Consider an incident normal shock wave that reflects from the end wall of a shock tube. The air in the driven section of the shock tube (ahead of the incident wave) is at p, = 0.01 atm and TI = 300 K. The pressure ratio across the incident shock is 1050. With the use of Eq. (7.23), calculate a. The reflected shock wave velocity relative to the tube b. The pressure and temperature behind the reflected shock

Final answer:

The accurate statement about magnetic field lines is that they form closed loops and never intersect, originating at the north pole and terminating at the south pole of a magnet.

Explanation:

The correct answer regarding magnetic field lines is B. Magnetic field lines form closed loops and never intersect. This observation can be explained through several key principles:

Therefore, options A, C, and D are not correct. The field lines start at the north pole and end at the south pole, not the other way around. Positive charges are affected by magnetic fields but not always along the magnetic field lines because the magnetic force is perpendicular to both the velocity of the charge and the magnetic field.

Answer:[tex]P_{bulb}=99.11\ kPa[/tex]

Explanation:

Given

When the bulb is squeezed then liquid level rises to height h i.e. pressure decreases inside the bulb which causes rises in the tube

for common elevation i.e. at liquid level pressure must be equal therefore

[tex]P_{bulb}+\rho gh=P_{atm}[/tex]

[tex]P_{bulb}=1.013\times 10^5-1490\times 9.8\times h[/tex]

for [tex]h=0.15\ m[/tex]

[tex]P_{bulb}=1.013\times 10^5-1490\times 9.8\times 0.15[/tex]

[tex]P_{bulb}=101.3\times 10^3-2.1903\times 10^3[/tex]

[tex]P_{bulb}=99.11\ kPa[/tex]

for [tex]h=0.1\ m[/tex]

[tex]P_{bulb}=101.3\times 10^3-1.4602\times 10^3[/tex]  

[tex]P_{bulb}=99.83\ kPa[/tex]

Answer:

99 dB

Explanation:

To find the new sound intensity level you calculate first the initial intensity by using the following formula:

[tex]\beta=10log(\frac{I}{I_o})\\\\10^{\frac{\beta}{10}}=10^{log(\frac{I}{I_o})}\\\\10^{\frac{\beta}{10}}=\frac{I}{I_o}\\\\I=I_o10^{\frac{\beta}{10}}[/tex]

where β is the sound level of 79dB and Io is the hearing threshold of 10^-12 W/m^2. By replacing you obtain:

[tex]I=(10^{-12}W/m^2)10^{\frac{79}{10}}=7.94*10^{-5}W/m^2[/tex]

The new sound intensity level is given by:

[tex]\beta'=10log(\frac{100I}{I_o})=10log(\frac{100(7.94*10^{-5}W/m^2)}{10^{-12}W/m^2})\\\\\beta'=99\ dB[/tex]

hence, the answer is 99 dB

Answer:

Explanation:

a ) Let angle of incidence and angle of refraction be i and r respectively .

submarine is 300 m away from the shore . The point where laser strikes the surface of sea is 90 m horizontally away .

Tan r = 90 / 120

= 3 / 4

.75

r = 37 degree

c ) sini / sin37 = 1.333

sini = .8

i = 53 degree

Tan 53 = 210 / h , h is height of the building .

h = 210 / tan 53

= 158 m

Answer:

Maximum Power = 144.3 D

The associated focal length of the lens = [tex]6.92*10^{-3} m[/tex]

Explanation:

According to the Lens maker's Formula:

[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )[/tex]

where;

[tex]n_1[/tex] = the refractive index of the medium

[tex]R_1[/tex] and [tex]R_2[/tex] = radius of curvature on each surface

For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;

[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )[/tex]

At maximum power

[tex]\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )[/tex]

= [tex]0.144 \ mm^{-1}[/tex]

This Implies

[tex]f = 6.92 mm\\f = 6.92*10^{-3} \ m[/tex]

Therefore; the power is given by the formula:

[tex]P_{max} = \frac{1}{f}[/tex]

[tex]P_{max}= \frac{1}{6.92*10^{-3}}[/tex]

= 144.3 D

The maximum power is 144.3 D and the associated focal length of the lens is 6.92 [tex]\rm \times 10^{-3}[/tex] m and this can be determined by using the lens maker's formula.

Given :

The lens maker's formula can be used in order to determine the maximum power and associated focal length of the lens.

The lens maker's formula is given below:

[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)[/tex]

where f is the focal length of the lens, n is the refractive index, and [tex]\rm R_1[/tex] and [tex]\rm R_2[/tex] are the radius of curvature on each surface.

The radius of curvature of the first surface is positive and the radius of curvature of the second surface is negative. So, the lens maker's formula becomes:

[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)[/tex]

Now, substitute the values of the known terms in the above formula.

[tex]\rm \dfrac{1}{f}=(1.430 - 1)\left(\dfrac{1}{6.50}+\dfrac{1}{5.50}\right)[/tex]

Now, simplify the above equation in order to determine the value of 'f'.

[tex]\rm \dfrac{1}{f}=0.144\;mm^{-1}[/tex]

f = 6.92 mm = 6.92 [tex]\rm \times 10^{-3}[/tex] m

Now, the maximum power is given by the formula:

[tex]\rm P_{max} =\dfrac{1}{f}[/tex]

[tex]\rm P_{max} = 144.3 \;D[/tex]

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Answer:

The power of the pump is 337.5 W.

Explanation:

Power is the rate of transfer of energy or the rate at which work is done. It is measured in watts (W).

                       Power = [tex]\frac{Energy}{Time}[/tex]

But the energy here is the kinetic energy (K.E), which is the energy possessed by a moving body or object.

           K.E = [tex]\frac{1}{2}[/tex] m[tex]v^{2}[/tex]

where m is the mass = 90 kg and v is the velocity = 15 m/s.

    K.E =  [tex]\frac{1}{2}[/tex] × 90 × [tex]15^{2}[/tex]

         = 10125 Joules

  K.E = 10125 J = 10.125 KJ

So that the power of the pump when t = 30 s is;

              Power = [tex]\frac{10125}{30}[/tex]

                         = 337.5 W

The power of the pump is 337.5 W.

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

Answer:[tex]\omega _f=1.185\ rad/s[/tex]

Explanation:

Given

mass of objects [tex]m=5\ kg[/tex]

Initially mass is at [tex]r=0.9\ m[/tex]

Initial angular speed [tex]\omega_i=0.66\ rad/s[/tex]

Moment of inertia of student and  stool is [tex]I_s=8\ kg-m^2[/tex]

Finally masses are at a distance of [tex]r_f=0.31\ m[/tex] from axis

[tex]I_i=I_p+I_m[/tex]

[tex]I_i=8+2\times 5\times (0.9)^2[/tex]

[tex]I_i=16.1\ kg-m^2[/tex]

Final moment of inertia of the system

[tex]I_f=I_s+I_m[/tex]

[tex]I_f=8+2\times 5\times (0.31)^2[/tex]

[tex]I_f=8+0.961=8.961\ kg-m^2[/tex]

As there is no external torque therefore moment of inertia is conserved

[tex]I_i\omega _i=I_f\omega _f[/tex]

[tex]\omega _f=\frac{16.1}{8.96}\times 0.66[/tex]

[tex]\omega _f=1.796\times 0.66[/tex]

[tex]\omega _f=1.185\ rad/s[/tex]

Final answer:

When the student pulls the objects closer, his angular speed increases to conserve angular momentum.

Explanation:

Angular momentum is conserved in this scenario. Initially, the student with the objects has a certain angular speed and moment of inertia. When he pulls the objects closer, his moment of inertia decreases, resulting in an increase in angular speed to conserve angular momentum.

Answer:

9.2 A

Explanation:

V = IR

115 V = I (12.5 Ω)

I = 9.2 A

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

[tex]\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}[/tex]     ..............3

put here value and we get  

[tex]\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}[/tex]  

[tex]\frac{m1+0.5}{m2+0.5}[/tex]   = 1.4

[tex]\frac{m1+0.5}{m2+0.5} = \frac{7}{5}[/tex]   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm

Using the formula for thin film interference, the minimum thickness of the plastic film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.

The question is asking for the minimum thickness of the film for which maximum light is reflected for the provided wavelengths. This is a classic example of thin film interference.

For constructive interference (maximum reflection), the thickness of the film (t) is given by the formula: t = mλ/2n. Here 'm' is the order of the bright fringe, 'λ' is wavelength, and 'n' is the index of refraction.

Considering the first-order maximum, we find the thickness for each wavelength:
t1 = (1)(477.1 x 10^-9 m)/2(1.58) = 1.51 x 10^-7 m (or 151 nm for 477.1 nm light)
t2 = (1)(668.0 x 10^-9 m)/2(1.58) = 2.12 x 10^-7 m (or 212 nm for 668.0 nm light)

Therefore, the minimum thickness of the film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.

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The correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.

Explanation

The moon is the most popular and well-known natural satellite on planet earth, it is a satellite widely studied by humans, they have even walked on the moon. However, from Earth, at night when you see the moon you always see the same face of the moon, which has caused people to wonder why this phenomenon. The answer to this phenomenon is that we always see the same face of the moon because it takes the same time to rotate once on itself as it does to go around the Earth (about 27 days). The result is that the same part of the moon always points towards the earth. According to the above, the correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.

Answer:

a) An equation for the x-component of the electric field.

Eₓ = (-15xy³ + 5.32xy⁴z²) N/C

b) An equation for the y-component of the electric field.

Eᵧ = (-22.5x²y² + 10.64x²y³z²) N/C

c) An equation for the z-component of the electric field.

Ez = (5.32x²y⁴z) N/C

d) At (-5.0, 2.0, 1.5) m, the electric field is given as

E = (-357.6î + 2,538ĵ + 3,192ķ) N/C

Magnitude of the electric field = 4,093.7 N/C

Explanation:

The electric field is given by the negative of the gradient of the electric potential,

E = −grad V

E = - ∇V

The electric potential is given as

V(x,y,z) = 3αx²y³ - 2γx²y⁴z²

α = 2.5 V/m⁵ and γ = 1.33 V/m⁸

V(x,y,z) = 7.5x²y³ - 2.66x²y⁴z²

grad = ∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)ķ

E = -grad V = -∇V

= -[(∂V/∂x)î + (∂V/∂y)ĵ + (∂V/∂z)ķ

E = -(∂V/∂x)î - (∂V/∂y)ĵ - (∂V/∂z)ķ

E = Eₓî + Eᵧĵ + Ez ķ

a) An equation for the x-component of the electric field.

Eₓ = -(∂V/∂x) = -(∂/∂x)(V)

= -(∂/∂x)(7.5x²y³ - 2.66x²y⁴z²)

= -(15xy³ - 5.32xy⁴z²)

= (-15xy³ + 5.32xy⁴z²)

b) An equation for the y-component of the electric field.

Eᵧ = -(∂V/∂y) = -(∂/∂x)(V)

= -(∂/∂y)(7.5x²y³ - 2.66x²y⁴z²)

= -(22.5x²y² - 10.64x²y³z²)

= (-22.5x²y² + 10.64x²y³z²)

c) An equation for the z-component of the electric field.

Ez = -(∂V/∂z) = -(∂/∂x)(V)

= -(∂/∂z)(7.5x²y³ - 2.66x²y⁴z²)

= -(0 - 5.32x²y⁴z)

= (5.32x²y⁴z)

d) E = Eₓî + Eᵧĵ + Ez ķ

E = (-15xy³ + 5.32xy⁴z²)î + (-22.5x²y² + 10.64x²y³z²)ĵ + (5.32x²y⁴z) ķ

At (-5.0, 2.0, 1.5) m

x = -5 m

y = 2 m

z = 1.5 m

Eₓ = (-15xy³ + 5.32xy⁴z²)

= (-15×-5×2³) + (5.32×-5×2⁴×1.5²)

= 600 - 957.6 = -357.6

Eᵧ = (-22.5x²y² + 10.64x²y³z²)

= (-22.5×(-5)²×2²) + (10.64×(-5)²×2³×1.5²)

= -2250 + 4788 = 2538

Ez = (5.32x²y⁴z) = (5.32×(-5)²×2⁴×1.5)

= 3192

E = -357.6î + 2,538ĵ + 3,192ķ

Magnitude = /E/ = √[(-357.6)² + 2538² + 3192²]

= 4,093.6763135353 = 4,093.7 N/C

Hope this Helps!!!!

The correct answers are as follows:

(a) The equation for the x-component of the electric field is[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The equation for the y-component of the electric field is [tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The equation for the z-component of the electric field is [tex]\[ E_z = 5.32x^2y^4z \][/tex]

(d) the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb is 14907.5 N/C.

To find the components of the electric field, we need to take the negative gradient of the electric potential function V(x,y,z). The gradient of a function is a vector field whose components are the partial derivatives of the function with respect to each variable. The electric field [tex]\( \vec{E} \)[/tex] is related to the electric potential [tex]\( V \)[/tex] by the equation:

[tex]\[ \vec{E} = -\nabla V \][/tex]

where [tex]\( \nabla \)[/tex] is the gradient operator.

(a) The x-component of the electric field [tex]\( E_x \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( x \)[/tex] :

[tex]\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_x = -3α(2xy^3) + 2γ(2xy^4z^2) \][/tex]

[tex]\[ E_x = -6αxy^3 + 4γxy^4z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_x = -6(2.5)xy^3 + 4(1.33)xy^4z^2 \][/tex]

[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The y-component of the electric field [tex]\( E_y \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( y \)[/tex] :

[tex]\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex] [tex]\[ E_y = -3αx^2(3y^2) + 2γx^2(4y^3z^2) \][/tex]

[tex]\[ E_y = -9αx^2y^2 + 8γx^2y^3z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_y = -9(2.5)x^2y^2 + 8(1.33)x^2y^3z^2 \][/tex]

[tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The z-component of the electric field

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( z \)[/tex] :

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -2γx^2y^4(-2z) \][/tex]

[tex]\[ E_z = 4γx^2y^4z \][/tex]

Substituting the given value of[tex]\( γ \)[/tex] :

[tex]\[ E_z = 4(1.33)x^2y^4z \]\\ E_z = 5.32x^2y^4z \][/tex]

(d) To calculate the magnitude of the electric field at the point[tex]\( P(-5.0, 2.0, 1.5) \)[/tex] m, we first substitute [tex]\( x = -5.0 \)[/tex] m, [tex]\( y = 2.0 \)[/tex] m, and [tex]\( z = 1.5 \)[/tex] m into the equations for[tex]\( E_x \)[/tex], [tex]\( E_y \)[/tex] , and [tex]\( E_z \)[/tex] :

[tex]E_x = -15(-5.0)(2.0)^3 + 5.32(-5.0)(2.0)^4(1.5)^2 \]\\ E_x = 1500 - 5.32(-5.0)(16)(2.25) \]\\ E_x = 1500 - (-851.88) \]\\ E_x = 1500 + 851.88 \]\\ E_x = 2351.88 \text{ N/C} \][/tex]

[tex]\[ E_y = -22.5(-5.0)^2(2.0)^2 + 10.64(-5.0)^2(2.0)^3(1.5)^2 \]\\ E_y = -22.5(25)(4) + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 4752 \]\\ E_y = 2552 \text{ N/C} \][/tex]

[tex]\[ E_z = 5.32(-5.0)^2(2.0)^4(1.5) \]\\ E_z = 5.32(25)(16)(1.5) \]\\ E_z = 5.32(600) \]\\ E_z = 3192 \text{ N/C} \][/tex]

Now, the magnitude of the electric field \( \vec{E} \) is given by:

[tex]\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{(2351.88)^2 + (2552)^2 + (3192)^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{5527641.64 + 6512644 + 10180416} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{22220502.64} \][/tex]

[tex]\[ |\vec{E}| \approx 14907.5 \text{ N/C} \][/tex]

Therefore, the magnitude of the electric field at the point [tex]\( P(-5.0, 2.0, 1.5) \)[/tex]m is approximately [tex]\( 14907.5 \)[/tex] N/C.

Answer:oxygen

Explanation:algae are plants that lives in aquatic habitat,with a few of them occurring in land. Algae have chlorophyll and as a result are autotrophic in nutrition. Algae uses carbon dioxide as a raw material for photosynthesis which is the process where they produce food. They also give off oxygen which results from the splitting of water by light.

This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.

Final answer:

Carbon dioxide is a metabolic waste of photosynthesis in algae that can be recycled back into the ecosystem through cellular respiration.

Explanation:

Among the metabolic wastes produced by algae during photosynthesis, carbon dioxide (CO₂) is a byproduct that can be recycled and used in cellular respiration. Photosynthesis is the process in which algae absorb light energy to create carbohydrates within chloroplasts, releasing oxygen as a byproduct. Alternatively, cellular respiration involves using oxygen to break down carbohydrates, primarily in the cytoplasm and mitochondria, releasing ATP and carbon dioxide. These two processes are synchronously linked in a biological cycle that allows organisms to harness energy from the sun, guide by electron transport chains that drive cellular reactions.

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

[tex]x=x_{0}+V t \\ x=\frac{1}{2}+V t[/tex]

The distance between the police car and the intersection is,

[tex]y=y_{0}+V t[/tex]

[tex]y=\frac{1}{2}-40 t[/tex]

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

[tex]z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })[/tex]

[tex]z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)[/tex]

The rate of change is,

[tex]2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)[/tex]

[tex]2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots[/tex]

Now finding [tex]z[/tex] when [tex]t=0,[/tex] from (1) we have

[tex]z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}[/tex]

[tex]z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071[/tex]

The officer's radar gun indicates 25 mph pointed at the other car then, [tex]\frac{d z}{d t}=25[/tex] when [tex]t=0,[/tex] from

From (2) we get

[tex]2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)[/tex]

[tex]2(0.7071)(25)=V+2 V^{2}(0)-40[/tex]

[tex]35.36=V-40[/tex]

[tex]V=35.36+40=75.36[/tex]

Hence the speed of the car is [tex]75.36 mph[/tex]

To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.

To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.

To calculate the speed of the other car, we can use the Pythagorean theorem. The police car is moving in a straight line, so the relative velocity is the hypotenuse of a right triangle. The northward velocity of the police car is one side of the triangle, and the eastward velocity of the other car is the other side. Using the equation a² + b² = c², where a is the northward velocity, b is the eastward velocity, and c is the relative velocity, we can solve for b. Rearranging the equation, we get b = √(c² - a²). Plugging in the known values, we have b = √(25² - 40²) = √(625 - 1600) = √(-975)

The square root of a negative number is not a real number, so it is not possible to determine the speed of the other car using only the given information.

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Answer:[tex]c=3.99\approx 4\ kJ/kg-K[/tex]

Explanation:

Given

Current [tex]I=300\ A[/tex]

Voltage [tex]V=270\ V[/tex]

Mass flow rate [tex]m=0.307\ kg/s[/tex]

Inlet temperature is [tex]T_i=15^{\circ}C\approx 288\ K[/tex]

[tex]T_{out}\leq 81^{\circ}C[/tex]

Here heat of Electromagnet is absorbed by liquid

Heat rate of Electromagnet [tex]\dot{Q}=VI[/tex]

[tex]\dot{Q}=270\times 300=81\ kJ/s[/tex]

Heat absorbed by liquid [tex]\dot{Q}=mc(\Delta T)[/tex]

[tex]\dot{Q}=0.307\times c\times (81-15)[/tex]

[tex]81\times 10^3=0.307\times c\times (81-15)[/tex]

[tex]c=3.99\ kJ/kg-K[/tex]

Answer:

The period is  [tex]T = 8.27 \ sec[/tex]

Explanation:

From the question we are told that

   The extension of the spring is  [tex]e_1 = 33 \ m[/tex]

   The  first weight  applied is  [tex]F_1 = 19 N[/tex]

     The second weight applied is  [tex]F_2 = W[/tex]

     The second extension is [tex]e_2 = 17 \ m[/tex]

The spring constant of the spring is mathematically evaluated as

         [tex]k = \frac{F_1}{e_1 }[/tex]

substituting values

       [tex]k = \frac{19}{33 }[/tex]

       [tex]k = 0.576[/tex]

We are told that

         19 N extended the spring to 33 m    

Then W N  will extended it by  17 m

Therefore     [tex]W = \frac{19 * 17}{33}[/tex]

                    [tex]W = 9.788 \ N[/tex]

Generally the period of the oscillation is mathematically represented as

           [tex]T = 2 \pi \sqrt{\frac{M}{K} }[/tex]

where M is the mass of the W which is mathematically evaluated as

         [tex]M = \frac{9.788}{9.8}[/tex]

          [tex]M = 1.0 \ kg[/tex]

substituting values

        [tex]T = 2 \pi \sqrt{\frac{1}{0.576} }[/tex]

       [tex]T = 8.27 \ sec[/tex]

The force required to stretch a spring is directly proportional to the amount the spring is stretched. The force constant can be calculated using the equation F = kx. By substituting the given values into the equation, we can find the unknown weight W.

The force required to stretch a spring is directly proportional to the amount the spring is stretched. This can be represented by the equation F = kx, where F is the force, k is the force constant, and x is the amount the spring is stretched. In this scenario, we are given that a 19 N weight stretches the spring by 33 m at equilibrium.

To find the force constant, we can use the equation:

k = F / x = 19 N / 33 m = 0.58 N/m

Next, we need to find the unknown weight W that will stretch the spring to a new equilibrium position 17 m below the position with no weight attached. Using the force constant we calculated, we can use the equation F = kx to find the force required:

F = kx = (0.58 N/m)(17 m) = 9.86 N

Therefore, the unknown weight W that will stretch the spring to the new equilibrium position is 9.86 N.

Answer:

This is because of the inertia of the object.

Explanation:

This is because of the inertia of the object.

When you push a static object you must overcome the static friction force of the object. Once you have overcome the static friction the inertia law demands that the object tend to conserve its motion. That is the reason why you need less force when the object is already in motion. In other words, the inertia gained by the object with the initial force, "helps" you with the work of moving the object. This is also the reason why the kinetic friction of an object in motion over a surface is lower than the static friction.

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.

Answer:

speed after one riveting operation = 230.5 rpm

mass moment of inertia = 9.90 kg

Explanation:

given data

torque energy  = 2.5 kW = 2500 W

mass = 125 kg

radius = 700 mm

energy = 1000 J

Speed of the flywheel N = 240 rpm

solution

we know here Speed of the flywheel N so here angular speed ω

ω = [tex]\frac{2\pi N}{60}[/tex]   .....................1

ω = [tex]\frac{2\pi 240}{60}[/tex]

ω = 25.13 rad/s

so here

change in energy is

ΔE = E1 - E2    ..............2

ΔE = 2500 - 1000

ΔE = 1500 J/sec

and

I = mr²    .........3

I = 125 × 0.7²

I = 61.25 kg-m²

and ΔE is express as here

ΔE = 0.5 × I × (ω² - ω1² )      ........4

put here value and we get

1500 =  0.5 × 61.25 × (25.13² - ω1² )

ω1 = 24.13 rad /s

and

reduction in speed after one riveting operation will be

N = [tex]\frac{24.13\times 60 }{2\pi }[/tex]

speed after one riveting operation = 230.5 rpm

and

for 35 rpm

ω1 = [tex]\frac{2\pi 35}{60}[/tex]

ω1 = 3.65

so ΔE  will be here

ΔE = 0.5 × mr² × (ω² - ω1² )    ....................5

put here value and we get m

1500 = 0.5 × m (0.7)² × (25.13² - 3.65² )

solve it we get

mass moment of inertia = 9.90 kg

Answer:

Explanation:

The star is revolving the black hole like earth revolves around the sun .so time period of rotation  T is given by the following relation

T² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex] , R is distance between black hole and star , M is mass of black hole

Given T = 4.8 hours

4.8² =  [tex]\frac{4\pi^2\times R^3}{GM }[/tex]

Using the same equation for earth sun system

24² =  [tex]\frac{4\pi^2\times (50R)^3}{GM_s }[/tex]  , Ms is mass of the sun and 50R is distance between the sun and the earth .

Dividing the equation

[tex](\frac{4.8}{24})^2[/tex] = [tex]\frac{M_s}{M}\times\frac{1^3}{50^3}[/tex]

[tex]\frac{M}{M_s}[/tex] = 2x 10⁻⁴

The mass of the black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.

Given information:

An X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours (T).

The star is orbiting around the black hole.

The distance between the centers of the star and the black hole is one-fiftieth of the distance between the centers of the Earth and our Sun.

Let R be the center to center distance between the black hole and the sun.

So, the time period T of the sun around the black hole will be,

[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\4.8^2=\dfrac{2\pi^2R^3}{GM}[/tex]

where M is the mass of the black hole.

The distance between the earth and our sun will be 50R.

So, the time period of the earth will be,

[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}[/tex]

where [tex]M_s[/tex] is the mass of our sun.

Now, compare the above two relations to get the mass of black hole in terms of mass of our sun as,

[tex]4.8^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}\\\dfrac{24^2}{4.8^2}=\dfrac{50^3}{1}\times \dfrac{M}{M_s}\\\dfrac{M}{M_s}=0.0002\\\dfrac{M}{M_s}=2\times 10^{-4}[/tex]

Therefore, the mass of black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.

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Answer:

Explanation:

Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli

= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴

=  13216 x 10⁻⁹ weber .

b ) When the coil becomes parallel to magnetic field  , flux through it will become zero.

c ) e m f induced = change in flux / time

= 13216 x 10⁻⁹ / 4.9 x 10⁻²

= 2697.14 x 10⁻⁷ V

= 269.7 x10⁻⁶

269.7 μV.

Answer:

a) [tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex], b) [tex]\Delta n = 7197.697\,rev[/tex], c) [tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex], d) [tex]a = 22.823\,\frac{m}{s^{2}}[/tex]

Explanation:

a) Constant angular acceleration is:

[tex]\ddot n = \frac{\dot n - \dot n_{o}}{\Delta t}[/tex]

[tex]\ddot n = \frac{0\,\frac{rev}{min} - 150\,\frac{rev}{min}}{(1.6\,h)\cdot \left(60\,\frac{min}{h} \right)}[/tex]

[tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex]

b) The amount of revolutions required to stop the flywheel is:

[tex]\Delta n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]

[tex]\Delta n = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(150\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-1.563\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]\Delta n = 7197.697\,rev[/tex]

c) The tangential acceleration of the particle is:

[tex]a_{t} = \left(1.563\,\frac{rev}{min^{2}} \right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}}\right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot (0.37\,m)[/tex]

[tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex]

d) The radial acceleration of the particle is:

[tex]a_{r} = \left[\left(75\,\frac{rev}{min} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\right]^{2}\cdot (0.37\,m)[/tex]

[tex]a_{r} = 22.823\,\frac{m}{s}[/tex]

The net linear acceleration is:

[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex]

[tex]a = \sqrt{\left(22.823\,\frac{m}{s^{2}} \right)^{2}+\left(1.009\times 10^{-3}\,\frac{m}{s^{2}} \right)^{2}}[/tex]

[tex]a = 22.823\,\frac{m}{s^{2}}[/tex]

Answer:

20m

Explanation:

Pressure = pgh

p = density of water 1000

kg/m^3

g = acceleration due to gravity 9.81 m/s^2

h is the depth of water

Pressure = 201 kPa = 201 x 10^3 Pa

201 x 10^3 = 1000 x 9.81 x h

201 x 10^3 = 9810h

h = 20.49 m

Approximately 20 m

Answer:

Explanation:

a ) angular frequency ω = [tex]\sqrt{\frac{k}{m} }[/tex]

k is spring constant and m is mass attached

ω = [tex]\sqrt{\frac{20}{1.5} }[/tex]

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

The frequency of the oscillation is 0.58 Hz and the displacement x varies as a cosine function. The maximum speed is 0.366 m/s and maximum acceleration is 0.427 m/s². The kinetic and potential energies are equal when the mass is 0.0707m away from the equilibrium point.

(a) The frequency of the oscillations is given by the formula f = 1/2π √(k/m), where k is the spring constant and m is the mass. Substituting the given values we get f = 1/2π √(20.0 N/m / 1.5 kg) = 0.58 Hz. The displacement as a function of time is given by the equation x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. Given the mass was released from rest, the phase angle is 0 and the equation becomes x(t) = (0.10 m) cos(2πft).

(b) The maximum speed is given by the formula vmax = Aω, substituting the values, vmax = (0.10 m)(2πf) = 0.366 m/s. The maximum acceleration is given by the formula amax = Aω² which is amax = (0.10 m)(2πf)² = 0.427 m/s².

(c) The kinetic energy and potential energy of the system are the same when the displacement is equal to the amplitude divided by √2: x = A/√2, that is x = 0.10m /√2 = 0.0707 m from the equilibrium point.

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Answer:

B(max) =  3.7971 × [tex]10^{-12}[/tex]  T

Explanation:

given data

radius R = 26 mm

plate separation d = 4.0 mm

potential difference Vm = 220 V

frequency f = 76 Hz

V = (220 V) sin[2π(76 Hz)t]

solution

we know that E will be

E = V ÷ d     ............1

put here value

E = [tex]\frac{220 \times sin(2\pi 76\times t)}{d}[/tex]  

and here we take as given r = R

so A = π R²    .................2

and

ФE  = E × A

ФE = [tex]\frac{\pi R^2 \times 220 \times sin(2\pi 76 \times t)}{d}[/tex]  .....................3

so use use here now  Ampere's Law that is

∫ B ds = [tex]\mu_o \times \epsilon_o \times \frac{d\Phi E}{dt} + \mu_o \times I_{encl}[/tex]      .....................4

and

here [tex]I_{encl}[/tex]  is = 0  and r = R

so

[tex]2B \times \pi \times R = \mu_o \times \epsilon_o \times \frac{d\Phi E}{dt}[/tex]      .....................5

and put here value we get

B =  [tex]\frac{\mu_o \times \epsilon_o \times \pi \times f \times R \times V_m cos(2\pi f t)}{d}[/tex]        .....................6

put here value  for B maximum cos(2πft) = 1

and we get B (max)

B(max) = [tex]\frac{\mu_o\times \epsilon_o\times \pi \times f\times R\times V_m}{d}[/tex]     ....................7

put here all value

B(max) = [tex]\frac{4\pi \times10^{-7} \times 8.85\times 10^{-12}\times \pi \times 76 \times 0.026\times 220 }{4\times 10^{-3}}[/tex]      

solve it we get

B(max) =  3.7971 × [tex]10^{-12}[/tex]  T

Answer:

 h ’= k  (h +L/2)²/ (2 M g )

Explanation:

For this exercise, one of the best methods to solve it is with energy conservation.

Starting point. Lower, spring with maximum compression

              Em₀ = Ke = ½ k x²

Final point. Higher after bounce

              = U = M g h ’

             Em₀ = Em_{f}

             ½ k x² = M g h’

the compressed distance is

          x = h+ L / 2

where h is the distance that compresses the spring by the height where it comes from and L/2 the additional compression

        h ’= ½ k x² / M g

we calculate

       h ’= k  (h +L/2)²/ (2 M g )

       h' = (k/2Mg) h2 (1 + L/2h)2

Answer:764.4N

Explanation:

Mass=78kg

Acceleration=9.8m/s^2

force=mass x acceleration

Force=78 x 9.8

Force=764.4

Force=764.4N

Answer:

867755.73 V/m

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

The final temperature of the ice is [tex]-40^{\circ}\ C[/tex].

Explanation:

It is given that,

Energy, Q = 10 cal

Mass of ice, m = 2 g

Initial temperature, [tex]T_1=-30^{\circ}\ C[/tex]

We need to find the final temperature of the ice. We know that the specific heat of ice is [tex]0.5\ cal\ g^{-1} ^{\circ} C^{-1}[/tex]

The heat added in terms of specific heat is given by :

[tex]Q=mc(T_1-T_2)[/tex]

[tex]T_2[/tex] final temperature of the ice

c is specific heat of ice

[tex]T_2=T_1-\dfrac{Q}{mc}\\\\T_2=(-30)-\dfrac{10}{2\times 0.5}\\\\T_2=-40^{\circ}[/tex]

So, the final temperature of the ice is [tex]-40^{\circ}\ C[/tex].

Answer: the correct answer is -20

Explanation:

Answer:

Option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.

Explanation:

Regarding Rayleigh's criterion, the angular resolution is given as below:

θ = 1.22λ/D

From this expression, it is observed that the larger the aperture size, the smaller will be the value of the angular resolution, and the better the device will be.

This signifies that at very high angular differences, the precision for distinguishing two points is high.

Therefore, option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.