Which of the following is an example of potential energy?

Solution :

We all know that a bar magnet have two poles, the north pole and the south pole. These poles interacts with each other. The ends of the magnets having similar poles will push each other away while the poles with like charges will pull each others towards it.

The compass needle is also a magnet having south polarity as well as north polarity. When the compass needle is close to the bar magnet, it is opposite to the poles or along the poles. The compass needle shows the direction or is pointed towards the north. So when the compass needle is placed near the north pole of the bar magnet, the pointer of the compass needle points towards the north, i.e. it gets deflected because of he like charges. And when it is placed near the south pole of the magnet, it gets attracted towards it and is pointed towards the pole.

Now as we move the compass needle from the poles to the region that is between the poles, the compass needle pointer points towards the north direction every time. It show a deflection always. If we place the magnetic lines, we will see that the magnetic lines will exit from the north poles and enters the south pole of the bar magnet.  

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

[tex]\lambda = 720 \ nm[/tex]

or,

  [tex]= 720\times 10^{-9} \ m[/tex]

[tex]D=0.85 \ m[/tex]

[tex]d = 0.22 \ mm[/tex]

or,

  [tex]=0.22 \times 10^{-3} \ m[/tex]

As we know,

Fringe width is:

⇒ [tex]\beta=\frac{\lambda D}{d}[/tex]

hence,

Separation between second and third bright fringes will be:

⇒ [tex]\theta=\beta=\frac{\lambda D}{d}[/tex]

       [tex]=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}[/tex]

       [tex]=2.78\times 10^{-3} \ m[/tex]

or,

       [tex]=2.78 \ mm[/tex]

Explanation:

Write what you know

Speed = Distance / Time

micro- = 10^-6

write your conversions as fractions

1 in / 0.0254 m

1 min / 60 sec

First convert time to regular seconds

7.4 x 10^-6 seconds

Use Velocity

1.49m / (7.4 x 10^-6) s

We've written our conversions in fractions because units cancel out just like numbers

[tex] \frac{1.49m}{7.4 \times {10}^{ - 6} } \times \frac{1in}{0.0254m} \times \frac{60sec}{1min} [/tex]

Multiply all the fractions accross and youll have your answer

The type of model shown here is an experimental model. The correct option is B.

Animals are used in experimental modeling to model the development and progression of diseases and to test new treatments before they are administered to humans.

This result stems from the distinction that, whereas experiments are versions of the real world captured within an artificial laboratory environment, models are artificial worlds constructed to represent the real world.

Theories are plausible explanatory propositions developed to connect potential causes to their effects.

Models are schematic representations of reality or one's view of a possible world that are built to improve one's understanding of the world and/or to make predictions.

Thus, the correct option is B.

For more details regarding model, visit:

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Your question seems incomplete, the probable complete question is:

Answer:

a)   w = 31.4 rad / s,  b)  a = 118.4 m / s²

Explanation:

a) let's reduce to the SI system

   w = 5 rev / s (2pi rad / 1 rev)

   w = 31.4 rad / s

b) the expression for the centripetal acceleration is

      a = v² / r

linear and angular variables are related

      v = w r

    we substitute

     a = w² r

     a = 31.4² 0.120

     a = 118.4 m / s²

the earth is on the month of June

Answer: [tex]115.52\ N[/tex]

Explanation:

Given

Length of plank is 1.6 m

Force [tex]F_1[/tex] is applied on the left side of plank

Force [tex]F_2[/tex] is applied 43 cm from the left end O.

Mass of the plank is [tex]m=13.7\ kg[/tex]

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

[tex]\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N[/tex]

Net vertical force must be zero on the plank

[tex]\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N[/tex]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = [tex]\frac{x-x_1}{t}[/tex]

we substitute the values

             v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = [tex]\frac{6600 - x_1}{4}[/tex]

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = [tex]\frac{6600 -128 a}{4}[/tex]

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

Answer:

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Answer:

Y = 5.03 x 10⁻³ m = 5.03 mm

Explanation:

Using Young's Double-slit formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = Fringe Spacing = Width of bright fringe = ?

λ =  wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = Screen distance = 3.1 m

d = slit width = 0.39 mm = 3.9 x 10⁻⁴ m

Therefore,

[tex]Y = \frac{(6.33\ x\ 10^{-7}\ m)(3.1\ m)}{3.9\ x\ 10^{-4}\ m}[/tex]

Y = 5.03 x 10⁻³ m = 5.03 mm

Answer:

amplitude

Explanation:

The loudness of a musical sound is a measure of the sound wave's ?

is amplitude explanation:- The loudness of a sound depends upon the amplitude.Loudness of a sound depends on the amplitude of the vibration producing that sound. Greater is the amplitude of vibration, louder is the sound produced by it. if you find this answer helpful please rate positive thank you so much.

Answer:

resonance is when a body is made to vibrate with the frequency of another body without touching it.

Explanation:

example of destructive: soldiers matching on a bridge unanimously can make the bridge collapse.

Answer:hi

i like ears

and lemon

and poetry

Explanation:

Answer:

If we convert a circuit into a current source with parallel load it is called source transformation

Answer:

B. They are inversely proportional to each other

[tex] \frac{momentum}{time} = force \\ \\ \frac{mass \times velocity}{time} = force \\ \\ \frac{mass \times velocity}{time} = mass \times acceleration[/tex]

Answer: The magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.

Explanation:

Given: Length = 0.50 m

No. of turns = 500

Current = 22 A

Formula used to calculate magnetic field is as follows.

[tex]B = \mu_{o}(\frac{N}{L})I[/tex]

where,

B = magnetic field

[tex]\mu_{o}[/tex] = permeability constant = [tex]4\pi \times 10^{-7} Tm/A[/tex]

N = no. of turns

L = length

I = current

Substitute the values into above formula as follows.

[tex]B = \mu_{o}(\frac{N}{L})I\\= 4 \pi \times 10^{-7} Tm/A \times (\frac{500}{0.5 m}) \times 22\\= 0.0276 T[/tex]

Thus, we can conclude that magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.

[tex]\sf{The~ different~ types~ of~ measuring~ instruments~ are:-}[/tex]

Answer:

a) The heat input per cycle is 2857.143 joules.

b) The temperature of the low-temperature reservoir is 49.655 °C.

Explanation:

a) The efficiency of the Carnot engine is defined by the following formula:

[tex]\eta_{th} = 1-\frac{T_{L}}{T_{H}} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (1)

Where:

[tex]T_{L}[/tex] - Low temperature reservoir, in Kelvin.

[tex]T_{H}[/tex] - High temperature reservoir, in Kelvin.

[tex]Q_{L}[/tex]  - Heat output, in joules.

[tex]Q_{H}[/tex] - Heat input, in joules.

[tex]\eta_{th }[/tex] - Engine efficiency, no unit.

If we know that [tex]\eta_{th} = 0.3[/tex] and [tex]Q_{L} = 2000\,J[/tex], the heat input of the Carnot engine is:

[tex]\eta_{th} = 1 - \frac{Q_{L}}{Q_{H}}[/tex]

[tex]\frac{Q_{L}}{Q_{H}} = 1 - \eta_{th}[/tex]

[tex]Q_{H} = \frac{Q_{L}}{1-\eta_{th}}[/tex]

[tex]Q_{H} = \frac{2000\,J}{1-0.3}[/tex]

[tex]Q_{H} = 2857.143\,J[/tex]

The heat input per cycle is 2857.143 joules.

b) If we know that [tex]T_{H} = 461.15\,K[/tex] and [tex]\eta_{th} = 0.3[/tex], then the temperature of the low-temperature reservoir:

[tex]\eta_{th} = 1 - \frac{T_{L}}{T_{H}}[/tex]

[tex]\frac{T_{L}}{T_{H}} = 1 - \eta_{th}[/tex]

[tex]T_{L} = T_{H}\cdot (1-\eta_{th})[/tex]

[tex]T_{L} = (461.15\,K)\cdot (1-0.3)[/tex]

[tex]T_{L} = 322.805\,K[/tex]

[tex]T_{L} = 49.655\,^{\circ}C[/tex]

The temperature of the low-temperature reservoir is 49.655 °C.

Answer:

I = 1.58 x 10⁻³ watt/m²

Explanation:

Here, we will use the following formula:

[tex]\beta = 10\ log_{10}(\frac{I}{I_o})[/tex]

where,

β = decibel level = 92 dB

I = Intenisty of sound in watt/m² = ?

I₀ = reference intensity = 10⁻¹² watt/m²

Therefore,

[tex]92\ dB =10\ log_{10}(\frac{I}{10^{-12}\ watt/m^2} )\\\\[/tex]

[tex]10^{9.2} = \frac{I}{10^{-12}}\ watt/m^2\\\\I = (1.58\ x\ 10^9)(10^{-12}\ watt/m^2)[/tex]

I = 1.58 x 10⁻³ watt/m²

“The magnification produced by a spherical mirror is -3”. List four information you obtain from this statement about the mirror/image.

Answer:

circuit breaker

Explanation:

A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.

Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.

Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.

Therefore, the correct answer is "circuit breaker"

Answer:

T = 17649.03 N = 17.65 KN

Explanation:

The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

[tex]T = W_A = m(g+a)\\[/tex]

where,

T = Tension in cable = ?

[tex]W_A[/tex] = Apparent weight

m = mass = 1603 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

[tex]T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\[/tex]

T = 17649.03 N = 17.65 KN

Answer:

Option D

Explanation:

From the question we are told that:

The attractive magnetic force per unit length as

 [tex]f = F/L[/tex]

Separation Distance [tex]x=2d[/tex]

Generally the equation for  Magnetic force between two current carrying wire is mathematically given by

[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]

[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]

Where

[tex]x=2r[/tex]

And

[tex]I_1=I_2=>2I[/tex]

Then

[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]

[tex]\frac{F}{\triangle l}=>2f[/tex]

Therefore s the force per unit length between the wires if their separation is 2d

[tex]\frac{F}{\triangle l}=>2f[/tex]

Option D

Answer:

acceleration is measured

This question is incomplete, the complete question is;

Suppose that the position of a particle is given by s=f(t)=5t³ + 6 t+ 9.

(a) Find the velocity at time t.

(b) Find the velocity at time t=3 seconds

Answer:

a) the velocity at time t is ( 15t² + 6 ) m/s

b) Velocity at time t=3 seconds is 141 m/s

Explanation:

Give the data in the question;

position of a particle is given by;

s = f(t) = 5t³ + 6t + 9

Velocity at t;

we differentiate with respect to t

so

V(t) = f'(t) = d/dt ( 5t³ + 6t + 9 )

V(t) = f(t) = 5(3t²) + 6(1) + 0 )

V(t) = f(t) = ( 15t²+6 ) m/s

Therefore, the velocity at time t is 15t²+6 m/s

b) Velocity at t = 3 seconds

V(t) = f(t) = ( 15t²+6 ) m/s

we substitute

V(3) = ( 15(3)² + 6 ) m/s

V(3) = ( (15 × 9) + 6 ) m/s

V(3) = ( 135 + 6 ) m/s

V(3) = 141 m/s

Therefore, Velocity at time t=3 is 141 m/s