How can magnetic levitation be improved?

According to the Newton's second law :- The acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Answer:

D) The bicycle is not in motion.

Explanation:

Study the position-time graph for a bicycle.

Which statement is supported by the graph?

A) The bicycle has speed but not velocity.

B) The bicycle is moving at a constant velocity.

C) The bicycle has a displacement of 3 m.

D) The bicycle is not in motion.

Solution:

Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.

Speed is the time rate of change of distance. It is the ratio of distance to time taken.

From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.

From the position-time data provided, it can concluded that the bicycle is not in motion.

Motion of a body involves a change in the position of that body with time.

A body in motion is constantly changing position or orientation as time passes.

The body may move with constant velocity/speed or changes in its velocity.

A position-time graph provides information about the motion of a body.

From the data provided:

The position of the bicycle remains the same for all time intervals.

Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.

Learn more about motion and position-time graph at: https://brainly.com/question/2356782

Answer:

v = 6491.94 m/s

Explanation:

We are given;

Frequency; f = 87 Hz

Wavelength;λ = 74.62 m

Formula for velocity(v) of waves from the wave equation is;

v = fλ

Thus;

v = 87 × 74.62

v = 6491.94 m/s

Answer:

the period of oscillation of the given object is 0.14 s

Explanation:

Given;

mass of the object, m = 3 kg

extension of the spring, x = 0.085 m

The spring constant is calculated as follows;

[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]

The angular speed of a 4 kg object is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]

Therefore, the period of oscillation of the given object is 0.14 s

Answer:

F' = 73.7 N

F = 8.749×10⁻³ N

a' = a =  9.83 m/s²

Explanation:

(a)

For the rock

Applying

F' = Gm'm/r²................... Equation 1

Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.

From the question,

Given: m' = 7.5 kg

Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m

Substitute these values into equation 1

F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²

F' = 7.37×10¹ N

F' = 73.7 N

Also, For the pebble,

F = GMm/r².............. Equation 2

Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth

Given: M = 8.9×10⁻⁴ kg,

Substitute into equation 2

F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²

F = 8.749×10⁻³ N

(b)

For the rock,

a' = F'/m'

Where a' = magnitude of the acceleration of the rock

a' = 73.7/7.5

a' = 9.83 m/s²

For the pebble,

a = F/M

Where a = acceleration of the pebble

a = (8.749×10⁻³)/(8.9×10⁻⁴)

a = 9.83 m/s²

M1 = 15/g = 15/9.8 = 1.53 kg = mass of object in air. M2 = 12/9.8 = 1.22 kg = mass of object immersed. M1-M2 = 1.53-1.22 = 0.31 kg lost by object = mass of water displaced. ... Do = 4.94 g/cm^3 = density of object.

Answer:Alternativa A.   Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.

Answer:

Moving electrons always create a magnetic field. Electrons moving along a wire make a magnetic field that goes in circles around the wire. When you bend the wire into a coil, the magnetic fields around each loop of the coil add up to make a long , thin magnet with north at one end and south at the other.

Explanation:

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = [tex]e^{kt}[/tex]

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = [tex]e^{-0.01t}[/tex]

V = V₀[tex]e^{-0.01t}[/tex]

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀[tex]e^{-0.01t}[/tex]

0.1V₀ = V₀[tex]e^{-0.01t}[/tex]

0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]

0.1 = [tex]e^{-0.01t}[/tex]

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

Answer:

the first answer

Explanation:

(32°F − 32) × 5/9 = 0°C

Answer:

Answer: A

Explanation:

d. An object did work on the environment.

Work is defined in many contexts. Some of these are;

i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.

ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.

In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).

Answer:

its c

Explanation:

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

6.28 m/s.

Explanation:

Given that,

The mass of the object, m = 0.2 kg

The radius of the circle, r = 0.5 m

It takes the object 0.500 s to complete one revolution.

We need to find the translational speed of the object. Let it is v. We know that,

[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]

So, the transalational speed of the object is 6.28 m/s.

Answer:

[tex]\lambda=5.46\times 10^{-8}\ m[/tex]

Explanation:

The hottest ordinary star in our galaxy has a surface temperature of 53,000 K.

We need to find the peak wavelength of its thermal radiation.

Using Wein's law,

[tex]\lambda T=2.898\times 10^{-3}\\\\\lambda=\dfrac{2.898\times 10^{-3}}{53000}\\\\=5.46\times 10^{-8}\ m[/tex]

So, the peak wavelength of its thermal radiation is equal to [tex]5.46\times 10^{-8}\ m[/tex].

Answer:

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),

Explanation:

Answer:

The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]

Explanation:

This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:

[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)

Where:

[tex]m[/tex] - Mass of the car, in kilogram.

[tex]v[/tex] - Initial velocity, in meters per second.

[tex]\mu[/tex] - Coefficient of friction, no unit.

[tex]s[/tex] - Travelled distance, in meters.

Then we derive an expression for the distance travelled by the vehicle:

[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]

[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]

As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]

Answer:

W = 16.4 kJ

Explanation:

Given that,

There are 135 steps from the ground floor to the sixth floor.

Each step is 16.6 cm tall.

The mass of a person, m = 73.5 kg

We need to find the work done by the person. We know that,

Work done = Fd

Where

d is the displacement, d = 135 × 0.166 = 22.41

So,

W = 73.5 × 10 × 22.41

= 16471.35  J

or

W = 16.4 kJ

So, 16.4 kJ is the work done by the person.

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

it's a form of government where people elect their representatives

Answer:

The word democracy itself means rule by the people.

Answer:

t = 6 minutes

Explanation:

Given that,

Force,[tex]F=6.8\times 10^5\ N[/tex]

Initial speed of the train, u = 0

Final speed of the train, v = 80 km/h = 22.22 m/s

The mass of the train, [tex]m=1.1\times 10^7\ kg[/tex]

We need to find the time taken by the train to come to rest. We know that,

F = ma

[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s[/tex]

or

t = 6 minutes (approx)

So, the required time is equal to 6 minutes.

Answer:

No of proton is 13 and nucleus is 13

Answer:

[tex]I_2=0.50 w/m^2[/tex]

Explanation:

From the question we are told that:

initial Intensity [tex]I_1=0.020 w/m^2[/tex]

Final Electric field [tex]E_2=5E[/tex]

Generally the equation for Relation ship between intensity and Electric field is mathematically given by

 [tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]

Therefore

 [tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]

 [tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]

 [tex]I_2=0.50 w/m^2[/tex]

Answer:

The angle is 4.1 rad.

Explanation:

The centripetal acceleration (α) is given by:

[tex] \alpha = \omega^{2} r [/tex]    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

[tex] a = \alpha r [/tex]      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

[tex] \omega^{2} r = 8.2\alpha r   [/tex]

[tex] \omega^{2} = 8.2\alpha [/tex]    (3)      

Now, we can find the angle with the following equation:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]

Where:

[tex] \omega_{f}[/tex]: is the final angular velocity                                                                              [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)

[tex]\Delta \theta[/tex]: is the angle

[tex] \omega^{2} = 2\alpha \Delta \theta [/tex]     (4)    

By entering equation (3) into (4) we can calculate the angle:

[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]

[tex] \Delta \theta = 4.1 rad [/tex]

Therefore, the angle is 4.1 rad.

I hope it helps you!