[3 marks]
A particle of mass 100 g executes simple harmonic motion about x = 0 with angular frequency ω = 10 s-1. Its total mechanical energy is Etot = 0.45 J. Find the displacement of
the particle when its speed is 2 m/s. [3 marks]​

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]

Answer:

2.00x 10 14th Hz

Explanation:

Answer:

2.99 x 10^14 Hz

Explanation:

E photon= hf (you have to solve for f)

f= E photon/h

f= 1.98 x 10^-19 J / 6.63 x 10^-34 J x s

f=2.99 x 10^14 Hz

Answer: 48,000 J

Explanation: i just did it

Answer:

acceleration is measured

Answer:

bvihobonlnohovicjfufufufucvkvkvvjcufufydyfuvi

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: 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.

Answer:

(a) I = 1650000 A

(b) 4.125 T

Explanation:

Magnetic field, B = 5.5 T

distance, r = 0.06 m

(a) Let the current is I.

The magnetic field due to a long wire is given by

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]

(b) Let the magnetic field is B' at distance r = 0.08 m.

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]

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

Answer:

a) the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Explanation:

Given the data in the question;

separation  between two slits  d = 0.36 mm = 0.00036 m

Separation between two adjacent fringes β = 5.5 mm = 0.0055 m

Distance of screen from slits D = 1.6 m

n = 2

a) the wavelength the light emits;

Using the formula;

β = (nD/d)λ

To find wavelength, we make λ the subject of formula;

βd = nDλ

λ = βd / nD

so we substitute

λ = ( 0.0055 m × 0.00036 m ) / ( 2 × 1.6 m )

λ = 0.00000198 / 3.2

λ = 6.1875 × 10⁻⁷ m

Therefore, the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes;

To find the distance between the two n = 1 dark fringes, we use the following formula;

y[tex]_m[/tex] = 2nλD / d

given that n = 1, we substitute

y[tex]_m[/tex] = ( 2 × 1 × ( 6.1875 × 10⁻⁷ m ) × 1.6 m ) / 0.00036 m

y[tex]_m[/tex] = 0.00000198 / 0.00036

y[tex]_m[/tex] = 0.0055 m

y[tex]_m[/tex] = 5.5 × 10⁻³ m

Therefore, the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Answer:

Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.

Explanation:

Answer:

its number 2 one but i am not sure hope its right

Answer:

b. 200N c. 100N

Explanation:

30.

the horizontal. The force needed to push the body up the plane is

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency​ reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, [tex]v_b[/tex] = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]

The frequency reflected by the stationary wall to the bat is calculated as;

[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]

[tex]f_s\approx 41 \ kHz[/tex]

Answer:

[tex]P_a=17.85psi[/tex]

Explanation:

From the question we are told that:

Tank Pressure [tex]P_t=30.0kpa[/tex]

Atmospheric Pressure [tex]P_a=13.5 psi[/tex]

Where

 [tex]1kpa=0.148psi[/tex]

Therefore

 [tex]30kpa=4.35psi[/tex]

Generally the equation for Absolute pressure [tex]P_a[/tex] is mathematically given by

 [tex]P_a=13.5+4.35[/tex]

 [tex]P_a=17.85psi[/tex]

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: [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]

Momentum = (mass) x (speed)

If you work the problem in the same units as the given data, then you get the momentum in units of kilogram-meters per second, and your horse has 1,440 of them.

Answer:

A

Explanation:

1440 kg*m/s

Answer:

[tex]v_{0,new} = v0\sqrt{}2[/tex]

Explanation:

Initial work done on the proton is given by, [tex]\DeltaW0 = q V_o[/tex]

we know that, [tex]\DeltaW = \DeltaK.E[/tex]

[tex]qV0 = (1/2) m v_0^2[/tex]

[tex]v_0 = \sqrt{}2 q V_0 / m[/tex]                                                        { eq.1 }

If it were accelerated instead through a potential difference of 2V0, then it would gain a speed will be given as :

using the above formula, we have

[tex]v_{0,new} = \sqrt{}2 q (2V0) / m[/tex]    

[tex]v_{0,new} = \sqrt{}4 q V0 / m[/tex]    

[tex]v_{0,new} = v0\sqrt{}2[/tex]

Answer:

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

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

The question is incomplete. The complete question is :

A long, straight wire lies along the z-axis and carries a 3.90-A current in the + z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mm segment of the wire centered at the origin.

A) x=2.00m,y=0, z=0

Bx,By,Bz = ? T

Enter your answers numerically separated by commas.

B) x=0, y=2.00m, z=0

C) x=2.00m, y=2.00m, z=0

D) x=0, y=0, z=2.00m

Solution :

The expression of the magnetic field using the Biot Savart's law is given by :

[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]

a). The position vector is on the positive x direction.

[tex]$\vec r = (2 \ m) \ \hat i$[/tex]

[tex]$|r| = 2 \ m$[/tex]

The magnetic field is  

[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]

[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat i }{4 \pi \times (2)^3}$[/tex]

[tex]$d \vec B=(5.85 \times 10^{-11} \ T)\hat j$[/tex]

The magnetic field is   [tex]$(0, \ 5.85 \times 10^{-11} \ T, \ 0).$[/tex]

b). The position vector is in the positive y-direction.

[tex]$\vec r = (2 \ m) \ \hat j$[/tex]

[tex]$|r| = 2 \ m$[/tex]

The magnetic field is

[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]

[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat j }{4 \pi \times (2)^3}$[/tex]

[tex]$d \vec B=(5.85 \times 10^{-11} \ T)(-\hat{i})$[/tex]

The magnetic field is   [tex]$(- 5.85 \times 10^{-11} \ T, \ 0, \ 0).$[/tex]

c). The position vector is :

[tex]$\vec r = (2)\hat i + (2)\hat j$[/tex]

[tex]$|\vec r| = \sqrt{(2)^2+(2)^2}$[/tex]

   [tex]$=2.828 \ m$[/tex]

The magnetic field is

[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]

[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times ((2)\hat i + (2) \hat j) }{4 \pi \times (2.828)^3}$[/tex]

   [tex]$=(4.13\times 10^{-11})\hat j+(4.13\times 10^{-11})(-\hat i)$[/tex]

The magnitude of the magnetic field is :

[tex]$|d\vec B|=\sqrt{(4.13\times 10^{-11})^2+(4.13\times 10^{-11})^2}$[/tex]

       [tex]$=5.84 \times 10^{-11} \ T$[/tex]

Therefore, the magnetic field is [tex]$(-4.13 \times 10^{-11}\ T, \ 4.13 \times 10^{-11}\ T, \ 0 )$[/tex]

d).  The position vector is in the positive y-direction.

[tex]$\vec r = (2 \ m) \ \hat k$[/tex]

[tex]$|r| = 2 \ m$[/tex]

The magnetic field is

[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]

[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat k }{4 \pi \times (2)^3}$[/tex]

   = 0 T

The magnetic field is (0, 0, 0)

Answer:

t = 4.08 s

Explanation:

if the body is thrown upward, it has negative gravity. Knowing through the International System that the earth's gravity is 9.8 m/s²

Data:

Use formula:

Replace and solve:

Time taken by it to reach max height is 4.08 seconds.

Greetings.

Answer:

the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

Explanation:

Given that;

speed of the first spacecraft from earth v[tex]_a[/tex] = 0.80c

speed of the second spacecraft from earth v[tex]_b[/tex] = -0.60 c

Using the formula for relative motion in relativistic mechanics

u' = ( v[tex]_a[/tex] - v[tex]_b[/tex] ) / ( 1 - (v[tex]_b[/tex]v[tex]_a[/tex] / c²) )

we substitute

u' = ( 0.80c - ( -0.60c)  ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )

u' = ( 0.80c + 0.60c ) /  ( 1 - ( -0.48c² / c² ) )

u' = 1.4c /  ( 1 - ( -0.48 ) )

u' = 1.4c /  ( 1 + 0.48 )

u' = 1.4c / 1.48

u' = 0.9459c ≈ 0.95c  { two decimal places }

Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

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

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

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