An inductor has an inductance of 0.053 H. The voltage across this inductor is 48 V and has a frequency of 270 Hz. What is the current in the inductor?

Answer:

A

Explanation:

So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.

Answer:

Time to move out of the way = 1.74 s

Explanation:

Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.

Time to move out of the way = 1.74 s

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

Answer:

a)  [tex]F=475.7Hz[/tex]

b)  [tex]F'=410.899Hz[/tex]

Explanation:

From the question we are told that:

Velocity of eagle [tex]V_1=35m/s[/tex]

Frequency of eagle [tex]F_1=440Hz[/tex]

Velocity of Black bird [tex]V_2=10m/s[/tex]

Speed of sound [tex]s=343m/s[/tex]

a)

Generally the equation for Frequency is mathematically given by

 [tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]

 [tex]F=440(\frac{343-10}{343-35})[/tex]

 [tex]F=475.7Hz[/tex]

b)

Generally the equation for Frequency is mathematically given by

 [tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]

 [tex]F'=440(\frac{343+10}{343+35})[/tex]

 [tex]F'=410.899Hz[/tex]

Consulta los dibujos adjuntos

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m[tex]_g[/tex]g = 0

where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m[tex]_g[/tex]g = 0

solve for m[tex]_g[/tex]

m[tex]_g[/tex] = ( Vρg - Mg ) / g

m[tex]_g[/tex] =  Vρg/g - Mg/g

m[tex]_g[/tex] =  ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m[tex]_g[/tex] / μ

where μ is the molecular mass of air

so

PV = (m[tex]_g[/tex]/μ)RT

solve for T

μPV = m[tex]_g[/tex]RT

T = μPV / m[tex]_g[/tex]R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T =  1405920 / 1979.442

T =  710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

The temperature required for the air to be heated is 710.26 K.

Given data:

The mass of a hot air-balloon is, m = 381 kg.

The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].

The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].

The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].

The condition where the hot air balloon is just about to take off is as follows:

[tex]F-mg - m'g =0[/tex]

Here,

m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,

[tex]F = V \times \rho \times g[/tex]

Solving as,

[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]

Now, apply the ideal gas law as,

PV = nRT

here, R is the universal gas constant and n is the number of moles and its value is,

[tex]n=\dfrac{m'}{M}[/tex]

M is the molecular mass of gas. Solving as,

[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]

Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,

[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]

Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.

Learn more about the ideal gas equation here:

https://brainly.com/question/18518493

Answer:

[tex]l=12916.5m[/tex]

Explanation:

Distance [tex]d=3600km[/tex]

Since

Density of steel [tex]\rho=7900kg/m^3[/tex]

Stress of steel [tex]\mu= 1*10^9[/tex]

Generally the equation for Stress on Cable is mathematically given by

[tex]S=\frac{F}{A}[/tex]

[tex]S=\frac{\rho Alg}{A}[/tex]

Therefore

[tex]l=\frac{s}{\rhog}[/tex]

[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]

[tex]l=12916.5m[/tex]

Answer:

Explanation:

Let the mass of objects be m₁ and m₂ .

Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²

Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4  =  1/2 ( m₁ + m₂ ) v² x .25

final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²

= 0.25

b )

Applying law of conservation of momentum to the system . Let m₁ > m₂

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁  -  m₂  = ( m₁ + m₂ )  / 2

2m₁ - 2 m₂ = m₁ + m₂

m₁ = 3m₂

m₁ / m₂ = 3 / 1

Answer:

(a) The ratio is 1 : 4.

(b) The ratio is 1 : 3.

Explanation:

Let the mass of each object is m and m'.

They initially move with velocity v opposite to each other.

Use conservation of momentum

m v - m' v = (m + m') v/2

2 (m - m')  = (m + m')

2 m - 2 m' = m + m'

m = 3 m' .... (1)

(a) Let the initial kinetic energy is K and the final kinetic energy is K'.

[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]

[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]

The ratio is

K' : K = 1 : 4

(b) m = 3 m'

So, m : m' = 3 : 1

Answer:

Quantum theory gives the concept of

Answer:

75

Explanation:

i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:

T=200-125

= 75

Answer:

In heterogeneous mixture do the physically distinct component parts each have distinct properties.

Answer:

C is the right answer

Explanation:

fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success

hope it was useful for you

Answer:

Initial velocity, u = 60 km/h = 16.7 m/s

Final velocity, v = 72 km/h = 20 m/s

time, t = 2 sec

From first equation of motion:

[tex]{ \bf{v = u + at}}[/tex]

Substitute the variables:

[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed,  v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]

Therefore, the maximum speed is 0.20 m/s

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

Answer:

The acceleration is in 2 D as in between east and south.

Explanation:

mass, m = 50 kg

acceleration, a = 0.25 m/s^2 horizontal

acceleration of elevator, a' = 1 m/s^2 downwards

When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:

Answer:

false?

Explanation:

The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.

Answer:

I think the answer is False.

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Explanation:

Given the data in the question;

wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m

Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

[tex]t_{min[/tex] = ( λ / 4n )

so we simply substitute in our given values;

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 5.4

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = 8.574 × 10⁻⁸ m

[tex]t_{min[/tex] = 85.74 × 10⁻⁹ m

[tex]t_{min[/tex] = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Answer:

[tex]r=200m[/tex]

Explanation:

From the question we are told that:

Charges:

[tex]Q_1=8.0mC[/tex]

[tex]Q_2=2.0mC[/tex]

[tex]Q_3=8.mC[/tex]

Distance [tex]d=300m[/tex]

Generally the equation for Force is mathematically given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Therefore

[tex]F_{32}=F_{31}[/tex]

[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]

[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]

[tex]r=2(300-r)[/tex]

[tex]r=200m[/tex]

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

Answer:

The correct answer is "121.36 meters and 2.40 MHz".

Explanation:

Given:

Red shift,

[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]

Wavelength,

[tex]\lambda = 125 \ meters[/tex]

The observed frequency will be:

⇒ [tex]f = \frac{c}{\lambda}[/tex]

      [tex]=\frac{3\times 10^8}{125}[/tex]

      [tex]=24\times 10^5[/tex]

      [tex]= 2.40\times 10^6[/tex]

      [tex]=2.40 \ MHz[/tex]

hence,

The Emitted wavelength will be:

⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]

   [tex]\frac{125}{\lambda_0}-1 =0.03[/tex]

         [tex]125=1.03 \ \lambda_0[/tex]

           [tex]\lambda_0=\frac{125}{1.03}[/tex]

                [tex]=121.36 \ m[/tex]

Answer:

Len's law

Explanation:

We can explain this exercise using Len's law

when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.

Answer:

In the given circuit, R

2

,R

6

and R

4

are in series. So,

R

1

′

=7+5+12=24Ω

Now R

1

′

and R

5

′

are in parallel. So,

R

2

′

1

=

8

1

+

24

1

=

24

3+1

=

24

4

=

6

1

R

2

′

=6ohm.

Now R

2

′

,R

1

and R

3

are in series. So,

R=R

2

′

+R

1

+R

3

=6+3+2=11ohm.

We know i=

R+r

E

=

11+1

6

=

12

6

=

2

1

i=0.5amp.

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Answer:

imma try and fail again and again

Answer:

12.74 V

Explanation:

We are given that

Current, I1=54 A

Potential difference, V1=9.18V

I2=2.10 A

V2=12.6 V

We have to find the battery's emf.

[tex]E=V+Ir[/tex]

Using the formula

[tex]E=9.18+54r[/tex] ....(1)

[tex]E=12.6+2.10r[/tex]  .....(2)

Subtract equation (1) from (2)

[tex]0=3.42-51.9r[/tex]

[tex]3.42=51.9r[/tex]

[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]

Using the value of r in equation (1)

[tex]E=9.18+54(0.0659)[/tex]

[tex]E=12.74 V[/tex]

It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is

∑ F = (force from engine) - (resistive forces) = 0

Answer:

[tex]F=202650N[/tex]

Explanation:

From the question we are told that:

Area [tex]a=50m^2[/tex]

Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]

Generally the equation for Force is mathematically given by

[tex]F=dP*A[/tex]

[tex]F=4053*50[/tex]

[tex]F=202650N[/tex]