Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R

Answer:

At the end of the handle farthest from the head of the hammer.

Explanation:

The force of the hammer is greatest the longer the radius is on a which would be the length of the handle. Simple mechanical advantage.

Answer:

The time taken is  [tex]\Delta t = 1.5 \ s[/tex]

Explanation:

From the question we are told that

   The mass of the ball is  [tex]m = 1.25 \ kg[/tex]

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  [tex]\theta = 1 rev = 2 \pi \ radian[/tex]

Generally the displacement for one  complete revolution is mathematically represented as

       [tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]

Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]

     [tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]

     [tex]\alpha = 0.9698 \ s[/tex]

Now the displacement for two  complete revolution is

         [tex]\theta_2 = 2 * 2\pi[/tex]

         [tex]\theta_2 = 4\pi[/tex]

Generally the displacement for two complete revolution is mathematically represented as  

     [tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]

=>   [tex]t^2 = 25.9187[/tex]

=>   [tex]t= 5.1 \ s[/tex]

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     [tex]\Delta t = t_2 - t[/tex]

substituting values

      [tex]\Delta t = 5.1 - 3.60[/tex]

     [tex]\Delta t = 1.5 \ s[/tex]

The time for the ball to complete the next revolution is 1.5 s.

The given parameters;

The constant angular acceleration of the ball is calculated as follows;

[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]

The time to complete the next revolution is calculated as follows;

[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]

[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]

Thus, the time for the ball to complete the next revolution is 1.5 s.

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Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have  

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, [tex]d_o[/tex] = 15.0 cm

image distance, [tex]d_i[/tex] = 5.0 cm

height of the object, [tex]h_o[/tex] = 9.0 cm

height of the image, [tex]h_i[/tex] = ?

Apply lens equation;

[tex]\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm[/tex]

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

Answer:

External force    W₁ = F L

Friction force    W₂ = - fr L

weight component   W₃ = - mg sin θ L

Y Axis   Force      W=0

Explanation:

When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.

For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular

let's write the equations of translational equilibrium in given exercise

X axis

        F - fr -Wₓ = 0

        F = fr + Wₓ

the components of the weight can be found using trigonometry

         Wₓ = W sin θ

         [tex]W_{y}[/tex] = W cos θ

let's look for the work of these three forces

          W = F x cos θ

External force

          W₁ = F L

since the displacement and the force have the same direction

Friction force

          W₂ = - fr L

since the friction force is in the opposite direction to the displacement

For the weight component

          W₃ = - mg sin θ L

because the weight component is contrary to displacement

Y Axis  

          N- Wy = 0

in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0

therefore work is worth zero

Answer:

The maximum height reached in the second trial is 16times the maximum height reached in the first trial.

Explanation:

The following data were obtained from the question:

First trial

Initial speed (u) = v

Final speed (v) = 0

Second trial

Initial speed (u) = 4v

Final speed (v) = 0

Next, we shall obtain the expression for the maximum height reached in each case.

This is illustrated below:

First trial:

Initial speed (u) = v

Final speed (v) = 0

Acceleration due to gravity (g) = 9.8 m/s²

Height (h₁) =.?

v² = u² – 2gh₁ (going against gravity)

0 = (v)² – 2 × 9.8 × h₁

0 = v² – 19.6 × h₁

Rearrange

19.6 × h₁ = v²

Divide both side by 19.6

h₁ = v²/19.6

Second trial

Initial speed (u) = 4v

Final speed (v) = 0

Acceleration due to gravity (g) = 9.8 m/s²

Height (h₂) =.?

v² = u² – 2gh₂ (going against gravity)

0 = (4v)² – 2 × 9.8 × h₂

0 = 16v² – 19.6 × h₂

Rearrange

19.6 × h₂ =16v²

Divide both side by 19.6

h₂ = 16v²/19.6

Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.

This is illustrated below:

Second trial:

h₂ = 16v²/19.6

First trial:

h₁ = v²/19.6

Second trial : First trial

h₂ : h₁

h₂ / h₁ = 16v²/19.6 ÷ v²/19.6

h₂ / h₁ = 16v²/19.6 × 19.6/v²

h₂ / h₁ = 16

h₂ = 16 × h₁

From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.

Answer:

The velocity is  [tex]v_h = 19.2 \ m/s[/tex]

Explanation:

From the question we are told that

   The speed of the roller coaster at ground level  is [tex]v = 26 \ m/s[/tex]

Generally we can define the roller coaster speed at ground level using the an equation of motion as

     [tex]v^2 = u^2 + 2 g s[/tex]

 u is  zero given that the roller coaster started from rest

      So

            [tex]26^2 = 0 + 2 * g * s[/tex]

So  

           [tex]s = \frac{26^2}{ 2 * g }[/tex]

=>       [tex]s = 37.6 \ m[/tex]

Now the displacement half way is mathematically represented as

    [tex]s_{h} = \frac{37.6}{2}[/tex]

     [tex]s_{h} = 18.8 \ m[/tex]

So

      [tex]v_h ^2 = u^2 + 2 * g * s_h[/tex]

Where  [tex]v_h[/tex] is the velocity at the half way point

=>  [tex]v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }[/tex]

=>   [tex]v_h = 19.2 \ m/s[/tex]

Answer:

Explanation:

To convert from °C to °F , the formula is

( F-32 ) / 9 = C / 5

F is reading fahrenheit scale and C is in centigrade scale .

F = 205 , C = ?

(205 - 32) / 9 = C / 5

C = 96°C approx .

Let us calculate the heat required .

Total heat required = heat required to heat up the ice at - 20 °C  to 0°C  + heat required to melt the ice + heat required to heat up the water at  0°C to

96°C.

=  5 x 2.04 x (20-0) +  5 x 336 + 5 x ( 96-0 ) x 4.2  kJ .

= 204 + 1680 + 2016

= 3900 kJ .

Complete Question    

The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by

[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]    in SI units.

Answer:

The  value  is  [tex]f = 1.98918*10^{5}\ Hz[/tex]

Explanation:

From the question we are told that

   The magnetic field is    [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]

 This above  equation can be modeled as

       [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]

So  

       [tex]w = \frac{10^7}{8}[/tex]

Generally the frequency is mathematically represented as

       [tex]f = \frac{w}{2 \pi}[/tex]

=>    [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]

=>    [tex]f = 1.98918*10^{5}\ Hz[/tex]

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

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Answer:

v = 345.6m/s

Explanation:

v = 384 x 0.9 = 345.6

v = 345.6m/s

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

Answer:

Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Explanation:

The length of the telescope is

         L = f_ocular + f_objetive

the magnification of the telescope is

         m = - f_objective / f_ocular

These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Answer:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;

[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = [tex]\frac{1}{(2/15)}[/tex]

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength [tex]\lambda[/tex] = 2m

Transverse speed [tex]v = f \lambda[/tex]

[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]

Hence, the transverse speed at that point is  15m/s

Answer:

Inertia! I hope this helps!

Answer:

inertia

Explanation

Inertia.

Answer:

The  speed of the water wave is [tex]v = 88 \ m/s[/tex]

Explanation:

From the question we are told that

      The  width of the tube is  [tex]L = 55 \ m[/tex]

     The fundamental  frequency is  [tex]f = 0.80 \ Hz[/tex]

Generally the fundamental frequency is mathematically represented as

      [tex]f = \frac{v}{2 * L }[/tex]

=>    [tex]v = f * 2 * L[/tex]

substituting values

       [tex]v = 0.8 * 2 * 55[/tex]

       [tex]v = 88 \ m/s[/tex]

The speed of the water wave will be 88 m/s.

Given information:

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center.

The frequency of the standing wave is [tex]f=0.8[/tex] Hz.

The width of the tub is [tex]w=55[/tex] m.

Let v be the speed of the standing wave.

The speed of the wave can be calculated as,

[tex]v=2wf\\v=2\times 55\times 0.8\\v=88\rm\; m/s[/tex]

Therefore, the speed of the water wave will be 88 m/s.

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Answer:

d.

Explanation:

Since, the capacitance( decreases )

therefore voltage between the plates(increases ).

Hence, option d is correct.

C =εA/d.

d is doubled, therefore  C decrease ( inverse relation).

D) The capacitance decreases and the voltage between the plates increases.

A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly, the capacitance decreases and the voltage between the plates increases.

The capacitance - (decreases)

The voltage between the plates- (increases ).

Thus, the correct answer is D.

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Answer:

Acceleration of the body is:

[tex]a=0.27\,\,m/s^2[/tex]

Explanation:

Use Newton's second Law to solve for the acceleration:

[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

Answer:

The net energy transferred between two objects

Explanation:

The physical property of matter that expresses hot or cold is called temperature. It demonstrates the thermal energy. A thermometer is used to measure temperature. It defines the rate to which the chemical reaction occurs. It tells about the thermal radiation emitted from an object.

The correct option that defines temperature is option C.

Answer:

A measure of the movement of atoms or molecules within an object

Explanation:

Process of elimination

Answer:

No

Explanation:

Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.

:-)

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

[tex]Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h[/tex]

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

[tex]h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}[/tex]

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

[tex]Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000[/tex]

Pb' = 122 KPa

(A)  The depth of the fluid is 2.14 m.

(B)  The new pressure at the bottom of container is 121972 Pa.

Given data:

The cross-sectional area of the container is, [tex]A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}[/tex].

The density of fluid is, [tex]\rho = 856 \;\rm kg/m^{3}[/tex].

The container pressure at bottom is, [tex]P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa[/tex].

The atmospheric pressure is, [tex]P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa[/tex].

(A)

The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,

[tex]P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}[/tex]

Here, h is the depth of fluid.

Solving as,

[tex]856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m[/tex]

Thus, the depth of the fluid is 2.14 m.

(B)

For an additional volume of [tex]2.35 \times 10^{-3} \;\rm m^{3}[/tex] to the liquid, the new depth is,

[tex]V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m[/tex]

Now, calculate the new pressure at the bottom of the container as,

[tex]P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa[/tex]

Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.

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Answer:

a

   [tex]F = 67867.2 \ N[/tex]

b

  [tex]\Delta L = 2.6 \ mm[/tex]

Explanation:

From the question we are told that

      The Young modulus is  [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]

      The stress is  [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]

      The  diameter is  [tex]d = 2.40 \ cm = 0.024 \ m[/tex]

The radius is mathematically represented as

       [tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]

The cross-sectional area is  mathematically evaluated as

        [tex]A = \pi r^2[/tex]

         [tex]A = 3.142 * (0.012)^2[/tex]

        [tex]A = 0.000452\ m^2[/tex]

Generally the stress is mathematically represented as

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

=>     [tex]F = \sigma * A[/tex]

=>    [tex]F = 1.50 *10^{8} * 0.000452[/tex]

=>    [tex]F = 67867.2 \ N[/tex]

Considering part b

      The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]

Generally Young modulus is mathematically represented as

           [tex]E = \frac{ \sigma}{ strain }[/tex]

Here strain is mathematically represented as

         [tex]strain = \frac{ \Delta L }{L}[/tex]

So    

       [tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]

        [tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]

=>     [tex]\Delta L = \frac{\sigma * L }{E}[/tex]

substituting values

       [tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]

       [tex]\Delta L = 0.0026[/tex]

Converting to mm

      [tex]\Delta L = 0.0026 *1000[/tex]

      [tex]\Delta L = 2.6 \ mm[/tex]

Answer:

The amplitude of the resultant wave is 12.93 cm.

Explanation:

The amplitude of resultant of two waves, y₁ and y₂, is given as;

Y = y₁ + y₂

Let y₁ = A sin(kx - ωt)

Since the wave is out phase by φ, y₂ is given as;

y₂ = A sin(kx - ωt + φ)

Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )

Given;

phase difference, φ = 45°

Amplitude, A = 7.00 cm

Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )

Y = 12.93 cm

Therefore, the amplitude of the resultant wave is 12.93 cm.

Answer:

[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]

Explanation:

Two forces are acting on the object.

Subtracting 2 N from both forces.

2 N → Object ← 5 N

- 2 N                 - 2N

0 N → Object ← 3 N

The force 3 N is pushing the object to the left side.

The mass of the object is 10 kg.

Applying formula for acceleration (Newton’s Second Law of Motion).

a = F/m

a = 3/10

a = 0.3

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

[tex] B = \frac{\mu_{0}NI}{L} [/tex]

[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]

Since each turn has length 2πr of wire, the total length is:

[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

Answer:

Thomas is correct that the zero displacements

Lilian is right that the distance is greater than zero.

Explanation:

In this problem we have to be clear about the difference between displacement and distance.

The displacement is a vector, that is, it has a modulation and direction, in this case we can draw a vector for the outward trip and another vector for the return trip, both will have the same magnitude, but their directions are opposite, so the resulting vector is zero.

The distance is a scalar and its value coincides with the modulus of the distance vector, in our case the distance is d for the outward journey and d for the return journey, so the total distance is 2d, which is different from zero.

The two students have some reason, but neither complete,

The displacement is zero because it is a vector and

the distance is different from zero (2d) because it is a scalar

Thomas is correct that the zero displacements

Lilian is right that the distance is greater than zero.

Therefore I agree with both, because each one has a 50% of the reason