\frac{V_S}{V_M}=\frac{1}{18*0.4}[/tex]

[tex]\frac{V_S}{V_M}= 0.139[/tex]

Thus, the velocity in the smaller arteries will be reduced by a factor of 0.139

Answer:

[tex]I = 37.5\ A[/tex]

Explanation:

Given,

Magnetic field,B = 0.5 x 10⁻⁴ T

distance,r= 15 cm = 0.15 m

Current = ?

Using Ampere's law of magnetic field

[tex]B = \dfrac{\mu_0I}{2\pi r}[/tex]

[tex]I= \dfrac{B (2\pi r)}{\mu_0}[/tex]

[tex]I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}[/tex]

[tex]I = 37.5\ A[/tex]

Current in the wire is equal to [tex]I = 37.5\ A[/tex]

The maximum current this wire can carry is equal to 37.5 Amperes.

Given the following data:

Scientific data:

In order to determine the maximum current, we would apply Ampere's law of magnetic field.

Mathematically, Ampere's law of magnetic field is given by this formula:

[tex]I=\frac{2B\pi r}{\mu_o }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]I=\frac{2 \pi \times 0.5 \times 10^{-4}\times 0.15}{4\pi \times 10^{-7} }\\\\I=\frac{0.5 \times 10^{-4}\times 0.15}{2\pi \times 10^{-7} }[/tex]

I = 37.5 Amperes.

Read more on magnetic field here: https://brainly.com/question/7802337

Answer:

Explanation:

Intensity of unpolarised light = I₀

intensity after passing through first polariser =  I₀ / 2

Angle between first and second polariser is  φ so

intensity after passing through second polariser

= (I₀ / 2) cos²φ

Now angle between second and third polariser

= 90 - φ

intensity after passing though third polariser

= (I₀ / 2) cos²φ cos²( 90 - φ)

= (I₀ / 2) cos²φ sin²φ

= (I₀ / 8) 4cos²φ sin²φ

= (I₀ / 8) sin²2φ

The intensity after the third polarizer will be:

I₃ = (I₀/8)*sin^2(2φ)

For non-polarized light that passes through any polarizer, we say that the intensity is reduced to its half.

Original intensity = I₀

After the first polarizer, the intensity will be:

I₁ = I₀/2.

Now, when it passes through a polarizer such that the difference in angles with the polarization is x, the new intensity will be:

I₂ = I₁*cos^2(x).

The angle between the second and the first polarizer is φ, then we have:

I₂ = (I₀/2)*cos^2(φ).

Now we also know that the first and the last polarizer are crossed (so there is an angle of 90°). Then if we define θ as the angle between the second and the third polarizer, we will have that:

φ + θ = 90°

then:

θ = 90° - φ

The intensity after the third polarizer will be:

I₃ = (I₀/2)*cos^2(φ)*cos^2(90° - φ)

And we know that:

cos(90° - φ) = sin(φ)

Then we can rewrite:

I₃ = (I₀/2)*cos^2(φ)*sin^2(φ)

But we want a single trigonometric function, then we use the relation:

cos^2(φ)*sin^2(φ) = sin^2(2φ)/4

And replacing that, we get:

I₃ = (I₀/2)*sin^2(2φ)/4 = (I₀/8)*sin^2(2φ)

If you want to learn more about light, you can read:

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Answer:

22.1 V

Explanation:

We are given that

[tex]A_1=0.7 cm^2=0.7\times 10^{-4} m^2[/tex]

[tex]A_2=2.5 cm^2=2.5\times 10^{-4} m^2[/tex]

[tex]A_3=2.2 cm^2=2.2\times 10^{-4} m^2[/tex]

[tex]A_4=3 cm^2=3\times 10^{-4} m^2[/tex]

Using [tex] 1cm^2=10^{-4} m^2[/tex]

We know that

[tex]R=\frac{\rho l}{A}[/tex]

In series

[tex]R=R_1+R_2+R_3+R_4[/tex]

[tex]R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}[/tex]

[tex]R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}[/tex]

Substitute the values

[tex]R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})[/tex]

[tex]R=\rho l(2.62\times 10^4)[/tex]

[tex]V=145 V[/tex]

[tex]I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}[/tex]

Voltage across the 2.5 square cm wire=[tex]IR=I\times \frac{\rho l}{A_2}[/tex]

Voltage across the 2.5 square cm wire=[tex]\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V[/tex]

Voltage across the 2.5 square cm wire=22.1 V

Answer:

Work = N x m

Explanation:

W=Fxd

Work equals force times the distance (displacement).

Answer:

(+2.093 × 10⁻⁷) C

Explanation:

Coulomb's law gives the force of attraction between two charges and it is given by

F = kq₁q₂/r²

where q₁ = charge on one of the two particles under consideration = charge on the balloon = - 3.75 × 10⁻⁸ C

q₂ = charge on the other body = charge on the rod = ?

k = Coulomb's constant = 1/(4 π ϵ₀) = 8.99 × 10⁹ N⋅m²/C²

r = distance between the two charges = d = 6.00 cm = 0.06 m

But for this question, the force of attraction between the charges was enough to lift the balloon and match its weight, Hence,

F = (kq₁q₂/d²) = - mg (negative because it's an attractive force)

m = mass of balloon = 2.00 g = 0.002 kg

g = acceleration due to gravity = 9.8 m/s²

(8.99 × 10⁹ × (-3.75 × 10⁻⁸) × q₂)/(0.06²) = 0.002 × 9.8

q₂ = (-0.002 × 9.8 × 0.06²)/(8.99 × 10⁹ × (-3.75 × 10⁻⁸)

q₂ = + 2.093 × 10⁻⁷ C