Trong thực nghiệm nhiễu xạ qua 1 khe hẹp, bề rộng khe là 2.10-5 m, màn quan sát đặt cách khe 0,5 m. Chiếu ánh sáng có bước sóng 480 nm xuyên qua khe và thấy có nền nhiễu xạ trên màn quan sát. Tính bề rộng của vân sáng liền kề vân sáng trung tâm.

Answer:

220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.

Answer:

B = 0.088 T

Explanation:

Given that,

The length of a solenoid, l = 0.4 m

No. of turns of wire, N = 356

Current, I = 79 A

We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.

[tex]B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T[/tex]

So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.

Answer:

a: true.

Explanation:

We can define an equipotential surface as a surface where the potential at any point of the surface is constant.

For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.

Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.

Now we want to see if the electric field is always perpendicular to these equipotential surfaces.

You can see that in the two previous examples this is true, but let's see for a general case.

Now suppose that you have a given field, and you have a test charge in one equipotential surface.

So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.

But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.

Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.

The correct option is a.

Answer:

The gauge pressure is equal to 147 kPa.

Explanation:

The pressure exerted by fluid is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is density of water

h is height

So, put all the values,

[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]

or

P = 147 kPa

So, the gauge pressure is equal to 147 kPa.

Answer:

The gauge pressure is 147000 Pa.

Explanation:

Height, h = 15 m

density of water, d= 1000 kg/m^3

gravity, g = 9.8 m/s^2

The gauge pressure is the pressure exerted by the fluid.

The pressure exerted by the fluid is given by

P  = h d g

P = 15 x 1000 x 9.8 =  147000 Pa

Answer:

height is 78.4m

Explanation:

h=u.t + 0.5.g.t^2

= 0 + 0.5x9.8x4^2

= 78.4m

Answer:

  f = 276.6 Hz

Explanation:

This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.

In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is

          L = λ/ 4

speed is related to wavelength and frequency

          v = λ f

          λ = v / f

we substitute

          L = v / 4f

          f = v / 4L

the speed of sound at 20ºC is

          v = 343 m / s

let's calculate

          f = [tex]\frac{343 }{4 \ 0.31}[/tex]

          f = 276.6 Hz

Answer:

Fq = k q Q / R^2

Fp = k q Q / (6 R)^2

The force exerted between S and p is 1 / 36 of that between S and q

or the force between S and q is 36 times that between S and p.

Answer:

the instantaneous velocity is 51 m/s

Explanation:

Given;

acceleration, a = 2 + 5t²

Acceleration is the change in velocity with time.

[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]

Therefore, the instantaneous velocity is 51 m/s

Answer:

The speed of car A before collision is 3.5 km/h.

Explanation:

Mass of car A = 690 kg eastwards

Mass of car B = 520 kg at 74 km/h west wards

Distance, s = 6 m

coefficient of friction = 0.75

Let the speed after collision is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]

Let the initial speed of car A is v'.

Use conservation of momentum

690 x v' - 520 x 74 = (690 + 520) x 33.8

690 v' + 38480 = 40898

v' = 3.5 km/h

Answer:

f = 106.3 N

Explanation:

The force applied on the sled must be equal to the static frictional force to move the sled:

Tension Force Horizontal Component = Static Frictional Force

[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]

where,

T = Tension = 120 N

θ = angle of rope = 37°

μ = coefficient of static friction = ?

m = mass of children plus sled = 55 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]

Now, the static friction acting on the mother will be:

[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 106.3 N

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

Answer:

8 cm

Explanation:

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

Initial velocity (u) = 20 cm/s

Height (h) = 80 cm

Horizontal distance (s) =?

Next, we shall determine the time taken for marble to get to the ground. This can be obtained as follow:

Height (h) = 80 cm

Acceleration due to gravity (g) = 1000 cm/s²

Time (t) =?

t = √(2h/g)

t = √[(2 × 80)/1000]

t = √(160/1000)

T = √0.16

t = 0.4 s

Finally, we shall determine the horizontal distance travelled by the marble. This can be obtained as illustrated below:

Initial velocity (u) = 20 cm/s

Time (t) = 0.4 s

Horizontal distance (s) =?

s = ut

s = 20 × 0.4

s = 8 cm

Thus, the horizontal distance travelled by the marble is 8 cm.

The horizontal distance traveled by the marble is 8 cm.

The given parameters;

The time of motion of the marble is calculated as follows;

[tex]h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.8}{9.8} }\\\\t = 0.4 \ s[/tex]

The horizontal distance traveled by the marble is calculated as follows;

[tex]X = v_0_x t\\\\X = (20 \times 0.4)\\\\X = 8 \ cm[/tex]

Thus, the horizontal distance traveled by the marble is 8 cm.

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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.

What is the magnitude of force required to accelerate a car of mass 1.7 × 10³ kg by 4.75 m/s²

Answer:

F = 8.075 N

Explanation:

Formula for force is;

F = ma

Where;

m is mass

a is acceleration

F = 1.7 × 10³ × 4.75

F = 8.075 N

In vacuum, going at 2.99×10^8 m/s.

Explanation:

Given that,

Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.

The rms speed for this collection is as follows :

[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]

The average speed of these molecules is :

[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]

So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.

Answer:

Where is the figure ?????

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)

[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

[tex]v_B=+5\ m/s[/tex]

Therefore, Carlos velocity in Bill's reference frame will be

[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]

So, the magnitude is 20 m/s and the direction is westward (negative sign).

Answer:

t = 2.26 x 10⁵ s

Explanation:

The energy supplied to the water will be equal to the heat required for the boiling of water:

E = ΔQ

Pt = mL

where,

P = Power = 10 W

t = time = ?

m = mass of water = 1 kg

L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]

t = 2.26 x 10⁵ s

This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.

Answer:

The velocity becomes [tex]v\sqrt 2[/tex].

Explanation:

The force acting on the bobber is centripetal  force.

The centripetal force is given by

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

when mass remains same, radius is doubled and the force is same, so the velocity is v'.

[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]

Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.    

Let's evaluate each tension.

Case T₃.

[tex] T_{3} - W_{3} = 0 [/tex]

For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.          

[tex] T_{3} = W_{3} [/tex]          

Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:                

[tex] T_{3} = W_{3} = mg [/tex]  (1)                                                                                                     Case T₂.

[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]    

[tex] T_{2} = T_{3} + W_{2} [/tex]   (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]

Case T₁.

[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]  

[tex] T_{1} = T_{2} + W_{1} [/tex]    (3)

Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

[tex] T_{1} = 3mg + 3mg = 6mg [/tex]        

Therefore, the correct option is d: T₁ > T₂ > T₃.

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I hope it helps you!

Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]

First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:

Mass m:

[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)

Mass 2m:

[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)

Mass 3m:

[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)

The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)

Answer:

 I = - m 16  the two impulses are the same,

Explanation:

The impulse is given by the relationship

         I = Δp

         I = p_f - p₀

in this case the final velocity is zero therefore p_f = 0

        I = -p₀

For driver A the steering wheel impulse is

        I = - m v₀

        I = - m 16

For driver B, the airbag gives an impulse

        I = - m 16

We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less

Hooke's law gives the relationship between the force applied and observed compression and expansion

(a) Hooke's law states that force applied is proportional to the deformation of an object

(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre

The reason for the above explanation are as follows;

(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F

Mathematically, we get;

F = k × ΔL

[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]

Given that the material cross sectional area = A, and the original length of the material = L, we get;

F/A = Stress = σ, ΔL/L = strain = ε

[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]

Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material

(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation

Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve

As the tyre straightens, the path of the stress change curve is given by the lower curve

The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening

Energy absorbed < Energy released

The balance energy is transformed into other forms of energy, including  heat energy, which is observed by the raising in temperature, warming, of the tyre

Energy absorbed = Energy released + Heat energy

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Answer:

a) Light that passes through the floor to reveal yourself (not shadow).

b) 2 rays of light that bounce between 2 transparent media.

c) I don't know what is Diffraction?

Answer:

Explanation:

Area of the loop = π x ( 55 x 10⁻³ )²

= 9.5 x 10⁻³ m²

Change in Magnetic flux dφ = 450 x 10⁻³ - 350 x 10⁻³ = 150x 10⁻³ Weber.

time dt =.10 s

emf induced = dφ / dt = 150x 10⁻³ Weber / .10 s

= 1.5 V .

b )

Magnetic field is directed outwards and it is increasing so according to Lenz's law , direction of induced current will be clockwise in the loop.

Answer:

(a) 9.5 mV

(b) clockwise

Explanation:

Radius, r = 55 mm

Time, t = 0.1 s

Change in magnetic field, B = 450 - 350 = 100 mT =0.1 T

(a) induced emf is given by

[tex]e = A \frac{dB}{dt}[/tex]

[tex]e = A \frac{dB}{dt}\\\\e=3.14\times 0.055\times0.055\times \frac{0.1}{0.1}\\\\e= 9.5 \times 10^{-3} V = 9.5 mV[/tex]

(b) According to the Lenz law, the direction of current is clockwise.

Answer:

[tex]T=8.1N[/tex]

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 [tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]

 [tex]\alpha=50 \textdegree[/tex]

Generally the equation for Tension is is mathematically given by

 [tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]

 [tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]

 [tex]T=8.1N[/tex]

Answer:

The required fraction is 0.023.

Explanation:

Given that

Mass of a car, m = 1030 kg

Mass of 4 wheels = 12 kg

We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.

The rotational kinetic energy due to four wheel is

[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]

Linear kinetic Energy of the car is:

[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]

Fraction,

[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]

So, the required fraction is 0.023.

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;

[tex]F_{min}[/tex] = m·g - m·a

∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N