A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answers

Answer 1

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

What's the expression of magnetic field inside a solenoid?Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×In = no. of turns per unit length and I = current through the solenoidWhat's is the magnetic field inside the solenoid here?Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

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Related Questions

A 5.0 kg box slides down a 4.0 m long ramp that makes a 25 angle with the ground. If the coefficient of kinetic friction is 0.65, how much thermal energy was produced?

Answers

The thermal energy was produced is 116J

What is the thermal energy produced?

Now we know that the frictional force produces the energy that is lost as heat as the body slides down the incline. The magnitude of the frictional force is obtained from;

Ff= μmgcosθ

Ff =  0.65 *  5.0 kg * 9.8 m/s^2 * cos 25

Ff = 29 N

Hence, the thermal energy is;

29 N * 4.0 m = 116J

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Calculate the gravitational force of the Earth and Moon. The Earth has a mass of 5.972x 1024 kg and the Moon has a mass of 7.348x 1022 kg. They are an average of 384 million meters apart.

Answers

Answer:

The force of gravitation is 1.98×10^(20) m.

Explanation:

Note: The value of m1, m2, r, and G is taken to be 5.97×10^(24) kg, 7.348×10^(22) kg, 3.84×10^(8) m, and 6.67×10^(-11) N m^2/ kg^2 respectively.

The force of gravitation F acting between two masses is calculated using the formula,

F=Gm1m2/ r^2

where G is the gravitational constant.

Hence force F is

F=6.67×10^(-11)×5.97×10^(24)×7.348×10^(22)/ (3.84×10^(8))^2

F=1.98×10^(20) N

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A typical muscle fiber is 2.0 cm long and has a cross-section area of 3.1 * 10-9 m2. When the muscle fiber is stimulated, it pulls with a force of 1.2 mN. What is the work done by the muscle fiber as it contracts to a length of 1.6 cm

Answers

= F.d

F = 1.2 x 10-3N

d= (2-1.6)10-2=0.004m

W=1.210-3(0.004)=4.8 x 10 -6 J

A typical muscle fiber is 2.0 cm long and has a cross-section area of 3.1 * 10-9 m2. When the muscle fiber is stimulated, it pulls with a force of 1.2 mN.

Please check the attached file for a brief answer.

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The work done by the muscle fiber is 4.80x 10 J

The work done by the muscle fiber as it contracts to a length is

W = FAL

= (1.20x103 N) (0.4x10-2 m)

= 4.80x 10 J

Muscle tissues contain something known as muscle fibers. Muscle fibers consist of a single muscle cellular.

They assist to control the physical forces in the body. When grouped collectively, they are able to facilitate organized movement of your limbs and tissues.

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A 18 kg box slides from rest down a ramp inclined at 25° to the horizontal onto a spring with a spring constant 740 N/m
as shown in the diagram. The spring is compressed 0.32 m before the box stops. Determine how far along the ramp the
box will slide before it stops. Assume friction is negligible.

Answers

The distance traveled by the box along the ramp is 0.51 m.

Distance traveled by the box

Apply the principle of conservation of energy;

Potential energy of the box at top of incline = Elastic potential energy at bottom

mgh = ¹/₂kx²

where;

h is the height of the incline m is mass of the boxx is the compression of the springk is spring constant

h = L sinθ

where;

L is length of the incline = distance traveled by the box

mg(L sinθ) = ¹/₂kx²

(18)(9.8)(L sin25) = ¹/₂(740)(0.32)²

74.55 L = 37.89

L = 37.89 / 74.55

L = 0.51 m

Thus, the distance traveled by the box along the ramp is 0.51 m.

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Two boys are playing with two different balls of masses m and 2 meter respectively. if first boy through vertically up and the second boy throws at an angle teta from vertical and balls remains same time in air what will be the ratio of height attend by the two balls

Answers

The ratio of the height attended by the object thrown upward to that thrown at some angle θ with the vertical is 1.  

Note: It is assumed that the mass of the first object is m and the mass of the second object is 2m. Also, the first object is thrown upward and the second object is thrown at some angle θ with the vertical.

Projectile motion: Any object that is thrown in the air is called a projectile and the motion described by it under gravity is called the projectile motion.  

If the air resistance is neglected, then the acceleration due to gravity g is the same for all objects irrespective of their masses. It is given that both objects remain in the air for the same time period. So first calculate the time period for the objects when they are in the air.

Time period of the first object: The first object is thrown in the air upwards so from the second kinematics equation,

h1=u1*t1-(1/2)*gt1^2

where h1 is the height, t1 is the time, and u1 is the initial velocity for the first object.

When the object is not in air h1=0, so

0=u1*t1-(1/2)*gt1^2

After solving the above quadratic equation, the values of t1 obtained are t1=0 which represents the initial time, and t1=2u1/g which represents the time period. So the time period of the first object is,

t1=2u1/g

Time period of the second object: The second object is thrown at some angle θ with the vertical as shown in the diagram. From the diagram, the initial velocity along the vertical direction is,

u2=uocos(θ)

where uo is the initial velocity and u2 is the initial velocity along the vertical direction.

From the second kinematic equation,

h2=u2*t2-(1/2)*gt2^2

where h2 is the height, t2 is the time, and u2 is the initial velocity for the second object along the vertical direction.

When the object is not in air h2=0, so using u2=uocos(θ),

0=uocos(θ)*t2-(1/2)*gt2^2

After solving the above quadratic equation, the values of t2 obtained are t2=0 which represents initial time, and t2=2uocos(θ)/g which represents  the time period. So the time period of the second object is,

t2=2uocos(θ)/g

Given that the time period is the same for both cases,

t1=t2

2u1/g=2uocos(θ)/g

u1=uo cos(θ)

Calculation of the ratio of the height of the object:

The maximum height is attained when the time of the object in the air is half of the total time period. At maximum height, velocity is zero.

From the third equation of motion.

v^2=u^2-2gh

h=u^2/2g

where h is the height and u is the initial velocity of an object.

Using it and u1=uo cos(θ) and u2=uo cos(θ), the ratio of h1 and h2 is,

h1/h2=  u1^2/2g÷u2^2/2g

h1/h2=(uo cos(θ))^2/(uo cos(θ))^2

h1/h2=1

The ratio of their height will be 1.

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A fluorescent light "hums" at
120 Hz. A bored grad student
hums at 118.4 Hz. What beat
frequency does he hear in his
head?
(Unit = Hz)
(This is on autobiographical problem buthe

Answers

The beat frequency he hears in his head will be 1.6 Hz

What is the frequency of the sound?

A sound pressure wave's frequency, also known as pitch, is the number of times it repeats itself every second.

The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.

Given data;

The frequency at which the fluorescent light "hums",f=120 Hz

The frequency at which the bored grad student "hums",f'=118.4 Hz

The beat frequency he hears in his head is,fₐ=?

The beat frequency he hears in his head is found as;

fₐ = f-f'

fₐ = 120 - 118.4 Hz

fₐ = 1.6 Hz

Hence the beat frequency he hears in his head will be 1.6 Hz.

At 120 Hz, a fluorescent light "hums." At 118.4 Hz, a Ph.D. student who is bored hums. He hears 1.6 Hz as the rhythm frequency in his brain.

Hence the beat frequency he hears in his head will be 1.6 Hz

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Use the graph to complete the sentences about two runners.

Runner A is
than runner B.

Runner
has a head start.

Answers

Runner A is faster than runner B and Runner B has a head start.

What is a velocity time graph?

The velocity time graph shows us how fast that an object can be. The slope of the graph is the acceleration.

If we look at the graph as shown in the image attached to this answer, we can see that Runner A is faster than runner B and Runner B has a head start.

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Answer:

Runner A is

✔ faster

than runner B.

Runner

✔ B

has a head start.

Explanation:


What are two forces on earth that could change the motion of the object?

Answers

Answer:

Force makes things move or, more accurately, makes things change their motion. Two natural forces that we have experienced are the force of gravity and magnetic forces magnetic forces. These two forces act at a distance and do not require direct contact between the objects to function..

I Explain why you feel cold when tap or well water in winter. state with reason.

Answers

Answer:

We feel cold when tap or well water in winter because heat flows from hot body to cold body.

Explanation:

Our body is in optimal status is a hot body and tap or well water is a cold body. Therefore we feel cold.

When we drink tap or well water in the winter, we feel chilled because heat transfers from a hot body to a cold body.

What is convection?

Convection, often known as a heat transfer mechanism, is a method of moving heat. When a fluid, such as air or water, is moving, this heat transfer takes place. Temperature differences within the fluid cause convection.

Due to the increased kinetic energy of the particles, a fluid loses density at the area nearest to the heat source as it is heated. The less dense portion of the fluid then rises, while the denser portion sinks.

The process keeps happening because the less dense fluids sink as they move away from their heat source while the denser fluids rise as they move toward the heat source. Convection currents are produced by this.

Therefore, When we drink tap or well water in the winter, we feel chilled because heat transfers from a hot body to a cold body.

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The primary coil of a transformer is connected to a 120 V wall outlet. The secondary coil is connected to a lamp that dissipates 60W. What is the rms current in the primary coil

Answers

The rms current in the primary coil of a transformer is 0.5 Ampere .

Energy/power is not gained or lost going through an ideal transformer.

Consequently, the transformer in this situation doesn't really matter. The entire device receives 60 watts of power from the wall outlet if the lamp consumes energy at a rate of 60 watts.

∴ Power = (voltage) x (current)

60 watts = (120 v) x (current)

Current = (60 watts) / (120 v)

Current = 0.5 Ampere

Therefore, the rms current in the primary coil of a transformer is 0.5 ampere.

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An unstoppable object is heading right toward an unmovable object. What's going to happen?

Answers

Answer:

In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.

Hope this helps.Good luck ✅.

A toy dart gun uses a spring with a spring constant of 35 N/m. To use the dart gun, you compress the spring by pushing in a dart of mass 0.002 kg.
a. If you compress the spring by 0.02 m, what force is the spring exerting on the dart?

b. With the spring compressed 0.02 m, how much elastic potential energy is stored in the spring?

c. If you release the spring, it pushes the dart forward. What is the kinetic energy of the dart when it reaches the natural length of the spring? Explain how you found the answer.

d. What is the speed of the dart when it reaches the natural length of the spring?

Answers

(a) The force is the spring exerting on the dart is 0.7 N.

(b) The elastic potential energy stored in the spring is 7 x 10⁻³J.

(c) The kinetic energy of the dart when it reaches the natural length of the spring is 7 x 10⁻³J.

(d) The speed of the dart when it reaches the natural length of the spring is 2.65 m/s.

Force the spring exerting on the dart

F = kx

where;

k is spring constantx is compression of the spring

F = 35 N/m x 0.02 m

F = 0.7 N

Energy stored in the spring

U = ¹/₂kx²

U = ¹/₂(35)(0.02)²

U = 7 x 10⁻³ J

Kinetic energy of the dart

The elastic potential energy of the spring will be converted into kinetic energy of the dart, with a magnitude of  7 x 10⁻³ J when it reaches natural length of the spring.

Speed of the dart

K.E = ¹/₂mv²

v² = (2K.E)/m

where;

v is the speed of the dartm is mass of the dart

v² = (2 x 7 x 10⁻³)/(0.002)

v² = 7

v = 2.65 m/s

Thus, the force is the spring exerting on the dart is 0.7 N.

The elastic potential energy stored in the spring is 7 x 10⁻³J.

The kinetic energy of the dart when it reaches the natural length of the spring is 7 x 10⁻³J.

The speed of the dart when it reaches the natural length of the spring is 2.65 m/s.

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On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance
does the bus cover in this time frame?
A. 64 m
B. 112 m
C. 80 m
D. 144 m

Answers

Answer B. 112 m



Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m

A spring is hung from the ceiling. A 2.0-kg mass suspended hung from the spring extends it by 6.0 cm. A downward external force applied to the mass extends the spring an additional 10 cm. What is the work done by the force

Answers

The work done by force on a spring hung from the ceiling will be 1.67 J

Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.

Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.

We need to find the work done by the force

Given mass is of 2 kg

So let,

F = 2 kg

x = 0.1 m

Stiffness of spring = k = F/x

k = 20/0.006 = 333 n/m

Now the formula to find the work done by force will be as follow:

Workdone = W = 0.5kx²

W = 0.5 x 333 x 0.1²

W = 1.67 J

Hence the work done by force on a spring hung from the ceiling will be 1.67 J

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Which of the following graphs represents the equation y-2 = 3(x-1)?

Answers

Answer:

Graph b

Explanation:

I used Desmond calculator which is a calculator where you can graph an equation! :)

Hello!

The slope-intercept form is :

y - 2 =  3(x - 1)

y - 2 = 3x - 3

y = 3x - 1

Two points that are found to be on the line :

(0, -1) and (1, 2)

∴ The graph which contains both points is Graph B.

lamp 1 and lamp 2 are in a series circuit
lamp 1 stops working
explain why lamp 2 also stops working​

Answers

Lamps 1 and 2 are connected in series. lamp 1 stops working because the circuit is not complete.

What is a series circuit?

In a series circuit, the amount of current flowing through any component is the same and the sum of the individual resistances equals the overall resistance of any series circuit.

The voltage in a series circuit is equal to the total of the individual voltage drops.

For the bulb to glow the circuit should be complete. If the circuit is not completed the bulb will not glow.

Lamps 1 and 2 are connected in series. lamp 1 stops working because the circuit is not complete.

Hence, lamp 1 stops working.

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An uncharged metal rod brought close to but not touching the cap of a charged electroscope caused decrease in the divergence of the leaf. Explain this observation.

Answers

-ItzKenzo

-CuteSiKenzo(Oo cute)

Alam kung wala kang pake tángina mo!!

Suppose that the engine of a 1,500 kg automobile has a maximum power output of 30 hp. What is the maximum grade (in percent) that the automobile can climb at 41 km/h if the drag force on it is 410 N

Answers

The maximum grade (in percent) is 14%.

Suppose angle be theta.

Power = force * velocity

35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)

sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139

grade = tan ( arcsin 0.139 )

= 0.140

= 14%

Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.

For instance drag on a ship moving in water or drag on an aircraft shifting in the air. therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.

Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be motion among the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid movements beyond a static stable item.

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The maximum grade (in percent) is 14%.

Suppose angle be theta.

Power = force * velocity

35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)

sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139

grade = tan ( arcsin 0.139 )

= 0.140

= 14%

Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.

For instance drag on a ship moving in water or drag on an aircraft shifting in the air. Therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.

Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be a move between the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid moves beyond a static stable item.

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What is typically responsible for forming volcanic island chains in the oceans?
A. Shield volcanoes
B. Fissures
Composite volcanoes
Hot spot volcanism

Answers

Hot spot volcanism is responsible for the formation of volcanic island chains in the oceans and is denoted as option D.

What is a Volcano?

This is a rupture in the earth crust thereby giving rise to the release of hot lava, gases etc from below the surface.

The pushing of the lava through the cracks partially melts the plate at a subduction zone thereby forming the volcanic island chains.

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A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its speed is _______ and its acceleration is _______.

Answers

When displacement is 0, an object is at its equilibrium position, at that moment, its speed is maximum and its acceleration is zero.

What is speed?

Speed is a scalar quantity used in physics to indicate the distance of change of an object that travels in any direction.

Conversely, displacement and acceleration are vector quantities because they express magnitudes but not a direction.

In conclusion, when an object is at its equilibrium position its speed is maximum and its acceleration is zero.

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A local FM radio station broadcasts at an energy of kJ/photon. () Calculate the frequency at which it is broadcasting. Frequency

Answers

The frequency is 94.2 Hz

We  are given the energy of a photon = 6.24x10-29 kJ

We  are asked to find the frequency (ν),

We use the equation E = hν where E is energy, h is Planck's constant and ν is the frequency

The quantity of events per unit of time that occur when an event repeats itself at regularly spaced intervals is known as its frequency.

6.24x10-29 kJ = 6.626x10-34 J-sec * ν

ν = 6.24x10-29 kJ x 1000 J/kJ  / 6.626x10-34 Jsec

ν = 6.24x10-26 J / 6.626x10-34 Jsec

ν = 0.942x108 s-1 = 9.42x107 s-1 = 9.42x107 Hz

9.42x107 Hz x 1 MHz/106 Hz = 94.2 MHz

Hence the frequency is 94.2MHz

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A toroid has a 4.63 cm square cross section, an inside radius of 15.0 cm, 527 turns of wire, and a current of 0.725 A. What is the magnetic flux through the cross section

Answers

Using the equation, B = (μoIN/2πr)

The inner radius is r = 16.2 cm,

so the field there is

B = (4π 10-7).(0.813).(535)/2π(0.162)

= 5.37 × 10-4 T

The outer radius is r = 16.2 + 5.2 = 21.4 cm.

The field there is B = (4π 10-7).(0.813).(535)/2π(0.214)

= 4.06 × 10-4 T.

A toroid having a square cross-section, 5.20cm on edge and an inner radius of 16.2cm has 535 turns and carries a current of 813mA to calculate the magnetic.

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A projectile is launched with a velocity of 13.2 m/s at an angle of 37.0° above the horizontal.

How high above the point from which it was fired will the projectile be 1.0 s after firing?

a.
3.0 m


b.
12.8 m


c.
5.6 m


d.
13.2 m

Answers

The height reached by the projectile from the point of projection is determined as 12.84 m.

Height reached by the projectile

H = vt + ¹/₂gt²

where;

v is vertical velocityt is time of motiong is acceleration due to gravity

H = (13.2 x sin37)(1) + (0.5)(9.8)(1)²

H = 12.84 m

Thus, the height reached by the projectile from the point of projection is determined as 12.84 m.

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A box having a mass of 7.3 kg is lifted 0.60 m from ground level to the top of a desk. What is the gravitational potential energy in J of the box on the desktop?

Answers

Considering the definition of potential energy, the potential energy of the box is 42.9678 J.

Potential energy

Potential energy is the energy that measures the ability of a system to do work based on its position. In other words, this is the energy that a body located at a certain height above the ground has.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the strength of gravity.

So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep= m×g×h

Where:

Ep is the potential energy in joules (J).m is the mass in kilograms (kg).h is the height in meters (m).g is the acceleration of fall in m/s² (approximately 9.81 m/s²).

Potential energy in this case

In this case, you know:

m= 7.3 kgg= 9.81 m/s²h= 0.60 m

Replacing in the definition of potential energy:

Ep= 7.3 kg × 9.81 m/s² × 0.60 m

Solving:

Ep= 42.9678 J

Finally, the potential energy of the box is 42.9678 J.

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A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed

Answers

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

402  =  u*5  + (1/2)*a*5^2

10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W  = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W  =  3987840. J

Therefore power rating of the dragster is given by,

P  ⇒  W/t. =  3987840/5 = 797568 watt.

P  ⇒ 797568/746 =  1069.1 hp.

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The answer is 1069.1 hp.  

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

 402  =  u*5  + (1/2)*a*5^2

 10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at^.

 130 = u + 5*a^

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work -energy theorem,

Work done = change in kinetic energy

 W  = (1/2)*m*(v^2 - u^2)

,  W = (1/2)*500*(130^2 - 30.8^2)

,  W  =  3987840. J

therefore Power rating of the dragster is given by,

P  =  W/t. =  3987840/5 = 797568 watt.

,  P  = 797568/746 =  1069.1 hp.        

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Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.

Answers

The deceleration is -8 m/s² and the time taken by the car is 2.5 s.

It is given that,

Speed, v = 8 m/s

Distance, d = 20m

We need to find the time taken by the car and deceleration.

As we know speed is defined as

v = d/t

Plugging in the values,

t =  20/8

t = 2.5s

Now from the deceleration formula,

a =  - v/ t

a is the deceleration, v is the speed and t is the time taken.

a = - 20/ 2.5

a = - 8 m/s²

Thus, the time taken by the car and deceleration is 2.5 s and -8 m/s² respectively.

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A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 m/s. (a) if the disk rolls up a 30.0o ramp, how far along the ramp will it move before it stops? (b) explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

Answers

(a) By using the law of conservation of energy, the distance moved by the solid disk along the ramp is 0.96 m.

(b) The answer does not depend on the mass and radius of the disc as these values are canceled when the law of conservation of energy is applied.

What is the law of conservation of energy?

The law of conservation of energy states that the total energy of an isolated system is conserved.

The total initial energy Ei of the disc rolling down an inclined plane is given by the formula,

Ei=1/2*mv^2 + 1/2*Iω^2

where m is the mass of the disc, v is the velocity of the disc, I is the moment of inertia and ω is the angular velocity of the disc.

For a solid disc, I=1/2mr^2, and since it is rolling without slipping, the rolling velocity of the disc will be equal to its translational velocity, that is,

v=ωr or ω=v/r

So using ω=v/r and I=1/2mr^2, it can be written,

Ei=1/2*m(v)^2 + 1/2*(1/2mr^2)(v/r)^2

Ei=1/2*m(v)^2*(1+1/2*)

Ei=3/4*mv^2

If the height covered by the disc is h before stopping, then its final total  energy Ef will be equal to the potential energy, that is,

Ef=m*g*h

From the law of conservation of energy, it can be written,

Ei=Ef

3/4*mv^2=m*g*h

h=3v^2/(4g)

The length of the ramp is then given by the formula,

l=hcosecθ

where θ is the inclination angle. So

l= 3v^2/(4g)*cosecθ

Here g=9.8 m/s^2, v=2.5 m/s  and θ=30.0 degree. Using these values,

l= 3*(2.5 )^2/(4*9.8)*cosec( 30)

l= 0.96 m

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The magnetic field at the center of two concentric current loops is zero. The smaller loop has a radius of 0.01 m and a current of 10.0 A. The larger current loop carries a current of 29.0 A. What is the radius of the larger loop

Answers

Answer:

B = μ I / (2 * R)     magnetic field at center of current carrying loop

I1 / R1 = I2 / R2        and we want to know R2

R2 = I2 / I1 * R1 = 29 / 10 * 01 = .029 m

Please answer number 3!! High-speed stroboscopic photographs show that the head of a 0.2kg golf club is traveling
at 55 m/s just before it strikes a 0.046kg golf ball at rest on a tee. After the collision, the
golf club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball
just after impact.

Answers

The velocity of the golf ball is 65.2 m/s

What is the principle of conservation of momentum?

The principle of conservation of momentum is that the momentum before collision is equal to the momentum after collision.

Given that;

(0.2kg *  55 m/s) + (0.046kg * 0 m/s) = (0.2kg * 40 m/s) + (0.046kg  * v)

11 = 8 + 0.046v

v = 11 - 8/0.046

v = 65.2 m/s

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A satellite orbits the earth at a speed of 8.0x 10^3m/s . Calculate the period of orbit of the satellite if the distance between the centre of the earth and the satellite is 6400km?​

Answers

Hi there!

The period of an orbit can be found by:

[tex]T = \frac{2\pi r}{v}[/tex]

T = Period (? s)
r = radius of orbit (6400000 m)

v = speed of the satellite (8000 m/s)

This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.

Let's plug in what we know and solve.

[tex]T = \frac{2\pi (6400000)}{8000} = \boxed{5026.55 s}[/tex]

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