How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A) 0.50c
B) 0.65c
C) 0.78c
D) 0.87c

Answers

Answer 1

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]

Option (D) is correct.


Related Questions

Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.

Answers

Answer:

A. 16 J

B. 16 J

C. 8 N

Explanation:

A. Determination of the kinetic energy.

Mass (m) = 0.5 Kg

Velocity (v) =. 8 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 8²

KE = ½ × 0.5 × 64

KE = 0.5 × 32

KE = 16 J

B. Determination of the Workdone by the force.

Kinetic energy (KE) = 16 J

Workdone =.?

Workdone and kinetic energy has the same unit of measurement. Thus,

Workdone = kinetic energy

Workdone = 16 J

C. Determination of the force.

Workdone (Wd) = 16 J

Displacement (s) = 2 m

Force (F) =?

Wd = F × s

16 = F × 2

Divide both side by 2

F = 16 / 2

F = 8 N

Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
95oC for 20 minutes.

Answers

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

[tex]Power = P = \frac{E}{t}[/tex]

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

[tex]P = \frac{2.927\ x\ 10^6\ J}{1200\ s}[/tex]

P = 2439.5 W = 2.439 KW

A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram

Answers

Answer:

b

Explanation:

An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)

Answers

Answer:

Explanation:

The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .

v² = u² + 2 g H .

v² = 0  + 2 x 9.8 x 4.7 m

= 92.12

v = 9.6 m /s

The fish's final velocity will have two components

vertical component = 9.6 m/s downwards

Horizontal component = 2.6 m /s  .

Resultant velocity = √ ( 9.6² + 2.6² )

= √ ( 92.16 + 6.76 )

= 9.9 m /s

Answer:

The speed of fish at the time of hitting the surface is 9.95 m/s.

Explanation:

Horizontal speed, u = 2.6 m/s

height, h = 4.7 m

Let the vertical velocity at the time of hitting to water is v.

Use third equation of motion

[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]

The net velocity with which the fish strikes to the water is

[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s.

Answers

This question is incomplete, the complete question is;

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s. a) θ = 30° V b) θ = 60° V c) θ = 90° V

Answer:

the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

Explanation:

Given the data in the question;

number of turns N = 70

Diameter of coil D = 11 cm

Radius r = D/2 = 11/2 = 5.5 cm = 0.055 m

magnitude of magnetic ΔB = 0.7T

Δt = 0.2 seconds

Now,

a)

For θ = 30°,

Angle of with area of vector θ' = 90° - 30° = 60°

so

emf  e = N( Δ∅ / Δt ) = N( ΔBAcosθ  / Δt )

hence

e =  NAcosθ'(ΔB / Δt )

where A is area ( πr² )

so we substitute

e =  70 × πr² × cos(60°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(60°) × ( 0.7 / 0.2 )

e = 1.16 V

b)

For θ = 60°,

Angle of with area of vector θ' = 90° - 60° = 30°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(30°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.01 V

c)

For θ = 90°,

Angle of with area of vector θ' = 90° - 90° = 0°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(0°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.33 V

Therefore, the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases uniformly to 36.5 rad/s in 5.75 seconds. What is the magnitude of the fan's angular acceleration

Answers

Answer: [tex]1.79\ rad/s^2[/tex]

Explanation:

Given

Initial angular speed is [tex]\omega_1=26.2\ rad/s[/tex]

Final angular speed is [tex]\omega_2=36.5\ rad/s[/tex]

Time period [tex]t=5.75\ s[/tex]

Magnitude of the fan's acceleration is given by

[tex]\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}[/tex]

Insert the values

[tex]\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2[/tex]

Thus, fan angular acceleration is [tex]1.79\ rad/s^2[/tex]

Answer:

The angular acceleration is given by 1.8 rad/s^2.

Explanation:

initial angular speed, wo = 26.2 rad/s

final angular velocity, w = 36.5 rad/s

time, t = 5.75 seconds

The first equation of motion is

[tex]w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2[/tex]

Question 15
Calculate the velocity of a body if its total energy is three times its rest energy
OA 0.54c
OB. 0.760
OC0.94c
OD.C
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Answers

Jjdjdjfjififififkjfndndj

Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?

Answers

Answer:

The correct answer is - low pitch

Explanation:

Now for the case it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

if frequency increases then pitch will be increase as well as pitch depends on frequency.

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.

As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

1. A child slide down an inclined plane of length 10 m at an angle of 45°. If the coefficient friction between the child and the plane is 0.1, evaluate The velocity just before touching the bottom of the plane.

Answers

Answer:

The speed at the bottom is 11.2 m/s.

Explanation:

length, s = 10 m

Angle, A = 45 degree

coefficient of friction = 0.1

let the velocity is v.

The acceleration is given by

[tex]a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2[/tex]

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s[/tex]

A certain heating element is made out of Nichrome wire and used with the standard voltage source of V=120 V. Immediately after the voltage is turned on, the current running through the element is measured at I1=1.28 A and its temperature at T1=25°C. As the heating element warms up and reaches its steady-state (operating) temperature, the current becomes I2=1.229 A.

Required:
Find this steady-state temperature T2.

Answers

Answer:

T₁ = 232.5 ºC

Explanation:

For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law

           V = i R

            R = V / i

i₀ = 1.28 A

            R₀ = 120 / 1.28

            R₀ = 93.75 ohm

i₁ = 1.229 A

             R₁ = 120 / 1.229

             R₁ = 97.64 or

Resistance in a metal is linear with temperature

            ΔR = α R₀ ΔT

where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹

            ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]

            ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]

            ΔT = 2,075 10² C

            ΔT = T₁-T₀ = 2,075 10²

            T₁ = T₀ + 207.5

             T₁ = 25+ 207.5

             T₁ = 232.5 ºC

The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.

Answers

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         [tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex]            (you has an mistake in the formula)

         [tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]

         [tex]\frac{1}{C_{eq1}}[/tex] = 0.1   10⁶

         [tex]C_{eq1}[/tex] = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          [tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]

          [tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]

          [tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶

          [tex]C_{eq2}[/tex] = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

           

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]

          U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]

          U₄ = 0.5 J

To get maximum current in a circuit, the resistance should be in _____
1)series
2)parallel

Answers

Answer:

no parallel is the correct answer

An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?

Answers

105 km/h ≈ 29.2 m/s

60.0 km/h ≈ 16.7 m/s

Before the collision the test car has an acceleration a of

a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²

During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is

a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²

i.e. with magnitude about 19.7 m/s².

Overall, the car has an average acceleration of

a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²

A tall cylinder contains 25 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder

Answers

Answer: [tex]377.3\ kPa[/tex]

Explanation:

Given

Water column height [tex]h=25\ cm[/tex]

After oil is poured, the total height becomes [tex]h'=40\ cm[/tex]

Pressure at the bottom will be the sum due to the water and oil column

Suppose the density of the oil is [tex]\rho=900\ kg/m^3[/tex]

Pressure at the bottom

[tex]\Rightarrow P=10^3\times g\times 25+900\times g\times 15\\\Rightarrow P=100g[250+135]\\\Rightarrow P=3773\times 100\ Pa\\\Rightarrow P=377.3\ kPa[/tex]

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop

Answers

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

[tex]V_i=50 \ m/h[/tex]

Now,

[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]

    [tex]=73.333 \ ft/sec[/tex]

Now,

Final velocity,

[tex]V_f=0[/tex]

Initial velocity,

[tex]V_{initial} = 73.333 \ ft/sec[/tex]

hence,

⇒ [tex]V_f^2=V_i^2+2aD[/tex]

By putting the values, we get

      [tex]0=(73.333)^2+2\times( -22) D[/tex]

  [tex]44D=(73.333)^2[/tex]

      [tex]D=\frac{(73.333)^2}{44}[/tex]

          [tex]=122.2211[/tex]

A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.

1. How many revolutions do the kids make before the constant operational speed is reached ?

2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.

3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?

Answers

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

Physics question on picture

Answers

Answer:

B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge

Answers

Answer:

I think it is of science is it true na i knew it bro dont take tension

A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with the axis. On the left and right faces, the field has a strength and , respectively. The field along the front and back faces has strengths and . The field at the bottom and top faces has strengths and , respectively. What is the total charge enclosed by the cube

Answers

Complete Question

Complete Question is attached below

Answer:

[tex]q=1.558*10^{-9}c[/tex]

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength [tex]E_l=784.75N/m[/tex]

Right field strength [tex]E_r=776.38 N/m[/tex]

Front field strength [tex]E_f=725.5 N/m[/tex]

Back field strength [tex]E_b=749.54 N/m[/tex]

Top field strength [tex]E_t=944.95 N/m[/tex]

Bottom field strength [tex]E_{bo}=1082.58 N/m[/tex]

Generally, the equation for  Charge flux is mathematically given by

[tex]\phi=EAcos\theta[/tex]

Where

Theta for Right,Left,Front and Back are at an angle 90

[tex]cos 90=0[/tex]

Therefore

[tex]\phi =0[/tex] with respect to Right,Left,Front and Back

Generally, the equation for  Charge Flux is mathematically also given by

[tex]\phi=\frac{q}{e_o}[/tex]

Where

[tex]Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2[/tex]

Therefore

[tex]Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t[/tex]

[tex]Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)[/tex]

[tex]Q_{net}=176N/C m^2[/tex]

Giving

[tex]q=\phi*e_0[/tex]

[tex]q=176N/C m^2*1.558*10^{-12}c[/tex]

[tex]q=1.558*10^{-9}c[/tex]

The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns

A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.

Answers

Answer:

C

Explanation:

Similarity

The reason why a teacher is more important then a farmer

Answers

Answer:

A teacher is more important than a famer.

Explanation:

A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.

The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars

Answers

Answer:

Explanation:

I’ll help

The index of refraction for a vacuum is 1.00000. The index of refraction for air is 1.00029. 1) Determine the ratio of time required for light to travel through 1000 m of air to the time required for light to travel through 1000 m of vacuum. (Express your answer to six significant figures.)

Answers

Answer:

 [tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029

Explanation:

The refractive index is defined

         n = c / v

         v = c / n

the speed of light per se wave is constant, so we can use the relations of uniform motion

         v = x / t

         t = x / v

we substitute

         t = x n / c

let's calculate the time

vacuum

        t₁ = 1000 1/3 10⁸

        t₁ = 3.333333 10⁻⁶ s

air

        t₂ = 1000 1.00029 / 3 10⁸

        t2 = 3.3343 10⁻⁶ s

the relationship between these times is

       t₂ / t₁ = 3.3343 / 3.3333333

       t₂ / t₁ = 1.00029

Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field

Answers

Answer:

Option (2) is correct.

The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Explanation:

An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.

The given problem is based on the concept and fundamentals of electromagnetic waves.  The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.

In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.

Learn more about the electromagnetic waves here:

https://brainly.com/question/25559554

Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel

across a 30 V potential difference to a light bulb.

a. Calculate the current delivered through the light bulb in the two cases.

b. Draw the circuit connection that will achieve the brightest light bulb.​

Answers

Explanation:

Given that,

Two resistors 4.5 Ω and 2.3 Ω .

Potential difference = 30 V

When they are in series, the current through each resistor remains the same. First find the equivalent resistance.

R' = 4.5 + 2.3

= 6.8 Ω

Current,

[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]

So, the current through both lightbulb is the same i.e. 4.41 A.

When they are in parallel, the current divides.

Current flowing in 4.5 resistor,

[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]

Current flowing in 2.3 ohm resistor,

[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]

In parallel combination, are brighter than bulbs in series.

NEED AN ANSWER QUICKLY PLEASE!!
If the length and number of turns of a solenoid are doubled strength of magnetic field will :
(a) Be doubled (b) become half (c) not change d) be four time​

Answers

Answer:

c). It wouldn't change.

Explanation:

[tex]{ \bf{F = \frac{ \ \gamma _{o}NI }{l} }}[/tex]

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.

Required:
a. What is the net force on the side of the container?
b. How long does it take and how much the water level will drop till water no longer comes out of the faucet?

Answers

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = [tex]\frac{P_o ( 1-0.850)}{\rho \ g}[/tex]

        h = [tex]\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}[/tex]

        h = 1.55 m

pha của dao động làm hàm

Answers

Answer:

pha của dao động là hàm bậc nhất của thời gian.

A measurement was made of the magnetic field due to a tornado, and the result was 13.00 nT to the north. The measurement was made at a position 8.90 km west of the tornado. What was the magnitude (in A) and direction of the current in the funnel of the tornado? Assume the vortex was a long, straight wire carrying a current.

Answers

Answer:

4

Explanation:

What is the current in milliamperes produced by the solar cells of a pocket calculator through which 9.00 C of charge passes in 8.50 h

Answers

Answer:

Current = 0.000294 A

Explanation:

Below is the given values:

Given the charge = 9.00 C

Time = 8.50 h

Use the below formula to find the current:

Current = Q / t

Now plug the values:

Current = 9 / (8.5 x 3600)

Current = 0.000294 A

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