It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide "normal" gravity?

Answer:

Explanation:

Attached is the evaluation

The given integral represents the volume of a solid bounded by specific surfaces and coordinates. To sketch this solid, imagine a cylindrical shape centered around the z-axis with varying radius and height. To evaluate the integral, integrate with respect to z, then θ, and finally r.

First, let's consider the given integral: ∫_0^6 ∫_0^(2π) ∫_4^24 r dz dθ dr. This triple integral represents the volume of a solid. The limits of integration indicate that the solid is bounded by the surfaces r = 4 and r = 24, the angle θ ranges from 0 to 2π, and the z-coordinate extends from 0 to 6.

To sketch this solid, imagine a cylindrical shape centered around the z-axis, with a varying radius (r) extending from 4 to 24, and a height (z) ranging from 0 to 6. The angle θ represents rotation around the z-axis.

To evaluate the integral, we can first integrate with respect to z, then θ, and finally r. This involves applying the limits of integration and performing the calculations step by step. The final result will give us the volume of the solid.

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Answer:

0.83 amps is what I believe is the answer.

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

        F = 1.6 10⁻¹⁹  1.344 10⁸ 0.4

       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

Answer:

R = 1.81 10² km

Explanation:

Let's start by looking for the power in the visible range emitted this is 10W, the energy of that power is one second is

         P = E₁ / t

         E₁ = P t

         E₁ = 10 J

 Let's find the energy of a photon with Planck's equation

          E = h f

          c = λ f

we substitute

         E = h c /λ

         E = 6.63 10⁻³⁴ 3 10⁸/580 10⁻⁹

         E = 3.42 10⁻¹⁹ J

we can use a direct proportions rule to find the number of photons in the energy E₁

          #_photon = E₁ / E

          #_photon = 10 / 3.42 10⁻¹⁹

          #_photon = 2.92 10¹⁹ photons

This number of photons is distributed on the surface of a sphere. Let's find what the distance is so that there are 500 photons in 3 mm = 0.003 m.

the area of ​​the sphere is

            A = 4π R²

area of ​​the circle is

           A´ = π r²

as the intensity is constant over the entire sphere

         P = #_photon / A = 500 / A´

       # _photon / 4π R² = 500 / π r²

       R² = #_photon r² / 4 500

       r = d / 2 = 0.003 / 2 = 0.0015 m

       R² = 2.92 10¹⁹ 0.0015 2/2000

       R = √ (3,285 10¹⁰)  

        R = 1.81 10⁵ m

       R = 1.81 10² km

Answer:

3241.35J

Explanation:

No. Of rods = 5

Mass = 2.89kg

Length (L) = 0.827m

W = 507rpm

Kinetic energy of rotation = ½I*ω²

For each rod, the moment of inertia (I) = ML² / 3

I = ML² / 3

I = [2.89*(0.827)²] / 3

I = 1.367 / 3 = 0.46kgm²

ω = 507 rev/min. Convert rev/min to rev/sec.

507 * 2Πrads/60s = 53.09rad/s

ω = 53.09rad/s

k.e = ½ I * ω²

K.E = ½ * 0.46 * (53.09)²

K.E = 648.27.

But there five (5) rods, so kinetic energy is equal to

K.E = 5 * 648.27 = 3241.35J

The rotational kinetic energy of a propeller with five blades, each modeled as a uniform rod, can be determined using formulas for moment of inertia and rotational kinetic energy.

The kinetic energy of rotating objects depends on their moment of inertia and angular velocity. Given that each propeller blade is modeled as a uniform rod rotating about its end, we can calculate the moment of inertia (I) of all five blades using the formula I=5*(1/3*m*L^2), where m is the mass and L is the length of each rod. After we find the moment of inertia, we can determine the angular velocity (ω) in radian per second, given that the propeller rotates at 507 rpm, by using the conversion factor ω=(507*2*pi)/60. Finally, we calculate the rotational kinetic energy (K) using the formula K=1/2*I*ω^2.

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Answer:

T1 = 112.07[lb]

T2 = 487.3 [lb]

Explanation:

To solve this problem we must perform a static balance analysis, for this we perform a free body diagram. In this free body diagram we use the angles mentioned in the description of the problem.

Performing a sum of forces on the X-axis equal to zero, we can find an equation that relates the tension of the T1 & T2 cables.

Then we perform a summation of forces on the Y-axis, in which we can find another equation. In this new equation, we replace the previous one and we can find the tension T2.

T1 = 112.07[lb]

T2 = 487.3 [lb]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The net force is  [tex]F_{net}= 6.44 *10^{-4} N[/tex]

Explanation:

Generally the net force is a force that come up due to the unequal centripetal force(A difference in centripetal force ) and it is mathematically represented  as

                     [tex]F_{net} = \Delta F_{cen}[/tex]

and the difference in centripetal force  [tex]\Delta F_{cen}[/tex] is mathematically represented as

               [tex]\Delta F_{cen} = \Delta m* rw^2[/tex]

Which the difference in mass multiplied by the centripetal acceleration

  Substituting 10 mg = [tex]10 *10^{-3}g[/tex] for [tex]\Delta m[/tex] , 12 cm = [tex]\frac{12}{100} = 0.12m[/tex] for radius

  and  70,000 rpm = [tex]70,000 *[\frac{2 \pi rad}{1 rev}][\frac{1 min}{60s} ] = 7326.7 rad/s[/tex]

              [tex]F_{net} = \Delta F_{cen} = 10*10^{-3} * 0.12 * 7326.7[/tex]

                     [tex]F_{net}= 6.44 *10^{-4} N[/tex]

Answer:

The magnitude of their velocities will be the same but their direction will be reversed.

Final answer:

In a completely elastic collision between two particles of identical mass and equal but opposite velocities, both particles simply swap their velocities as a result of the collision, conserving both momentum and kinetic energy.

Explanation:

In a completely elastic head-on collision between two particles with identical mass and equal but opposite speeds, the outcome is quite straightforward. Since the collision is elastic, both conservation of momentum and conservation of kinetic energy apply.

Before the collision, let's assume particle 1 with mass m has velocity v, and particle 2 also with mass m has velocity -v. The total momentum before collision would be m*v + m*(-v) = 0, and the total kinetic energy would be (1/2)*m*v² + (1/2)*m*(-v)².

For identical masses and equal and opposite velocities, the particles simply swap velocities after collision. Therefore, the first particle will have a velocity of -v and the second a velocity of v post-collision, as momentum and kinetic energy are conserved. They essentially 'bounce' off each other and move in the opposite directions with the same speed they had before the collision.

The momentum of the box at t = 1.90 s  is  - 2.586 i + 2.85j

And, the x and y component is - 2.586 and 2.85

Since we know that

impulse = force * time

Here impulse means the change in momentum

Now we can write as  in different way like

change in momentum = force * time

Also, Force = F = .285 t -.46t²

Due to variable force

change in momentum = ∫ F dt

So,

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

Sine t = 1.9

So,

change in momentum = .285 * 1.9² /2 i  -  .46 * 1.9³ / 3 j

= .514i - 1.05 j

And,

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

So, finally

x component = - 2.586

y component = 2.85

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the momentum of the box at t = 1.90 s is approximately (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^.

To find the momentum of the box at t = 1.90 s, we can integrate the net force over time to get the change in momentum.

The net force is given as:

F⃗ = (0.285 N/s)ti^ + (-0.460 N/s^2)t^2j^

To find the change in momentum, we integrate this force with respect to time from t = 0 to t = 1.90 s:

Δp⃗ = ∫(0 to 1.90) F⃗ dt

Δp⃗ = ∫(0 to 1.90) (0.285ti^ - 0.460t^2j^) dt

Now, we integrate each component separately:

Δpx = ∫(0 to 1.90) (0.285t) dt

Δpx = 0.285 * (t^2/2)|[0, 1.90]

Δpx = 0.285 * (1.90^2/2 - 0)

Δpx = 0.2565 N·s

Δpy = ∫(0 to 1.90) (-0.460t^2) dt

Δpy = -0.460 * (t^3/3)|[0, 1.90]

Δpy = -0.460 * (1.90^3/3 - 0)

Δpy = -2.4993 N·s

Now, we need to add the change in momentum to the initial momentum to find the momentum at t = 1.90 s:

p⃗(t = 1.90 s) = p⃗(t = 0 s) + Δp⃗

p⃗(t = 1.90 s) = (-3.10 kg·m/s)i^ + (3.90 kg·m/s)j^ + (0.2565 N·s)i^ + (-2.4993 N·s)j^

Now, add the x and y components separately:

p⃗(t = 1.90 s) = (-3.10 + 0.2565)i^ + (3.90 - 2.4993)j^

p⃗(t = 1.90 s) = (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^

So, the momentum of the box at t = 1.90 s is approximately (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^.

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Answer:

Maximum speed for a point on the string at anti node will be 22.6 m/sec

Explanation:

We have given length of string L = 3.5 m

For 7th harmonic length of the string [tex]L=\frac{7\lambda }{2}[/tex]

So [tex]\lambda =\frac{2L}{7}[/tex]

Speed of the wave in the string is 150 m/sec

Frequency corresponding to this wavelength [tex]f=\frac{v}{\lambda }=\frac{7v}{2L}[/tex]

So angular frequency will be equal to [tex]\omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec[/tex]

Maximum speed is equal to [tex]v_m=A\omega =0.024\times 942=22.60m/sec[/tex]

So maximum speed for a point on the string at anti node will be 22.6 m/sec

a)

i) Negative

ii) 160 J

b) 1280 N/m

c) [tex]32 m/s^2[/tex]

d) 1.27 Hz

e)

i) See attached plot

ii) See attached plot

Explanation:

a)

i) The work done by a force is given by

[tex]W=Fx cos \theta[/tex]

where

F is the force

x is the displacement of the object

[tex]\theta[/tex] is the angle between the direction of the force and the direction

Here we have:

- The force that the spring exerts on the box is to the left (because the box is moving to the right, trying to compress the spring)

- The displacement of the box is to the right

So, F and x have opposite direction, and so [tex]\theta=180^{\circ}[/tex] and [tex]cos \theta=-1[/tex], which means that the work done is negative.

ii)

According to the work-energy theorem, the work done by the spring is equal to the change in kinetic energy of the box:

[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

where

[tex]K_i[/tex] is the initial kinetic energy of the box

[tex]K_f[/tex] is the final kinetic energy

m = 20 kg is the mass of the box

u = 4.0 m/s is its initial speed

v = 0 m/s is the final speed (the box comes to rest)

Therefore,

[tex]W=\frac{1}{2}(20)(0)^2-\frac{1}{2}(20)(4.0)^2=-160 J[/tex]

So, the magnitude is 160 J.

b)

The elastic energy stored in a spring when it is compressed is given by

[tex]U=\frac{1}{2}kx^2[/tex]

where

k is the spring constant

x is the stretching/compression of the spring

Due to the law of conservation of energy, the kinetic energy lost by the box is equal to the elastic energy gained by the spring, so:

[tex]|W|=U=\frac{1}{2}kx^2[/tex]

We have

[tex]|W|=160 J[/tex]

x = 0.50 m is the maximum compression of the spring

Solving for k:

[tex]k=\frac{2U}{x^2}=\frac{2(160)}{(0.50)^2}=1280 N/m[/tex]

c)

The magnitude of the force exerted on the box is given by

[tex]F=kx[/tex]

where

k = 1280 N/m is the spring constant

x = 0.50 m is the compression of the spring

Substituting,

[tex]F=(1280)(0.50)=640 N[/tex]

Now we can find the maxmum acceleration using Newton's second law:

[tex]a=\frac{F}{m}[/tex]

where

F = 640 N is the maximum force

m = 20 kg is the mass of the box

So,

[tex]a=\frac{640}{20}=32 m/s^2[/tex]

d)

The frequency of oscillation of a spring-mass system is given by

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where

k is the spring constant

m is the mass

Here we have:

k = 1280 N/m is the spring constant of this spring

m = 20 kg is the mass of the box

So, the frequency of this system is:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{1280}{20}}=1.27 Hz[/tex]

e)

i)

Here we want to sketch the kinetic energy of the box as a function of the position, x: find this graph in attachment.

In a spring-mass oscillating system, the kinetic energy is zero when the system is at the extreme position, i.e. when the spring is maximum compressed/stretched. In this problem, this happens when x = - 0.50 m and x = +0.50 m (we called x = 0 the position of equilibrium of the spring). In these positions in fact, the mass has zero speed, so its kinetic energy is zero.

On the other hand, the box has maximum speed when x = 0 (because it's the moment where all the elastic energy is converted into kinetic energy, which is therefore maximum, and so the speed is also maximum).

ii)

Here we want to plot the acceleration of the box as a function of the position x: find the graph in attachment.

In a spring-mass system, the acceleration is proportional to the negative of the displacement, since the restoring force

[tex]F=-kx[/tex]

By rewriting the force using Newton's second Law, we have

[tex]ma=-kx \\a=-\frac{k}{m}x[/tex]

Which means that acceleration is proportional to the displacement, but with opposite sign: so, this graph is a straight line with negative slope.

Answer:

They are equal

Explanation:

Newton 3rd Law of motion states that for every force applied or action there is usually an equal and opposite force.

Impulse = Force * time

It is measured in Newton seconds.

The force and time of collision is the same which translates to an equal impulse by both scenarios.

The impulse imparted to the smaller mass by the larger mass is equal in magnitude to the impulse imparted to the larger mass by the smaller one, due to the conservation of momentum and Newton's third law. Both experience equal and opposite momentum transfers, ensuring the total momentum of the system remains constant.

In the context of collisions, impulse is defined as the change in momentum of an object when it is subjected to a force over a period of time. According to Newton's third law, 'For every action, there is an equal and opposite reaction,' meaning that the impulse imparted to the smaller mass by the larger mass is exactly equal in magnitude to the impulse imparted to the larger mass by the smaller one, although the direction of the impulses will be opposite. When considering the conservation of momentum, the total momentum before the collision must equal the total momentum after the collision if no external forces are acting on the system (assuming a closed system). Therefore, if two cars collide, such as described in the provided text, regardless of their masses, the momentum transfer will be the same for both, thus the total momentum of the system remains constant.

Explanation:

Given that,

Mass of the hockey puck, m₁ = 0.45 kg

Initial peed of the hockey puck, u₁ = 5.25 m/s (east)

Mass of other puck, m₁ =  0.85 kg

Initial speed of other puck, u₂ = 0 (at rest)

Let v₁ and v₂ are the final speeds of both pucks after the collision respectively. Using the conservation of momentum as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1v_1+m_2v_2=0.45\times 5.25+0.85\times 0\\\\m_1v_1+m_2v_2=2.36\\\\0.45v_1+0.85v_2=2.36.........(1)[/tex]

The coefficient of restitution for elastic collision is equal to 1.          

[tex]C=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{5.25-0}\\\\v_2-v_1=5.25.......(2)[/tex]      

On solving equation (1) and (2) we get :

[tex]v_1=-1.611\ m/s\\\\v_2=3.63 m/s[/tex]

Hence, this is the required solution.

Final answer:

Using the conservation of momentum for an inelastic collision, Ender's mass is calculated to be approximately 51.43 kg when the given velocities and Shen's mass are considered.

Explanation:

The subject of this question is physics, specifically dealing with the concept of conservation of momentum during collisions. We are asked to find Ender's mass given the velocities and mass involved in a collision in space. Since the collision here is inelastic (Ender and Shen hold on to each other), the total momentum before the collision must be equal to the total momentum after the collision.

According to the conservation of momentum, m1 × v1 + m2 × v2 = (m1 + m2) × v3, where m1 and m2 are the masses of Ender and Shen respectively, and v1, v2, and v3 are their velocities.

We know Shen's mass (m2) is 45 kg, Ender's velocity (v1) is 12 m/s, Shen's velocity (v2) is 9 m/s, and the combined velocity after the collision (v3) is 6.4 m/s. Using the momentum conservation formula, we can solve for Ender's mass (m1) as follows:

m1 × 12 m/s + 45 kg × 9.0 m/s = (m1 + 45 kg) × 6.4 m/s

Expanding this and rearranging the terms, we get:

12m1 = 6.4m1 + 6.4 × 45

12m1 - 6.4m1 = 6.4 × 45

m1(12 - 6.4) = 6.4 × 45

m1 = (6.4 × 45) / (12 - 6.4)

m1 = 288 / 5.6

m1 = 51.43 kg

Therefore, Ender's mass is approximately 51.43 kg.

Answer:

The angular velocity is 4.939rad/sec

Explanation:

θ = 61°

N = 118 rev/min

The formula for angular velocity is given as;

w = θ/t

where;

w = angular velocity

θ = angle

t = time in rad/sec

1 rev/ min = 0.1047 radian/second.

Therefore 118 rev/min = 118*0.1047rad/sec

                                     =12.35rad/sec

Substituting into the formula, we have

w = 61/12.35

w = 4.939rad/sec

Answer:

a) Seems true at first glance, but on further inspection, the statement is false.

b) True

c) False

d) True

e) True

f) False

g) False

Explanation:

Taking the statements one by one

a.) The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body.

The Clausius Statement denies the possibility of heat transfer from a cooler to a hotter body without extra work. It does not outrightly state that there is no possibility of heat transfer from a cooler to a hotter body.

For example, an Air conditioner or refrigerator rejects heat from a cold reservoir to a hot reservoir.

So, this statement is false.

b) The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs.

The Coefficient of Performance of a reversible cycle is the maximum efficiency possible. It is the efficiency of a Carnot Engine.

Hence, it is greater than or equal to the Coefficient of Performance of an irreversible cycle.

COP(reversible) ≥ COP(irreversible)

This statement is true.

c. Mass, energy, and temperature are the examples of intensive properties.

Intensive properties are properties of thermodynamic systems that do not depend on the extent of the system. They are the same for a particular size of substance and stay the same if the size of the substance is doubled or halved. Examples include temperature, specific capacity, specific volume, every specific property basically, etc.

Extensive properties depend on the extent of the system. They double or half when the size of the extent doubles or halves respectively.

Mass and Energy are Extensive properties.

Temperature is the only intensive property among these options.

This statement is false.

d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is

COPHP = COPR + 1.

The coefficients of performance for reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs are indeed related through

COP(HP) = COP(R) + 1

COP of a heat pump = COP(HP) = (Qh/W)

COP of a refrigerator = COP(R) = (Qc/W)

But, Qh = Qc + W

Divide through by W

(Qh/W) = (Qc/W) + (W/W)

COP(HP) = COP(R) + 1 (Proved!)

This Statement is true.

e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11.

The COP is given as (1/efficiency).

Efficiency = 1 - (Tc/Th)

Tc = temperature of cold thermal reservoir in Kelvin = 350°C = 623.15 K

Th = temperature of hot thermal reservoir in Kelvin = 550°C = 823.15 K

Efficiency = 1 - (623.15/823.15)

= 1 - 0.757 = 0.243

COP = (1/Efficiency) = (1/0.243) = 4.11

This statement is true.

f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%.

In the absence of friction and other irreversibilities and for a heat engine to have 100% efficiency, the temperature of its cold reservoir has to be 0 K or the tempersture of its hot reservoir has to be infinity.

Efficiency = 1 - (Tc/Th)

For efficiency to be 1,

(Tc/Th) = 0; that is, Tc = 0 or Th = infinity

These two aren't physically possible and for 100% efficiency to happen, the heat engine will have to violate the Kelvin-Planck's statement of the second law of thermodynamics.

According to Kelvin-Planck's statement of the second law of thermodynamics, net amount of work cannot be produced by exchanging heat with single reservoir i.e. there will be another reservoir to reject heat.

Hence, a heat engine cannot have an efficiency of 100%.

This statement is false.

g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.

The COP of a refrigerator is given as

COP = (Qcold)/W

Qcold = Heat rejected from the cold reservoir = Heat rejected from refrigerated space = 60 KJ/min

W = work done on the system = electrical power consumed = 50 KJ/min

COP = (60/50) = 1.2

This system does not violate the first law of thermodynamics.

This statement is false.

Hope this Helps!!!

Answer:

The tension T₁ in string 1 at the moment that the monkey is halfway between the ends of the bar is 34.68 N

Explanation:

Given;

mass of the uniform horizontal bar, m₁ = 2.5 kg

length of the bar, L = 3.0 m

mass of the monkey, m₂ = 1.25 kg

distance from the right end of the second string, d = 0.74 m

For a body to remain in rotational equilibrium, the net external torque acting on it due to applied external forces must be equal.

∑τ = 0

For vertical equilibrium of bar-string system, in which T₁ and T₂ are the tension on both ends of the string;

T₁ + T₂ = (m₁ + m₂)g

T₁ + T₂ = (2.5 + 1.25) 9.8

T₁ + T₂ = 36.75 N

For rotational equilibrium when the monkey is halfway between the ends of the bar, take moment about the left end of the string.

(m₁ + m₂) L /2 = T₂(L - d)

(m₁ + m₂)0.5L = T₂( L - d)

(2.5 + 1.25)0.5 x 3 = T₂ ( 3 - 0.74)

4.6875 = T₂ (2.26)

T₂ = (4.6875) / (2.26)

T₂ = 2.074 N

Thus, T₁ = 36.75 N - T₂

T₁ = 36.75 N  - 2.074 N

T₁ = 34.68 N

The linear charge density on the wire is approximately 0.002 N/Cm.

The linear charge density on the wire can be calculated using the formula:

λ = E × r / 2πk

where λ is the linear charge density, E is the electric field, r is the distance from the wire, and k is the Coulomb's constant. Plugging in the known values of E = 20.0 N/C and r = 2.00 cm = 0.02 m, we can solve for λ:

λ = (20.0 N/C) × (0.02 m) / (2π × 8.99 × 10^9 Nm^2/C^2) ≈ 0.002 N/Cm

Therefore, the linear charge density on the wire is approximately 0.002 N/Cm.

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The work done by the force of gravity on the projectile is given by W = mgh. The change in kinetic energy of the projectile is given by ΔKE = (1/2)mv0^2. The final kinetic energy of the projectile as it hits the water is given by KEf = (1/2)mv0^2.

(a) Work done by the force of gravity:

The work done by the force of gravity on the projectile during its flight can be calculated using the formula:

W = mgh

where W is the work done, m is the mass of the sphere, g is the acceleration due to gravity, and h is the height above the water.

(b) Change in kinetic energy:

The change in kinetic energy of the projectile from the time it was fired until it hits the water can be calculated using the formula:

ΔKE = (1/2)mv0^2

where ΔKE is the change in kinetic energy, m is the mass of the sphere, and v0 is the launch speed.

(c) Final kinetic energy:

The final kinetic energy of the projectile as it hits the water can be calculated using the same formula as in part (b):

KEf = (1/2)mv0^2

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Answer:

49 N

Explanation:

The diagram of the bar is obtained online and attached to this solution.

The free body diagram is also attached.

Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by

τ = mgx

where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m

τ = (7)(9.8)(0.3686) = 25.29 Nm

The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)

= 25.29 ÷ 0.5162 = 48.99 N = 49 N

Hope this Helps!!!

Answer:

The force supported by the pin at A is 69.081 N

Explanation:

The diagram is in the figure attach. The angular acceleration using the moment expression is:

[tex]-mg(\frac{Lcos\theta }{2} )=I\alpha \\\alpha =\frac{-3g}{2L} cos\theta[/tex]

Where

L = length of the bar = 900 mm = 0.9 m

[tex]\alpha =\frac{-3*9.8cos35}{2*0.9} =-13.38rad/s^{2}[/tex]

The acceleration in point G is equal to:

[tex]a_{G} =a_{A} +\alpha kr_{G/A} -w^{2} r_{G/A}[/tex]

Where

aA = acceleration at A = 0

w = angular velocity of the bar = 3 rad/s

rG/A = position vector of G respect to A = [tex]\frac{L}{2} cos\theta i-\frac{L}{2} cos\theta j[/tex]

[tex]a_{G} =(\frac{L}{2}\alpha sin\theta -\frac{w^{2}Lcos\theta }{2} )i+(\frac{L}{2}\alpha cos\theta +\frac{w^{2}Lsin\theta }{2} )j=(\frac{0.9*(-13.38)*sin35}{2} -(\frac{3^{2}*0.9*cos35 }{2} )i+(\frac{0.9*(-13.38)*cos35}{2} +\frac{3^{2} *0.9*sin35)}{2} )j=-6.76i-2.61jm/s^{2}[/tex]

The force at A in x is equal to:

[tex]A_{x} =ma_{G} =7*(-6.76)=-47.32N[/tex]

The force at A in y is:

[tex]A_{y} =ma_{G} +mg=(7*(-2.61))+(7*9.8)=50.33N[/tex]

The magnitude of force A is equal to:

[tex]A=\sqrt{A_{x}^{2}+A_{y}^{2} } =\sqrt{(-47.32^{2})+50.33^{2} } =69.081N[/tex]

The speed of waves on the string is 32 m/s.

The speed of waves on a string can be calculated using the formula:

velocity = frequency × wavelength

In this case, the frequency is given as 80.0 Hz and the distance between adjacent antinodes is 20.0 cm. Since the wavelength is twice the distance between adjacent antinodes, we have:

wavelength = 2 × 20.0 cm = 40.0 cm = 0.4 m

Plugging these values into the formula, we get:

velocity = 80.0 Hz × 0.4 m = 32 m/s

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Answer:

1117.51N/C

Explanation:

The magnitude of the electric force is given by:

[tex]|\vec{F}|=|k\frac{q_1q_2}{r^2}|[/tex]

k: Coulomb's constant = 8.98*10^9Nm^/2C^2

r: distance between the charges = 0.15m

By replacing the values of q1, q2, k and r you obtain:

[tex]|\vec{F}|=|(8.98*10^9Nm^2/C^2)\frac{(-4.0*10^{-5}C)(7.0*10^{-5}C)}{(0.15m)^2}|=1117.51\frac{N}{C}[/tex]

hence, the force between the charges is 1117.51N/C

Answer:

The force of attraction between the charges is 1120 N

Explanation:

Given:

positive charge of the particle, q1 = 7 x 10^-5 C

negative charge of the particle, q2 = -4 x 10^-5 C

distance between the charges, r = 0.15 m

The force of attraction between the charges will be calculated using Coulomb's law:

F = (k|q1q2|) / r^2

Where:

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 Nm^2/c^2

|q1| is magnitude of charge 1

|q2| is magnitude of charge 2

F = (9 x 10^9 x 7 x 10^-5 x 4 x 10^-5) / (0.15 x 0.15)

F = 1120 N

Thus, the force between the charges is 1120 N

Answer:

[tex]2.08\cdot 10^9 A[/tex]

Explanation:

The magnetic dipole moment of a circular coil with a current is given by

[tex]\mu = IA[/tex]

where

I is the current in the coil

[tex]A=\pi r^2[/tex] is the area enclosed by the coil, where

[tex]r[/tex] is the radius of the coil

So the magnetic dipole moment can be rewritten as

[tex]\mu = I\pi r^2[/tex] (1)

Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius

[tex]r=3500 km = 3.5\cdot 10^6 m[/tex]

Here we also know that the Earth's magnetic dipole moment is

[tex]\mu = 8.0\cdot 10^{22} J/T[/tex]

Therefore, we can re-arrange eq (1) to find the current that the charges produced:

[tex]I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A[/tex]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance [tex]R_e[/tex] defined by the formula:

[tex]\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}[/tex]

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

[tex]\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\[/tex]

and when this is combined with the third resistor in series, the equivalent resistance ([tex]R''_e[/tex]) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

[tex]R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}[/tex]

The problem states that the difference between the equivalent resistances in both circuits is given by:

[tex]R''_e=R_e+630 \,\Omega[/tex]

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

[tex]\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega[/tex]

The answers of the all the questions are given as 1-a,2-b,3-b,4-c,5-c,6-a,7-b,8-c,9-b,10-b or c,11-b,

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

A whole branch of physics, thermodynamics, deals with how heat is transferred between different systems and how work is done in the process.

Almost every transfer of energy that takes place in real-world physical systems does so with an efficiency of less than 100% and results in some thermal energy.

This energy is usually in the form of low-level thermal energy. Here, low-level means that the temperature associated with the thermal energy is close to that of the environment.

It is only possible to extract work when there is a temperature difference, so low-level thermal energy represents 'the end of the road' of energy transfer. No further useful work is possible; the energy is now 'lost to the environment.

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Answer:

Coefficient of static friction will be equal to 0.642  

Explanation:

We have given acceleration [tex]a=6.3m/sec^2[/tex]

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

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to [tex]a=\mu g[/tex], here [tex]\mu[/tex] is coefficient of static friction and g is acceleration due to gravity

So [tex]\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642[/tex]

So coefficient of static friction will be equal to 0.642

The mass of the ion is 5.96 X 10⁻²⁵ kg

Explanation:

The electrical energy given to the ion Vq will be changed into kinetic energy [tex]\frac{1}{2}mv^2[/tex]

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force [tex]\frac{mv^2}{r}[/tex].

So,

[tex]Vq = \frac{1}{2}mv^2[/tex]

and

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

Right from these eliminating v, we can derive

[tex]m = \frac{B^2r^2q}{2V}[/tex]

On substituting the value, we get:

[tex]m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\[/tex]

m = 5.96 X 10⁻²⁵ kg.

The mass of the ion is found using the principles of kinetic and magnetic forces, is approximately 9.65 x 10⁻²⁷ kg.

To find the mass of the ion, we use the principles of energy and magnetic force. When a singly charged positive ion is accelerated through a voltage (V), it gains kinetic energy equal to the electrical potential energy provided by the voltage:

Kinetic Energy (KE) = qV

Since the ion starts from rest, this kinetic energy is also given by:

Kinetic Energy (KE) = (1/2)mv²

Equating these two expressions for kinetic energy gives:

qV = (1/2)mv²

Solving for the velocity (v), we get:

v =√(2qV/m)

When the ion enters the magnetic field (B) and bends into a circular path, the centripetal force required is provided by the magnetic force:

Magnetic Force = qvB

The centripetal force is also given by:

Centripetal Force = mv²/r

Equating these two expressions for the forces, we get:

qvB = mv²/r

Solving for the mass (m) of the ion:

m = qBr/v

We already have v = √(2qV/m), substituting this into the equation above to eliminate v, we get:

m = (qBr) / √(2qV/m)

Square both sides to simplify and solve for m:

m² = (q²B²r²) / (2qV)

m = (qB²r²) / (2V)

Using the given values (q = 1.602 x 10⁻¹⁹ C as charge of a singly charged positive ion, V = 2000 V, B = 0.400 T, and r = 0.305 m), we get:

m = (1.602 x 10⁻¹⁹ C x (0.400 T)² x (0.305 m)²) / (2 x 2000 V)

Calculating this gives the mass of the ion approximately equal to:

m ≈ 9.65 x 10⁻²⁷ kg

Answer:

a) The work done on the astronaut by the force from the helicopter is [tex]W_{h}=16087.68\ J[/tex].

b) The work done on the astronaut by the gravitational force is  [tex]W_{g}=-15082.2\ J[/tex] .

Explanation:

We are told that the mass of the astronaut is [tex]m=81\ kg[/tex], the displacement is [tex]\Delta x=19\ m[/tex], the acceleration of the astronaut is  [tex]|\vec{a}|=\frac{g}{15}[/tex]  and the acceleration of gravity is [tex]g=9.8\ \frac{m}{s^{2}}[/tex] .

We suppose that in the vertical direction the force from the helicopter [tex]F_{h}[/tex] is upwards and the gravitational force [tex]F_{g}[/tex] is downwards. From the sum of forces we can get the value of [tex]F_{h}[/tex]:

                                               [tex]F_{h}-F_{g}=m.a[/tex]

                                              [tex]F_{h}-mg=m.\frac{g}{15}[/tex]

                                              [tex]F_{h}=mg(1+\frac{1}{15})[/tex]

                              [tex]F_{h}=(\frac{16}{15}).81\ kg.\ 9.8\ \frac{m}{s^{2}}\ \Longrightarrow\ F_{h}=846.72\ N[/tex]

We define work as the product of the force, the displacement of the body and the cosine of the angle [tex]\theta[/tex] between the direction of the force and the displacement of the body:

                                                 [tex]W=F.\Delta x.\ cos(\theta)[/tex]

a) The work done on the astronaut by the force from the helicopter

                                                  [tex]W_{h}=F_{h}.\Delta x[/tex]

                                             [tex]W_{h}=846.72\ N.\ 19\ m[/tex]

                                                  [tex]W_{h}=16087.68\ J[/tex]

b) The work done on the astronaut by the gravitational force

                                                   [tex]W_{g}=-F_{g}.\Delta x[/tex]

                                                    [tex]W_{g}=-mg\Delta x[/tex]

                                            [tex]W_{g}=-81\ kg.\ 9.8\ \frac{m}{s^{2}}.\ 19\ m[/tex]

                                                  [tex]W_{g}=-15082.2\ J[/tex]

Answer:

a) Work done on the astronaut by the force from the helicopter = 16.104 kJ

b) Work done on the astronaut by the gravitational force = -15.082 kJ

Explanation:

mass of the astronaut, m = 81 kg

height, h = 19 m

acceleration of the astronaut, a = g/15

Since the astronaut is lifted up, using the third law of motion:

T - mg = ma

T = mg + ma

T = (81*9.81) + 81*(9.81/15)

T = 847.584 N

Work done on the astronaut by the helicopter

Work done = Tension * height

W = T* h

W = 847.584 * 19

Work done, W = 16104.096 Joules

W = 16.104 kJ

b) Work done on the astronaut by the gravitational force on her

[tex]W = -f_{g} h[/tex]

[tex]f_{g} = mg = 81 * 9.8\\f_{g} = 793.8 N[/tex]

[tex]W = -793.8 * 19\\W =- 15082.2 J[/tex]

W = -15.082 kJ

Answer:

(d) Zero

Explanation:

Given:

Side of the square, = L

Magnetic field, = B

According to Faraday, the net force acting on a conductor is equal to product of magnetic field, current, length of conductor and sine of the angle between the field and current.

F = BILsinΦ

Where:

B is magnetic field

I is current in the loop

L is length of the loop

Φ is the angle between the magnetic field and current

Φ = 0, since the current and magnetic field are directed in opposite directions.

F = BILsin(0)

F = 0

Therefore, the net magnetic force acting on the loop is zero.

The correct answer is (d) Zero

The correct option is Option d (zero). The net magnetic force acting on the loop is zero because the forces on the sides of the loop cancel each other out. Thus, the correct answer is d. Zero.

In a uniform magnetic field, the net force on a current-carrying loop in a magnetic field can be determined using the equation F = I \times B. We need to apply this equation to each side of the loop.

Step-by-Step Explanation:

Therefore, the net magnetic force acting on the loop is zero, making the correct answer d. Zero.

Answer:

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as a fishing

Explanation:

A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.  

Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.  

Answer:

loss of areas for tourism

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as fishing