A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.

a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above

Answers

Answer 1

Answer:

None of the above

Explanation:

The formula of the magnetic field of a point next to a wire with current is:

B = 2×10^(-7) × ( I /d)

I is the intensity of the current.

d is the distance between the wire and the point.

● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T

A 1.2-m Length Of Wire Centered On The Origin Carries A 20-A Current Directed In The Positive Y Direction.

Related Questions

A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200

Answers

Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!                              

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.

Required:
How far away did the impact occur?

Answers

Answer:

The distance is [tex]d = 193.6 \ m[/tex]

Explanation:

From the question we are told that

   The time interval between the sounds is  k[tex]t_1 = k + t_2[/tex] =  0.50 s

    The  speed of sound in air is  [tex]v_s = 343 \ m/s[/tex]

    The  speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]

 

Generally the distance where the collision occurred is  mathematically represented as

          [tex]d = v * t[/tex]

Now from the question we see that d is the same for both sound waves

 So

        [tex]v_c t = v_s * t_1[/tex]

Now  

So [tex]t_1 = k + t[/tex]

      [tex]v_c t = v_s * (t+ k)[/tex]

=>     [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]t = 0.0645 \ s[/tex]

So

     [tex]d = 3000 * 0.0645[/tex]

     [tex]d = 193.6 \ m[/tex]

       

     

front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire

Answers

Answer:

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

Explanation:

This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.

Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.

Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?

Answers

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

What is pressure?

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

To learn more about the pressure refer to the link;

https://brainly.com/question/356585

The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.

Answers

Answer:

A) it transforms a small force acting over a large distance into a large force acting over a small distance.

Explanation:

The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.

Pressure transmitted P = F/A

where F is the force applied

and A is the area over which the force is applied.

This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.

F/A = f/a

where the left side of the equation is for the output, and the right side is for the input.

The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that

volume V = (area A) x (distance d)

this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.

From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.

A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?

Answers

Answer:

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction

Answers

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that

Answers

Answer:

A) the moment of inertia of the system decreases and the angular speed increases.

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

[tex]I_{1} w_{1} = I_{2} w_{2}[/tex]    ....1

where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.

and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

[tex]I = mr^{2}[/tex]    ....2

where [tex]m[/tex] is the mass of the rotating body,

and [tex]r[/tex] is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

The molecules in Tyler are composed of carbon and other atoms that share one or more electrons between two atoms, forming what is known as a(n) _____ bond.

Answers

Answer:

covalent

Explanation:

covalent bonds share electrons

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?

Answers

Answer:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?

Answers

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

The amount of charge after 5 seconds is:

5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

Convert 76.2 kilometers to meters?

Answers

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?

Answers

Answer:

Explanation:

According to Equations of Projectile motion :

[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

[tex]Horizontal Range = vcosx * t[/tex]

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

An electric train operates on 800 V. What is its power consumption when the current flowing through the train's motor is 2,130 A?

Answers

Answer:

1704 kW

Explanation:

To solve for the power consumed by the trains motor we have to employ the formula for power which is

Power= current * voltage

Given that

voltage V= 800 V

current I= 2130 A

Substituting in the formula for power we have

Power= 2130*800=  1704000 watt

Power = 1704 kW

This is the amount of energy consumed, transferred or converted per unit of time

Hence the power consumed  by the trains motor is 1704 kW

A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.

Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.

E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.

Answers

Answer:

A. 990v/m

B.330x10^-8T

C.2.19x10^-6J/m³

D.1.45x10^-11J

Explanation:

See attached file

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish

Answers

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

The apparent depth of the fish is 60 cm.

To calculate the apparent depth of the fish, we use the formula below.

Formula:

R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1

Where:

R.F = Refractive index of water

Make D' The subject of the equation.

D' = D/R.F................... Equation 2

From the question,

Given:

D = 80 cmR.F = 1.333

Substitute these values into equation 2

D' = 80/1.33D' = 60.01D' = 60 cm

Hence, the apparent depth of the fish is 60 cm

Learn more about apparent depth here: https://brainly.com/question/24319677

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?

Answers

Answer:

6000 counts per second

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC

Answers

Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

PLEASE HELP ANSWER FAST As the vibration of molecules decreases, the _____ of the substance decreases. 1.temperature 2.internal energy 3.kinetic energy 4.all of the above

Answers

I think 1 and 3 is absolutely right but im not sure about number 2.
I think the answer is 4 all of the above because as the vibration decrease automatically the kinetic energy decrease and the temperature is decrease because when the vibration of molecules decrease thats mean the substances is slightly become a solid and you can get a solid cube of liquid if you freeze them

Convert 7,348 grams to kilograms

Answers

0.00735 kilograms is your answer. I’m sorry If I didn’t explain it right :(,

PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.

Answers

Answer:

1.The temperature of each sample will increase by the same amount

Explanation:

This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.

This is shown by the calculation below

Q = mcΔT

ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.

Since Q, m and c are the same for both substances, thus ΔT will be the same.

So, the temperature of each sample will increase by the same amount

Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion​

Answers

Last point, Newton’s first law of motion is the correct answer.

Hope this helps ya

Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors

Answers

Answer:

it is going to D. all of these are resistors

Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.

Answers

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)

Answers

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______

Answers

Answer:

-z axis

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T

Answers

Answer:

The speed of the proton is 4059.39 m/s

Explanation:

The centripetal force on the particle is given by;

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

The magnetic force on the particle is given by;

[tex]F = qvB[/tex]

The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

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

where;

r is the radius of the circular path moved by both electron and proton;

⇒For electron;

[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]

⇒For proton

The speed of the proton is given by;

[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]

Therefore, the speed of the proton is 4059.39 m/s

A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)

Answers

Answer:

Change in the frequency (in Hz) = 104.96 Hz

Explanation:

Given:

Speed of sound in air (v) = 343 m/s

Speed of car (v1) 36 m/s

Frequency(f) = 500 Hz

Find:

Change in the frequency (in Hz)

Computation:

Frequency hear by the observer(before)(f1) = [f(v+v1)] / v

Frequency hear by the observer(f1) = [500(343+36)] / 343

Frequency hear by the observer(f1) = 552.48 Hz

Frequency hear by the observer(after)(f2) = [f(v-v1)] / v

Frequency hear by the observer(f2) = [500(343-36)] / 343

Frequency hear by the observer(f2) = 447.52 Hz

Change in the frequency (in Hz) = f1 - f2

Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz

Change in the frequency (in Hz) = 104.96 Hz

The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circular orbit as well (mean distance is 19,600 km ).
Earth Moon Pluto Charon
Mass (kg) 5.97 x 10^24 0.07342 x 10^24 0.0146 x 10^24 0.00162 x 10^24
Equatorial radius (km) 6378.1 1738.1 1185 604
Which object exhibits the longest orbital period? Hint: perform order of magnitude analysis.
a. Moon around Earth
b. Charon around Pluto
c. About the same for both

Answers

Answer:

a. Moon around Earth.

Explanation:

Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.

The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.

The object that should exhibit the longest orbital period is option a. Moon around Earth.

What is Charon's orbit?

Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.

learn more about orbit here: https://brainly.com/question/25404554

If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b

Answers

Answer:

Will be equal to alpha x r; less than UsN

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