Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity is 6.20 m/s2, and atmospheric pressure is 7.00 ✕ 104 Pa. The atmospheric temperature and pressure on this planet causes the density of the liquid methane ocean to be 415 kg/m3. (a) What force (in N) is exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of the ocean? N (b) What is the weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m? (Enter your answer in N.) N (c) What is the pressure (in Pa) at a depth of 10.0 m in the methane ocean?

Answer:

[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Explanation:

It is given that,

Distance between two infinitely long wires, d = 1 m

Current flowing in both of the wires, I = 1 A

The magnetic field in a wire is given by :

[tex]B=\dfrac{\mu_oI}{2\pi d}[/tex]

The force per unit length acting on the two infinitely long wires is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}[/tex]

[tex]\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 1^2}{2\pi \times 1}[/tex]

[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

So, the force acting on the parallel wires is [tex]2\times 10^{-7}\ N/m[/tex]. Hence, this is the required solution.

The force per unit length between the two wires is [tex]\( 2 \times 10^{-7} \)[/tex] newtons per meter.

The force per unit length (F/L) between two infinitely long wires carrying current can be calculated using Ampere force law, which is given by:

[tex]\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r} \][/tex]

 Plugging in the values, we get:

[tex]\[ \frac{F}{L} = \frac{(4\pi \times 10^{-7} \text{ N/A}^2) \times (1 \text{ A}) \times (1 \text{ A})}{2 \pi \times (1 \text{ m})} \][/tex]

Simplifying the expression by canceling out [tex]\( \pi \)[/tex] and multiplying the numerical values, we have:

[tex]\[ \frac{F}{L} = \frac{(4 \times 10^{-7} \text{ N/A}^2) \times (1 \text{ A})^2}{2 \times (1 \text{ m})} \][/tex]

[tex]\[ \frac{F}{L} = \frac{4 \times 10^{-7} \text{ N}}{2 \text{ m}} \][/tex]

[tex]\[ \frac{F}{L} = 2 \times 10^{-7} \text{ N/m} \][/tex]

 Therefore, the force per unit length between the two wires is [tex]\( 2 \times 10^{-7} \)[/tex] newtons per meter.

The answer is: [tex]2 \times 10^{-7} \text{ N/m}[/tex]

Answer:

The number of capillaries is

[tex]N=2.625x10^9[/tex] Capillaries

Explanation:

[tex]v_{aorta}=42cm/s[/tex], [tex]r_{aorta}=1.1 cm[/tex], [tex]v_{cap}=0.064cm/s[/tex], [tex]r_{cap}=5.5x10^{4}cm[/tex],

To find the number of capillaries in the human body use the equation:

[tex]N_{cap}=\frac{v_{aorta}*\pi*r_{aorta}^2}{v_{cap}*\pi*r_{cap}^2}[/tex]

So replacing numeric

[tex]N_{cap}=\frac{42cm/s*\pi*(1.1cm)^2}{0.064cm/s*\pi*(5.5x10^{-4}cm)^2}[/tex]

Now we can find the number of capillaries

[tex]N=26250000000[/tex]

[tex]N=2.625x10^9[/tex] Capillaries

Answer: I = k^2m.. equa1

I = moment of inertia

M = mass of skater

K = radius of gyration.

When her angular momentum is conserved we have

Iw = I1W1... equ 2

Where I = with extended arm, w = angular momentum =40rads/s, I1 = inertia when hands her tucked in, w1 = angular momentum when hands are tucked in.

Substituting equation 1 into equ2 and simplifying to give

W = (k/k1)^2W..equation 3

Where'd k= 64cm, k1 = 32cm, w = angular momentum when hands is tucked in= 40rad/s

Substituting figures into equation 3

W1 = 10rad/s

Explanation:

Assuming a centroidal axis of the skater gives equation 1

Final answer:

When the moment of inertia is doubled, the angular velocity will decrease by half due to the conservation of angular momentum.

Explanation:

When the moment of inertia is doubled, the angular velocity will decrease by half due to the conservation of angular momentum. In this case, Kristen's initial angular velocity is 40 rad/s, and her initial moment of inertia is 32 cm. After doubling her radius of gyration to 64 cm, her final moment of inertia is 128 cm. Using the conservation of angular momentum equation, we can calculate her final angular velocity:

Initial Angular Momentum = Final Angular Momentum

Initial Angular Velocity * Initial Moment of Inertia = Final Angular Velocity * Final Moment of Inertia

Substituting the values: 40 rad/s * 32 cm = Final Angular Velocity * 128 cm

Simplifying the equation: Final Angular Velocity = 10 rad/s

Therefore, Kristen's angular velocity about her longitudinal axis after increasing her radius of gyration is 10 rad/s.

Answer:

the maximum distance of rotation can he stand without sliding is 1.77 m

Explanation:

given information:

angular velocity , ω = 1.58 rad/s

static friction, μs = 0.45

now we calculate the vertical force

N - W = 0, N is normal force and W is weight

N = W

   = m g

next, for the horizontal force we only have frictional force, thus

F(friction) = m a

μs N = m a

μs m g = m a

a = μs g,

now we have to find the acceleration which is both translation and cantripetal.

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

[tex]a_{t} ^{2}[/tex] is the acceleration for translation

[tex]a_{t} ^{2}[/tex] = 0

[tex]a_{c} ^{2}[/tex] is centripetal acceleration

[tex]a_{c} ^{2}[/tex] = ω^2r

therefore,

a = [tex]\sqrt{a_{c} ^{2}  }[/tex]

  = [tex]a_{c} ^{2}[/tex]

  = ω^2r

Now, to find the radius, substitute the equation into the following formula

a = μs g

ω^2r = μs g

r = μs g / ω^2

 = (0.45 x 9.8) / (1.58)

 = 1.77 m

Answer:

Thermal energy will be equal to 784 J

Explanation:

We have given that mass of the child m = 40 kg

Height h = 2 m

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We have to find the thermal energy '

The thermal energy will be equal to potential energy

And we know that potential energy is given by

[tex]W=mgh=40\times 9.8\times 2=784J[/tex]

So the thermal energy will be equal to 784 J

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = [tex].250\times 10^{- 3}\ m[/tex]

Now,

Applying the equilibrium condition at the left end for torque:

[tex]T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}[/tex]

[tex]T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}[/tex]

[tex]T_{C} = 341.357\ Nm[/tex]

The weight of the wire balances the tension in both the wires collectively:

[tex]W = T_{Al} + T_{C}[/tex]

[tex]531 = T_{Al} + 341.357[/tex]

[tex]T_{Al} = 189.643\ Nm[/tex]

Now,

The fundamental frequency is given by:

[tex]f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]

where

[tex]\mu = A\rho = \pi R^{2}\rho[/tex]

(a) For the fundamental frequency of Aluminium:

[tex]f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}[/tex]

[tex]f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}[/tex]

where

[tex]\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}[/tex]

[tex]f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz[/tex]

(b)  For the fundamental frequency of Copper:

[tex]f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}[/tex]

[tex]f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}[/tex]

where

[tex]\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}[/tex]

[tex]f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz[/tex]

Answer:

the heat released by the magnesium is 72 kJ

Explanation:

the heat exchanged will be

Q = m*c*(T final - T initial)  

where Q= heat released, c= specific heat capacity, T initial= initial temperature of water, T final = final temperature of water

Assuming the specific heat capacity of water as c= 1 cal/g°C=4.186 J/g°C

replacing values

Q = m * c* (T final - T initial) = 1910 g * 4.186 J/g°C*(33.10 °C - 24°C) =  72 kJ

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string [tex]T=5 N[/tex]

[tex]2\pi r=2[/tex]

[tex]r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m[/tex]

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is [tex]90^{\circ}[/tex]

[tex]W=\int\vec{F}\cdot \vec{r}[/tex]

[tex]W=\int Fdr\cos 90 [/tex]

[tex]W=0[/tex]

The work done by the ball as it travels once around the circular string is 0.

The given parameters;

The work-done by a body is the dot product of applied force and displacement.

For one complete rotation around the circumference of the circle, the displacement of the object is zero.

The work-done by the ball when its  makes a complete rotation around the circle is calculated as;

Work-done = F x r

where;

Work-done = 5 x 0 = 0

Thus, the work done by the ball as it travels once around the circular string is 0.

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Answer:

0.639 V

Explanation:

The volatge of the cell containing both Ag/AgCl reference electrode and

[tex]Fe^{2+}/Fe^{3+}[/tex] electrode = 0.693 V

Thus,

[tex]E_{cathod}-E_{anode} =0.693 V[/tex]

E_{anode}=0.197 V

Note: potential of the silver-silver chloride reference electrode (0.197 V)

⇒E_{Cthode}= 0.693+0.197 = 0.890V

To calculate the voltage of the cell containing both the calomel reference

electrode and [tex]Fe^{2+}/Fe^{3+}[/tex] electrode as follows

Voltage of the cell = [tex]E_{cathod}-E_{anode}[/tex]

E_{anode}= calomel electrode= 0.241 V

Voltage of the cell = 0.890-0.241 = 0.639 V

Therefore, the new volatge is = 0.639 V

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁  - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

T - m g = m a

T =  m ω² r  + m g

T + m g = m a

T=  m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

Final answer:

The tension in the cord connected to a ball whirled in a vertical circle is greatest at the bottom of the circle because the tension must overcome gravity and provide the centripetal force to keep the ball in motion.

Explanation:

The question asks, "A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?" The correct answer is c. At the bottom of the circle.

When the ball is at the bottom of the circle, the tension in the string is highest because it must counteract both the gravitational force pulling downwards and provide enough centripetal force to keep the ball moving in a circular path. At this point, the sum of the gravitational force and the centripetal force dictates the necessary tension, making it greater than at any other point in the ball's circular trajectory. This is in contrast to the top of the circle, where the tension is lowest since gravity assists in providing the centripetal force needed, sometimes reducing the tension in the string to nearly zero if the ball moves at the minimum speed required to continue in circular motion.

Answer:

Total internal reflection is going on. The refractive index of water is about 1.3, so sin 90/sin r=1/sinr=1.3. So the fish can only see objects outside the water within about 50 degrees of the vertical

Answer:

4341.44763 kg/m³

Explanation:

[tex]\rho'[/tex] = Actual density of cube = 1800 kg/m³

[tex]\rho[/tex] = Density change due to motion

v = Velocity of cube = 0.91c

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

Relativistic density is given by

[tex]\rho=\frac{\rho'}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-\frac{0.91^2c^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-0.91^2}}\\\Rightarrow \rho=4341.44763\ kg/m^3[/tex]

The cube's density as measured by an experimenter in the laboratory is 4341.44763 kg/m³

Answer:

(a) [tex]\displaystyle s(t)= \frac{t^3}{3}-\frac{t^2}{2}-20t+C\ \ \ \ \forall\ \  3\leqslant t\leqslant 9[/tex]

(b) 78 m

Explanation:

Physics' cinematics as rates of change.

Velocity is defined as the rate of change of the displacement. Acceleration is the rate of change of the velocity.

[tex]\displaystyle v=\frac{ds}{dt}[/tex]

Knowing that

[tex]\displaystyle v(t)= t^2 - t - 20[/tex]

(a) To find the displacement we need to integrate the velocity

[tex]\displaystyle \frac{ds}{dt}=t^2 - t - 20[/tex]

[tex]\displaystyle ds=(t^2 - t - 20)dt[/tex]

[tex]\displaystyle s(t)= \int(t^2 - t - 20)dt=\frac{t^3}{3}-\frac{t^2}{2}-20t+C\ \ \ \ \forall \ \ \ 3\leqslant t\leqslant 9[/tex]

(b) The displacement can be found by evaluating the integral

[tex]\displaystyle d=\int_{3}^{9} (t^2 - t - 20)dt[/tex]

[tex]\displaystyle d=\left | \frac{t^3}{3}-\frac{t^2}{2}-20t \right |_3^9=\frac{45}{2}+\frac{111}{2}=78\ m[/tex]

To find displacement, we integrate the velocity function from 3 to 9. The displacement is the integral of the velocity function. For distance, we integrate the absolute value of the velocity function from 3 to 9 because distance is a scalar quantity and includes total path length.

To solve the problem, we need to find the displacement and distance traveled by the particle. For part (a), the displacement for the given time period is obtained by integrating the velocity function from 3 to 9. The displacement is the integral of the velocity function v(t) = t2 - t - 20 from 3 to 9, which gives us the value s(9) - s(3).

For part (b), to calculate the distance traveled, we need to integrate the absolute value of the velocity function because the distance is always positive. Negative values would represent backward motion, but the distance traveled includes total path length and does not care about direction.

So, the distance traveled from time 3 to 9 would be ∫ |t2 - t - 20| dt from 3 to 9. The calculation of this integral will give the distance traveled.

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Answer:

True A and B

Explanation:

Let's propose the solution of the exercise before seeing the affirmations.

We use the law of refraction

      n₁ sin θ₁ = n₂. Sin θ₂

Where n₁ and n₂ are the refractive indices of the two means, θ₁ and θ₂ are the angles of incidence and refraction, respectively

      sin θ₂ = (n1 / n2) sin θ₁

Let's apply this equation to the case presented. The index of refraction and airs is 1 (n1 = 1)

     Sin θ₂ = (1 / n2) sin θ₁

 the angle  θ₂ which is the  refracted angle is less than the incident angle

Let's analyze the statements time

A. False. We saw that it deviates

B. True Approaches normal (vertical axis)

C. False It deviates, but it is not parallel to normal

D. False It deviates, but approaching the normal not moving away

E. True. Because its refractive index is higher than air,

When a beam of light with an angle of incidence less than 90° enters a denser medium like a glass slab, it bends closer to the normal, bends away from the normal, and slows down, dependent on the refractive indices of the two media. Here options B, D, and E are correct.

When a beam of light travels from air into a denser medium, such as a glass slab, it undergoes refraction. Refraction is the bending of light as it passes from one medium to another with a different optical density.

The angle of incidence, the angle formed between the incident ray and the normal (a line perpendicular to the surface at the point of incidence), plays a crucial role in determining the behavior of the refracted light.

These statements are correct. The degree to which the light bends depends on the refractive indices of the two media. In this case, as light enters the glass slab, it slows down due to the higher refractive index of glass compared to air.

The bending of light towards the normal and slowing down are characteristic behaviors of light when it travels from a less dense to a denser medium. Here options B, D, and E are correct.

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Answer:

 i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

    2 L = n λ           n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

     v = λ f

     λ = v / f

Let's replace in the first equation

    2 L = 1 (v / f₁)

For the shortest length L = L-l

   2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

    v = 2L f₁

    v = 2 (L-l) f₂

    2L f₁ = 2 (L-l) f₂

    L- l = L f₁ / f₂

    l = L - L f₁ / f₂

    l = L (1- f₁ / f₂)

.

Let's calculate

    l / L = (1- 309/463)

    i / L = 0.3326

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

[tex]f_i = f_0 \frac{c}{c+v}[/tex]

Where

[tex]f_h[/tex]= Approaching velocities

[tex]f_i[/tex]= Receding velocities

c = Speed of sound

v = Emitter speed

And

[tex]f_h = f_0 \frac{c}{c+v}[/tex]

Therefore using the values given we can find the velocity through,

[tex]\frac{f_h}{f_0}=\frac{c-v}{c+v}[/tex]

[tex]v = c(\frac{f_h-f_i}{f_h+f_i})[/tex]

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

[tex]v = 353(\frac{2.4-1}{2.4+1})[/tex]

[tex]v = 145.35m/s[/tex]

Therefore the cars goes to 145.3m/s

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)=[tex](2.4-x^2)\hat{i}N[/tex]

Initial position of the block=x=0

Initial velocity of block=[tex]v_i=0[/tex]

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy=[tex]K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0[/tex]

Work energy theorem:

[tex]K_f-K_i=W[/tex]

Where [tex]K_f=[/tex]Final kinetic energy

[tex]K_i[/tex]=Initial kinetic energy

[tex]W=Total work done[/tex]

Substitute the values then we get

[tex]K_f-0=\int_{0}^{2}F(x)dx[/tex]

Because work done=[tex]Force\times displacement[/tex]

[tex]K_f=\int_{0}^{2}(2.4-x^2)dx[/tex]

[tex]K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}[/tex]

[tex]K_f=2.4(2)-\frac{8}{3}=2.13 J[/tex]

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy =[tex]K=2.4x-\frac{x^3}{3}[/tex]

When the kinetic energy is maximum then [tex]\frac{dK}{dx}=0[/tex]

[tex]\frac{d(2.4x-\frac{x^3}{3})}{dx}=0[/tex]

[tex]2.4-x^2=0[/tex]

[tex]x^2=2.4[/tex]

[tex]x=\pm\sqrt{2.4}[/tex]

[tex]\frac{d^2K}{dx^2}=-2x[/tex]

Substitute x=[tex]\sqrt{2.4}[/tex]

[tex]\frac{d^2K}{dx^2}=-2\sqrt{2.4}<0[/tex]

Substitute x=[tex]-\sqrt{2.4}[/tex]

[tex]\frac{d^2K}{dx^2}=2\sqrt{2.4}>0[/tex]

Hence, the kinetic energy is maximum at x=[tex]\sqrt{2.4}[/tex]

Again by work energy theorem , the  maximum kinetic energy of the block between x=0 and x=2.0 m is given by

[tex]K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx[/tex]

[tex]k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}[/tex]

[tex]K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J[/tex]

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

Answer:C

Explanation:

Given

Four masses are attached to the wire such that distance between two mass is L

therefore the Length of wire is 4 L

and the center of mass is at 2L

moment of inertia is distribution of mass from its rotational axis

thus moment of Inertia I is given by

[tex]I=m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2+m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2[/tex]

[tex]I=2\times m\times (\frac{L}{2})^2+2\times m\times (\frac{3L}{2})^2[/tex]

[tex]I=\frac{2mL^2}{4}+\frac{18mL^2}{4}[/tex]

[tex]I=5mL^2[/tex]

The rotational inertia of the four-point mass system about its center of mass, with equal spacing between adjacent masses, would be 2mℓ^2.

The rotational inertia of a system around any particular axis is given by sum of the product of the mass of each particle and square of perpendicluar distance from the axis of rotation. In the case of our four-point mass system, the masses are arranged such that they are all equidistant from the center. Therefore, the total inertia, Icm, is the sum of the inertias of each individual mass.

Assuming the center of rotation is halfway between the second and third mass, there will be two masses at distance ℓ/2 and two at 3ℓ/2. Therefore, Icm= 2* m*((ℓ/2)^2) + 2* m*((3ℓ/2)^2) = 2mℓ^2. Hence, the correct answer is E.) 2mℓ^2.

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Answer: 75ºC

Explanation:

Assuming that the Styrofoam is perfectly adiabatic, and neglecting the effect of the sugar on the system, the heat lost by the tea, can only be transferred to the spoon, reaching all the system to a final equilibrium temperature.

If the heat transfer process is due only to conduction, we can use this empirical relationship for both objects:

Qt = ct . mt . (tfn – ti)

Qs = cs . ms . (ti – tfn)

If the cup is perfectly adiabatic, it must be Qt = Qs

Using the information provided, and solving for tfinal, we get:

tfinal = (83,600 + 1,800) / (90 + 1045) ºC  

tfinal = 75º C

Answer:

[tex]I_{edge} = 6.67 kg.m^2[/tex]

Explanation:

given,

mass of rod = 5 Kg

Length of rod = 2 m

R = 1 m

moment of inertial from one edge of the rod = ?

moment of inertia of rod through center of mass

[tex]I_{CM}= \dfrac{1}{12}M(L)^2[/tex]

using parallel axis theorem

[tex]I_{edge} = I_{CM} + MR^2[/tex]

[tex]I_{edge} =\dfrac{1}{12}ML^2+ M(\dfrac{L}{2})^2[/tex]

[tex]I_{edge} =\dfrac{1}{12}ML^2+ \dfrac{ML^2}{4}[/tex]

[tex]I_{edge} =\dfrac{1}{3}ML^2[/tex]

now, inserting all the given values

[tex]I_{edge} =\dfrac{1}{3}\times 5 \times 2^2[/tex]

[tex]I_{edge} = 6.67 kg.m^2[/tex]

The velocity of Ned as measured by Pam is the interpretation of v.

Answer: Option D

Explanation:

According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:

The Ned reference framework : (x, t)  

The Pam reference framework :  [tex]\left(x^{\prime}, t^{\prime}\right)[/tex]

From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.  

At that point, [tex]v^{\prime}[/tex] is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.

In the context of relativity, the velocity denoted by v' represents the velocity of Ned as measured by Pam. Primed quantities typically refer to measurements in the moving reference frame, making option (d) correct.

The question is asking about the interpretation of the notation v', which is used in a relativistic physics context. The correct interpretation of v' in the context of relativity would be d. The velocity of Ned as measured by Pam. This is because primed quantities (e.g. x', t', v') typically refer to measurements made in the moving reference frame relative to the unprimed frame. Since Pam's frame is the one moving with velocity v in Ned's frame, v' would be the velocity Ned appears to have when observed from Pam's frame. Accordingly, option (d) is the correct choice.

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The x-coordinate of the center of mass of the lidless shoebox's base, which is a rectangle, will be at half its length. Therefore, the x-coordinate is 24 cm.

The center of mass is the coordinate that averages the distribution of mass in a physical object. To calculate the center of mass, one must consider the dimensions of the object. For a rectangular object like the shoebox's base with length, L = 48 cm, and width, W = 25 cm, the x-coordinate of the center of mass will be half the length. Therefore, the x-coordinate of the center of mass of the shoebox's base is L/2 = 48 cm / 2 = 24 cm. This is because in this rectangular coordinate system, the center of mass is located at the middle along each dimension (L, W).

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The x-coordinate of the center of mass of the shoebox's base is half its length which is 24 cm.

The center of mass of an object is basically the average position of all its parts. Given that the shoebox's base lies on the xy-plane, the x-coordinate for its center of mass will simply be half the length of the base. Hence, using the length L = 48 cm, the x-coordinate for the center of mass is: L/2 = 48/2 = 24 cm. Therefore, the center of mass of the shoebox's base along the x-axis lies at 24 cm.

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Answer:

T_t = Ts  (1-A[tex])^{1/4}[/tex] √ (Rs/D)

Explanation:

The black body radiation power is given by Stefan's law

       P = σ A e T⁴

This power is distributed over a spherical surface, so the intensity of the radiation is

     I = P / A

Let's apply these formulas to our case. Let's start by calculating the power emitted by the Sun, which has an emissivity of one (e = 1) black body

    P_s = σ A_s 1 T_s⁴

This power is distributed in a given area, the intensity that reaches the earth is

     I = P_s / A

    A = 4π R²

The distance from the Sun Earth is R = D

     I₁ = Ps / 4π D²

     I₁ = σ (π R_s²) T_s⁴ / 4π D²

     I₁ = σ T_s⁴ R_s² / 4D²

Now let's calculate the power emitted by the earth

     P_t = σ A_t (e) T_t⁴

     I₂ = P_t / A_t

     I₂ = P_t / 4π R_t²2

     I₂ = σ (π R_t²) T_t⁴ / 4π R_t²2

     I₂ = σ T_t⁴ / 4

The thermal equilibrium occurs when the emission of the earth is equal to the absorbed energy, the radiation affects less the reflected one is equal to the emitted radiation

     I₁ - A I₁ = I₂

     I₁ (1 - A) = I₂

Let's replace

    σ T_s⁴  R_s²/4D²   (1-A)  = σ T_t⁴ / 4

    T_s⁴ R_s² /D²   (1-A) = T_t⁴

    T_t⁴ = T_s⁴  (1-A)  (Rs / D) 2

    T_t = Ts  (1-A[tex])^{1/4}[/tex] √ (Rs/D)

Answer:

a) Fh = 25.23 N

b) The direction of the force exerted by the hand is pointed to the downward (negative) direction.

c) Fs = 37.97 N

d) The direction of the force exerted by the shoulder is pointed to the upward (positive) direction.

Explanation:

Given data

Distance from the small end of the bat to the shoulder: d = 23.5 cm

Distance from the small end of the bat to the center of mass: r = 70.0 cm

Mass of the bat: m = 1.30 Kg

This situation can be seen in the pic.

a) We can apply

∑τA = 0   ⇒  + (23.5cm)*Fh - (70 - 23.5)cm*m*g = 0

⇒   + (23.5cm)*Fh - (70 - 23.5)cm*(1.30 Kg)*(9.8 m/s²) = 0

⇒   Fh = 25.23 N (↓)

b) The direction of the force exerted by the hand is pointed to the downward (negative) direction.

c) We apply

∑Fy = 0  (↑)

⇒      - Fh + Fs - m*g = 0

⇒     Fs = Fh + m*g

⇒     Fs = 25.23 N + (1.30 Kg)*(9.8 m/s²)

⇒     Fs = 37.97 N (↑)

d) The direction of the force exerted by the shoulder is pointed to the upward (positive) direction.

I examine the same second hand on the clock. Again, there are two points called A and B on the clock, with A farther from the center than B. "The angular acceleration for both points is 0" is true.

Answer: Option C

Explanation:

The clock is in rotator motion. All the three hands of clock, move in same direction, but different speeds. And hence, we count hours, minutes and seconds.  And, when we take each hand, they move related to the centre of the clock, where all the three are attached.

So, there is a centripetal acceleration which depends upon the velocity.  But, the motion is uniform everywhere in the circle. The hands have no tangential acceleration. And hence, there is no angular acceleration, which is derived from tangential one. So, at any point, the angular acceleration is zero.

To solve this problem it is necessary to apply the expressions related to the calculation of angular acceleration in cylinders as well as the calculation of linear acceleration in these bodies.

By definition we know that the angular acceleration in a cylinder is given by

[tex]\alpha = \frac{2\mu_k g cos\theta}{r}[/tex]

Where,

[tex]\mu_k[/tex] = Coefficient of kinetic friction

g = Gravitational acceleration

r= Radius

[tex]\theta[/tex]= Angle of inclination

While the tangential or linear acceleration is given by,

[tex]a = g(sin\theta-\mu_k cos\theta)[/tex]

ANGULAR ACCELERATION, replacing the values that we have

[tex]\alpha = \frac{2\mu_k g cos\theta}{r}[/tex]

[tex]\alpha = \frac{2(0.4)(9.8) cos(30)}{10*10^{-2}}[/tex]

[tex]\alpha = 67.9rad/s[/tex]

LINEAR ACCELERATION, replacing the values that we have,

[tex]a = (9.8)(sin30-(0.4)cos(30))[/tex]

[tex]a = 1.5m/s^2[/tex]

Therefore the linear acceleration of the solid cylinder is [tex]1.5 m/s^2[/tex]

To solve this problem it is necessary to use three key concepts. The first of these is the Centripetal Force which is in the upward direction and would be described as

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

Where,

m= mass

v= Velocity

r = Radius

The second is the voltage generated which is given by 1400N. Finally the third is the force generated by the weight and that would be described by Newton's second law as

F =mg

Where,

m = Mass

g = Acceleration gravity

F = 80*9.8

F = 784N

For balance to exist, the sum of Force must be equal to the Centripetal Force, therefore

[tex]\sum F = F_c[/tex]

[tex]T - mg = \frac{mv^2}{r}[/tex]

Replacing we have

[tex]1400 - 784 = \frac{(80kg)v^2}{5,5}[/tex]

[tex]v^2 = \frac{ 616*5.5}{80}[/tex]

[tex]v = \sqrt{42.35}[/tex]

[tex]v=6.5m/s[/tex]

Therefore the maximum speed he can tolerate at the lowest point of his swing is 6.5m/s

One of the scientific characteristics that can make the difference between the Earth and the moon is the so-called Rayleigh dispersion effect. This concept is identified as the dispersion of visible light or any other electromagnetic radiation by particles whose size is much smaller than the wavelength of the dispersed photons. Our atmosphere allows even our 'shadows' to be clear.

On the moon under the absence of the atmosphere or any other mechanism that allows absorbing or failing to re-irradiate sunlight towards the area in its shadows, which makes the shadows on the moon look darker.

Answer:

a) F = 2.7 10⁻¹⁴ N , b)  a = 2.97 10¹⁶ m / s²  c) θ = 14º

Explanation:

The magnetic force on the electron is given by the expression

     F = q v x B

Which can be written in the form of magnitude and the angle found by the rule of the right hand

     F = q v B sin θ

where θ is the angle between the velocity and the magnetic field

a) the maximum magnitude of the force occurs when the velocity and the field are perpendicular, therefore, without 90 = 1

     F = e v B

     F = 1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²

     F = 2.73 10⁻¹⁴ N

     F = 2.7 10⁻¹⁴ N

b) Let's use Newton's second law

    F = m a

    a = F / m

    a = 2.7 10⁻¹⁴ / 9.1 10⁻³¹

    a = 2.97 10¹⁶ m / s²

The actual acceleration (a1) is a quarter of this maximum

    a1 = ¼ a

    a1 = 7.4 10¹⁵ m / s²

With this acceleration I calculate the force that is executed on the electron

     F = ma

    e v b sin θ= ma

    sin θ = ma / (e v B)

    sin θ = 9.1 10⁻³¹ 7.4 10¹⁵ / (1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²)

    sin θ = 6.734 10⁻¹⁵ / 27.26 10⁻¹⁵

    sin θ = 0.2470

    θ = 14.3º