The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.6 cm apart with a 20 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.a)What is the electric field strength between the plates?

b)With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: the exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big.

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.05.

Explanation:

Given that,

Weight of the hockey puck, W = mg = 3.5 N

Acceleration of the hockey puck, [tex]a=0.45\ m/s^2[/tex]

The frictional force opposing the motion of the puck is given by

[tex]F=\mu N[/tex]

This force is balanced by the force due to motion of the hockey puck. So,

[tex]\mu N=ma\\\\\mu mg=ma\\\\\mu=\dfrac{a}{g}\\\\\mu=\dfrac{0.45}{9.8}\\\\\mu=0.045[/tex]

or

[tex]\mu=0.05[/tex]

So, the coefficient of kinetic friction between the puck and the ice is 0.05. Hence, this is the required solution.

Answer:

72 Nm²

Explanation:

work = F . d

work = (12i ,0j) . (6i ,-8j )

work = 72 J

The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.

It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j

When a force vector F acts in a nonparallel direction d, the work is given by:

W = |F|*|d|*cos(theta) where theta is the angle between the vectors.

alternatively you can use the dot product of the two vectors to get:

W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²

Final answer:

The work done on the block by the constant force of 12 N in the +x direction is 24 J.

Explanation:

The work done on the block can be calculated by multiplying the force applied to the block by the displacement of the block in the direction of the force. In this case, the force applied is 12 N in the +x direction and the displacement is (6 - 8) m = -2 m. Since the force and displacement are in opposite directions, the work done will be negative. The formula for work is given by:

Work = Force * Displacement * cos(theta)

where theta is the angle between the force and the displacement.

In this case, the angle between the force and the displacement is 180 degrees, so cos(theta) = -1. Substituting the values into the formula:

Work = 12 N * -2 m * (-1) = 24 J

The pilot's maximum endurance speed in a circle of 85 m diameter without blacking out can be calculated using the centripetal acceleration formula, taking into account the acceleration limit of 7 g, where 1 g equals 9.80 m/s².

The question requires the application of concepts from physics, specifically the idea of centripetal force and acceleration when a body moves in a circular path.

Given that the maximum acceleration a pilot can endure without blacking out is 7 g, and knowing that 1 g is equivalent to 9.80 m/s², we can calculate the maximum speed under these conditions using the formula for centripetal acceleration, which is `a = v² / r`, where `a` is the acceleration, `v` is the speed, and `r` is the radius of the circle.

The diameter of the circle is given as 85 meters, which means the radius `r` will be half of that, 42.5 meters. The maximum acceleration of 7 g translates to 7 × 9.80 m/s², which is 68.6 m/s². We can rearrange the formula to solve for `v`, leading to `v = √(a × r)`.

Substituting the values provided, we get `v = √(68.6 m/s² × 42.5 m)

v = √2923.175 m²/s² ≈ 54.1 m/s

Therefore, the maximum speed the pilot can tolerate while spinning in the horizontal circle is approximately 54.1 m/s, which is equivalent to about 192 km/h.

Answer:

The negative electrons moves to the left While the positively charged current moves in the opposite direction to the right when an external field is applied.

The positive atomic nuclei remains almost motionless.

In an external field, electrons and atomic nuclei move differently. In an electric field, electrons tend towards the positive field direction, whereas nuclei tend towards the negative. In a magnetic field, their paths are perpendicular to the field lines.

The motion of negative electrons and positive atomic nuclei under the influence of an external field depends on the nature of the external field. This field could be of many types such as electric, magnetic, or electromagnetic. In an electric field, electrons move towards the positive field direction while the nuclei move towards the negative field direction. In a magnetic field, the motion of electrons and nuclei is governed by the right-hand rule, which means they move in a path perpendicular to the magnetic field lines.

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Answer:

the magnitude and direction of the magnetic field in the region through which the current passes is 0.008 T and +z direction.

Explanation:

given information:

current, I = 20 A

magnetic force per unit length, F/L = 0.16 N/m

the conductor in the negative y-direction

θ = 90° (perpendicular)

as we know the formula to calculate magnetic force is

F = B I L sin θ

B = F/(I L sin θ)

   = (F/L) (1/I sin θ)

   = 0.16 (1/15 sin 90)

   = 0.008 T

since F is in the negative y direction, based of the right hand rule the magnetic field is in positive z direction

Answer:

Explanation:

Given:

current, I = 20 A

Magnetic force per unit length, F/L

= 0.16 N/m

Conductor in the negative y-direction, therefore θ = 90° (perpendicular)

For a magnetic field,

F = B I L sin θ

B = F/(I L sin θ)

= 0.16 × (1/15 sin 90)

= 0.008 T

The field is in the +ve z - direction.

Answer:

Explanation:

Given that,

Sphere of radius R

At radial distance ¼R it has an electric field of Eo

Magnitude of electric field E at r=2R

Solution

electric field is given as

E=KQ/r²

Now at R,

E=KQ/R²

When, at r=R/4 inside the sphere is given as

Eo=kQ/R³ • R/4

Eo=kQ/4R²

When, r =2R

Then,

E=kQ/(2R)²

E=kQ/4R²

Comparing this to Eo

Then, E=Eo

Then, the electric field at r=2R is equal to the electric field at r=R/4

Answer:

77.88 lbm/ft³

Explanation:

Given,

Specific gravity, SG = 1.25

Density of water, ρ = 62.30 lbm/ft³

density of the fluid =

   = S.G x ρ_{water}

   = 62.30 x 1.25

   = 77.88 lbm/ft³

Density of the fluid is equal to 77.88 lbm/ft³

Answer:

  B = -0.215 s⁻¹,  A₀ = 1.537º

Explanation:

To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values

           A = A₀ [tex]e^{-Bt}[/tex] + C

The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is

            (A-1) = A₀  e^{-Bt}

We do the logarithm

           Log (A-1) = log Ao –Bt log e

Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let's perform this calculation

              Ln (A-1) = Ln A₀ - Bt

To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are

              θ’= ln (θ -1)

θ(º)    t(s)        θ'(º)

6.2      2     1.6487

4.7      4     1.3083

3.2      6    0.7885

2.4      8    0.3365

1.8     10   -0.2231

We can use a graph paper and graph on the axis and the primary angle (θ') and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations

To find the line described by the equation

              y -y₀ = m (x -x₀)

             m = (y₂ -y₁) / (x₂ -x₁)

Where m is the slope of the graph e (x₀, y₀) is any point, let's start as the first point of the series

            (x₀, y₀) = (2, 1.65)

           (x₂, y₂) = (2, 1.65)

           (x₁, y₁) = (6, 0.789)

We use the slope equation

           m = (1.65 - 0.789) / (2-6)

           m = -0.215  

The equation is

           y - 1.65 = -0.215 (x- 2)

           y = -0.215 x +0.4301

We buy the two equations and see that the slope is the constant B

            B = -0.215 s⁻¹

The independent term is

            b = ln A₀

            A₀ = [tex]e^{b}[/tex]

           A₀ = [tex]e^{0.43}[/tex]

           A₀ = 1.537º

The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.

When a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.

This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.

The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.

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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.

A child is standing on the edge of a merry-go-round that is rotating with frequency f.

As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.

This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.

Angular momentum (L) is given by the formula:

where I is the moment of inertia and ω is the angular velocity.

complete question:

A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?

Answer:

0.080625 m/s

Explanation:

Horizontally, the total momentum must be conserved, meaning the momentum before and after the throw must be the same.

The total momentum before the throw is 0, and so is after the throw:

mv + MV = 0

where m, M are the masses of the ball and the quarterback, respectively. v and V are the velocities of the ball and the quarterback, respectively.

0.43*15 + 80V = 0

80V = -6.45

V = -6.45 / 80 = -0.080625 m/s

So he will be moving in the opposite direction with the ball at the rate of 0.080625 m/s

Final answer:

To determine the quarterback's movement after throwing the football, one would need to use conservation of momentum principles. Initial velocity, time in the air, and the maximum height of the football once thrown can be calculated using the equations of projectile motion.

Explanation:

The scenario presented in the question can be analyzed using the principles of conservation of momentum. Since the situation described involves horizontal motion and the exchange of momentum between the quarterback and the football, we will apply the concepts of momentum conservation and the horizontal aspect of projectile motion.

Conservation of Momentum

For part of the scenario where we determine the quarterback's movement after throwing the ball, we use the conservation of momentum which states that the total momentum of a system is conserved if there are no external forces acting on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity. Just before the quarterback throws the ball, both he and the ball are part of the same system, and their combined momentum is zero if he jumps straight up in the air. After throwing the ball, the quarterback will have a backward velocity to conserve the total momentum of the system; we calculate this by equating the momentum of the thrown ball and the momentum of the moving quarterback.

Projectile Motion

In a situation where the football is thrown in a certain manner, the motion of the football can be separated into horizontal and vertical components. The initial velocity of the football can be calculated based on the given range, height, and angle of throw using kinematic equations. The time it takes for the ball to reach the receiver and its maximum height can also be determined through these equations.

Final answer:

The maximum height a wrench can be thrown on the Moon with the same starting speed as on Earth is 6 times higher.

Explanation:

To determine how many times higher an astronaut could jump on the Moon compared to Earth, we need to compare the gravitational accelerations of both locations. The gravitational acceleration on the Moon is about 1/6 of the acceleration on Earth, denoted as g.

When an object is thrown vertically upward, its maximum height is determined by the initial velocity and the gravitational acceleration. Since the astronaut gives the wrench the same starting speed in both places, the maximum height it can reach on the Moon will be 6 times higher than on Earth. This is because the weaker gravitational acceleration on the Moon allows the object to reach a greater height before being pulled back down.

Therefore, if the astronaut can throw the wrench 15.0 m vertically upward on Earth, they could throw it about 90.0 m (15.0 m * 6) on the Moon with the same starting speed.

Answer:23 N

Explanation:

Given

Weight of box [tex]W=60\ N[/tex]

Coefficient of static friction is [tex]\mu _s=0.5[/tex]

Applied force [tex]F=23\ N[/tex]

When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.

Thus Static friction [tex]F_s[/tex]=applied force

[tex]F_s=23\ N[/tex]

Although it maximum value can go up to [tex](F_s)_{max}=\mu _sN[/tex]

[tex]F_s=0.5\times 60[/tex]

[tex]F_s=30\ N[/tex]

Answer:

210,600πft-lb

Explanation:

Force is a function F(x) of position x then in moving from

x = a to x= b

Work done = [tex]\int\limits^b_a {Fx} \, dx[/tex]

Consider a water tank conical in shape

we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w

we will have ,

[tex]\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a[/tex]

weight of slice under construction

weight = volume × density × gravitational constant

[tex]weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb[/tex]

Now we can find work done

[tex]W = \int\limits \, dw\\[/tex]

[tex]\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,[/tex]

= [tex]62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3[/tex]

= 210,600πft-lb

Answer:

Explanation:

Length, l = 35.5 cm = 0.355 m

y = 2 x

Slope, y / x = 2

tan θ = 2

θ = 63.4°

i = 22.5 A

B = 0.318 i Tesla

[tex]\overrightarrow{l}=0.355\widehat{k}+0.355 Cos63.4\widehat{i}+0.355 Sin63.4\widehat{j}[/tex]

[tex]\overrightarrow{l}=0.16\widehat{i} + 0.32\widehat{j}+0.355\widehat{k}[/tex]

[tex]\overrightarrow{B}=0.318\widehat{i}[/tex]

The magnetic force is given by

[tex]\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})[/tex]

[tex]\overrightarrow{F}=22.5\left ( 0.16\widehat{i}+0.32\widehat{j}+0.355\widehat{k} \right )\times 0.318\widehat{i}[/tex]

[tex]\overrightarrow{F}=2.54\widehat{j}-2.29\widehat{k}[/tex]

Magnitude of force

[tex]F=\sqrt{2.54^{2}+2.29^{2}}[/tex]

F = 3.42 N

The angle is Ф

tanФ = -2.29/2.54

Ф = 42° below y axis

The total force experienced by the wire in the magnetic field has a magnitude of 145 N and is directed -30 degrees relative to the z-axis, according to the right-hand rule and Lorentz force law.

In order to determine the direction of the total force on the wire, you would need to use the right-hand rule. First, you know the current is flowing down the z-axis and out along the y=2x line in the xy-plane. You also know that the magnetic field, B, is given by (.318 i)T which lies along the x-axis.

According to the Lorentz force law, the force exerted on a current-carrying conductor in a magnetic field is given by F = I * (L x B), where I represents the current, L is the length vector of the wire, and x is the cross product. Because the wire is bent at a right angle, the force will have two components: one for each section of the wire.

The force along the z-axis is given by F = I*LB*sin(theta), where theta is the angle between L and B. Here, L = 35.5/2 = 17.75 cm, B = 0.318 T, I = 22.5 A, and theta = 90 degrees (since B is along x-axis and L is along z-axis). Calculating gives Fz = 22.5 A * 17.75 cm * 0.318 T * sin(90) = 126.6 N.

The force along the y-axis is similar, but with the length vector along y=2x in the xy plane and B still along the x-axis, so theta = 180 degrees - arctan(2). Substituting the values we get Fy = - 22.5 A * 17.75 cm * 0.318 T * sin(180 - arctan(2)) = -73.8 N.

Therefore, the total force is given by the vector sum of Fz and Fy, which has magnitude sqrt(Fz^2 + Fy^2) = 145 N, in the direction of atan2(Fy, Fz) = -30 degrees relative to the z-axis.

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Answer:

Option D is correct.

Explanation:

Bmax = Emax / c

The general form for electromagnetic wave equation is

E = jEmax ×cos(kx-wt)

We were given

(360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ].

So from the equation above

Emax = 360V/m

Bmax = 360/(3×10⁸) = 1.2 ×10‐⁶ T.

Answer

Option D

Amplitude of Magnetic field = B = 1.2×10⁻⁶ T

Explanation:

The relationship between electric field and magnetic field of an electromagnetic wave is given by

B = E/c

Where B is the amplitude of magnetic field and E is the amplitude of electric field and c is the speed of light

The amplitude of electric field is given as 360 V/m

B = (360 V/m)/(3×10⁸ m/s)

B = 1.2×10⁻⁶ V.s/m²

Since 1 Tesla is equal to 1 V.s/m²

B = 1.2×10⁻⁶ T

Therefore, option D is correct

Answer:

5.02 m/s

Explanation:

given,

frequency of sound emitted by bat, [tex]f_1[/tex] = 30 kHz = 30000 Hz

beat frequency detected  by bat[tex]f_b[/tex] = 900 Hz

Speed of sound, v = 340 m/s

[tex]f_b = f_1 -f_2[/tex]

[tex]900 = 30000-f_2[/tex]

[tex]f_2 = 30900[/tex]

Using Doppler's effect formula

[tex]f_2=f_1(\dfrac{v+v_b}{v-v_b})[/tex]

[tex]v_b = velocity\ of\ bat[/tex]

[tex]30900=30000(\dfrac{340+v_b}{340-v_b})[/tex]

[tex]1.03 =(\dfrac{340+v_b}{340-v_b})[/tex]

[tex]v_b = 5.02 m/s[/tex]

Hence, the speed of bat is equal to 5.02 m/s

D. Torque

Explanation:

Torque is a pattern of the force that can make a victim to revolve on an axis. The torque's direction vector based on the force's direction on the axis. The SI unit for torque is the Newton-meter. Torque can be both static or dynamic.

A static is one that does not generate an angular acceleration.The  Torque's magnitude  vector ζ for a torque generated by a given force F is

ζ = F .r sin(θ)

where,

r is the width of the moment arm

θ is the angle within the force vector and the moment arm.

In rotational kinematics, torque is measured as

ζ = Iα

Where,

α is the angular acceleration

I is the rotational inertia

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To solve this problem we will apply the concepts related to Power, such as the product of voltage and current. Likewise, Power is defined as the amount of energy per unit of time, therefore, using this relationship we will solve the second part.

PART A)

We know that Power is,

[tex]P = VI[/tex]

Substitute [tex]25.0V[/tex] for V and 12.0 A for I in the expression,

[tex]P=(25.0V)(12.0A)[/tex]

[tex]P = 300W[/tex]

Therefore the power deliver to the body is 300W

PART B) Converting the time to SI,

[tex]t = 3ms = 0.003s[/tex]

Therefore if we have that the expression for energy is,

[tex]P = \frac{E}{t} \rightarrow E = Pt[/tex]

Here,

E = Energy,

P = Power,

t = Time,

Replacing,

[tex]E = (300W)(0.003s)[/tex]

[tex]E = 0.9J[/tex]

Therefore the energy transferred is equal to 0.9J

This question involves the concepts of electrical power and energy.

(a) The power delivered by the defibrillator is "300 W".

(b) The energy trasferred is "0.9 J".

The electrical power delivered by the defibrillator can be given by the following formula:

[tex]P=IV[/tex]

where,

Therefore,

[tex]P=(12\ A)(25\ V)[/tex]

P = 300 W

The transferred energy can be given by the following formula:

[tex]P=\frac{E}{t}\\\\E=Pt[/tex]

where,

Therefore,

[tex]E=(300\ W)(3\ x\ 10^{-3}\ s)\\[/tex]

E = 0.9 J

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Answer:

(a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s - 6106 J

(b) he is then lifted at the constant speed of 4.70 m/s - 5488 J

(c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage - 4869 J

Explanation:

knowing

d = 10 m

m = 56 kg

The work done by the applied force to pull the spelunker is given by

Wa + Wg = Kf - Ki

Wg = -mgd

First

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = (56*9.8*10) + (0.5*56*[tex]4.7^{2}[/tex])

Wa = 6106 J

Second

Kf = Ki

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = 56*9.8*10

Wa = 5488 J

Third

Kf = 0

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = (56*9.8*10) - (0.5*56*[tex]4.7^{2}[/tex])

Wa = 4869 J

Answer:

Φ = 2.26 10⁶ N m² / C

Explanation:

The electric flow is

           Ф = E .ds = [tex]q_{int}[/tex] / ε₀

Since we have two components for this flow. The uniform electric field and the flux created by the point charge.

The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface

The flow created by the point load at the origin is

              Φ = q_{int} /ε₀

Let's calculate

              Φ = 2.00 10⁻⁶ /8.85 10¹²

              Φ = 2.26 10⁶ N m² / C

Answer: [tex]v \approx 18.37 \frac{m}{s}[/tex]

Explanation:

Let assume that system is conservative. From application of the Principle of Energy Conservation, it is noticed that initial linear kinetic energy must be equal to the gravitational energy at the top of the circle. That is to say:

[tex]K_{1} = U_{2}\\[/tex]

[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (2\cdot R) \\v = \sqrt{4 \cdot g \cdot R}[/tex]

Where [tex]g = 9.81 \frac{m}{s^{2}}[/tex].

[tex]v = \sqrt{4 \cdot (9.81 \frac{m}{s^2})\cdot(8.6 m)} \\v \approx 18.37 \frac{m}{s}[/tex]

The cylinder will descend a distance of f/k before coming momentarily to rest.

When the cylinder is released, it will initially accelerate downwards due to the force of gravity. As it accelerates, the spring attached to it will begin to stretch, exerting an upward force on the cylinder. The cylinder will continue to descend until the force exerted by the spring is equal in magnitude to the force of kinetic friction between the cylinder and the walls of the tube. At this point, the net force on the cylinder will be zero, and it will come momentarily to rest.

To find the distance the cylinder descends before coming to rest, we can equate the force exerted by the spring with the force of kinetic friction:

kx = f

Where k is the force constant of the spring and x is the distance the cylinder has descended. Solving for x gives:

x = f/k

Therefore, the vertical distance the cylinder descends before it comes momentarily to rest is given by x = f/k.

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Answer:

Th = 50°C

Explanation:

See the attachment below.

K = 7.75

Tc = 13.5°C = 273+13.5 = 286.5K

Answer:

292.984 K or 19.98 °C

Explanation:

Using,

η = Tc/(Th-Tc)................ Equation 1

Where η = coefficient of performance, Tc = cold temperature of the refrigerator, Th = hot temperature of the refrigerator.

Make Th the subject of the equation

η (Th-Tc) = Tc

η Th-η Tc = Tc

Th = (Tc+ηTc)/η ....................... Equation 2

Given: Tc = -13.5°C+273 = 259.5 °C, η  = 7.75

Substitute into equation 2

Th = (259.5+7.75×259.5)/7.75

Th = (259.5+2011.125)/7.75

Th = 2270.625/7.75

Th = 292.984 °C or 19.98 °C

Hence the maximum hot reservoir temperature = 292.984 °C or 19.98 °C

Final answer:

The magnitude of the electric force on the second object remains the same when the first object is moved to a distance 2r from it.

Explanation:

To find the magnitude of the electric force on the second object (B) when the first object (A) is moved to a distance 2r from B, we can use Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Given that the charge of A is Q and the charge of B is 4Q, the force exerted on B when A is at distance r is F. Now, when A is moved to a distance 2r from B, the new distance between A and B is 2r. Using Coulomb's law and the proportionalities mentioned earlier, the magnitude of the new electric force on B can be calculated as:

F' = (k * Q * 4Q) / (2r)^2

Where k is the constant in Coulomb's law. Simplifying the equation gives:

F' = (4 * F) / 4 = F

Therefore, the magnitude of the electric force on the second object B remains the same when the first object A is moved to a distance 2r from B.

Incomplete question.The complete one is here

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?

Answer:

[tex]T=5900s=99min[/tex]

Explanation:

Given

[tex]r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\[/tex]

To find

orbital period of a spacecraft T

Solution

An the initial calculating is computing the angular velocity of satellite :

[tex]w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s[/tex]

Computing T

[tex]T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min[/tex]

Explanation:

Given data:

G = 11.5×10⁶psi

d₂ = 2.0 inch

d₁ = 1.5 inch

ε[tex]_{max}[/tex] = 170 × 10⁻⁶

Y[tex]_{max}[/tex] = 2ε

T/J = τ[tex]_{max}[/tex] /R

[tex]\frac{Td_{2} }{2J}[/tex] = τ[tex]_{max}[/tex]         (1)

τ[tex]_{max}[/tex] = G Y

from 1 and 2

[tex]\frac{T d_{2} }{2J} = G Y_{max}[/tex]

T = [tex]\frac{2 G Y_{max}J }{d_{2} }[/tex]

[tex]\frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4} }{2* 0.0254}[/tex]

  = 474.14 Nm

Answer:

Explanation:

Since , resistances are joined in parallel , potential difference across either of R₁ or R₂ will be same , and it will be equal to emf of the cell provided no internal resistance exists in the cell.

Since potential difference across R₁ is equal to emf and resistance is R₁ , current through it too will be same as that in the previous case.

The potential difference across resistor R1 remains the same when R2 is added in parallel, and the current through R1 also remains unchanged because the voltage across it does not change.

When R2 is added in parallel to R1, the potential difference (voltage) across R1 remains the same because in a parallel circuit, all components have the same potential difference across them. However, the current i1 through R1 remains the same as previously as well since the potential difference across it hasn't changed and Ohm's Law (I = V/R) dictates that the current through a resistor will depend only on the potential difference across it and its resistance, which remains unchanged.

Adding R2 in parallel with R1 has the effect of decreasing the overall resistance of the circuit as found using the equation 1/Rp = 1/R1 + 1/R2. This means that the total current drawn from the battery will increase, but it will be divided among R1 and R2. Since the voltage across R1 remains constant, the current through R1 is unaffected by the addition of R2.

Answer:

a) The statement is true, b)  [tex]K_{tr.atom} = 6.073 \times 10^{-21} \frac{J}{atom}[/tex],[tex]v_{rms} \approx 477.904 \frac{m}{s}[/tex]

Explanation:

a) The standard conditions are 273.15 K and 1 atm. Let consider that gas behaves ideally, whose equation of state is:

[tex]P \cdot V = n \cdot R_{u} \cdot T[/tex]

The volume is cleared out:

[tex]V = \frac{n \cdot R_{u} \cdot T}{P}[/tex]

By replacing terms:

[tex]V = \frac{(1 mole)\cdot (0.082 \frac{L \cdot atm}{mole \cdot K} )\cdot (273.15 K)}{1 atm}\\V = 22.399 L[/tex]

Therefore, the statement is true.

b) The average kinetic energy per atom is:

[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{R_{u}\cdot T}{N_{A}}[/tex]

Where [tex]N_{A}[/tex] is the Avogadro constante and is equal to [tex]6.022 \times 10^{23} \frac{atoms}{mole}[/tex].

[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{(8.317 \frac{J}{mole\cdot K} )\cdot (293.15 K)}{6.022 \times 10^{23} \frac{atoms}{mole} } \\K_{tr,atom} = 6.073\times 10^{-21} \frac{J}{atom}[/tex]

The rms speed is determined by the following formula:

[tex]v_{rms} = \sqrt{\frac{3 \cdot k \cdot T}{n \cdot M_{O} \cdot \frac{1}{N_{A}} } }\\v_{rms} = \sqrt{\frac{3\cdot(1.38 \times 10^{-23} \frac{J}{K} )\cdot(293.15 K)}{(2 moles) \cdot (16 \frac{g}{mole} )(\frac{1 kg}{1000 g} ) \cdot (\frac{1 mole}{6.022 \times 10^{23} atoms} )}} \\v_{rms} \approx 477.904 \frac{m}{s}[/tex]

Answer:

1.7 m

Explanation:

Draw a free body diagram of the ladder.  There are 5 forces:

Normal force N pushing up at the base of the ladder.

Friction force Nμ pushing right at the base of the ladder.

Weight force mg pushing down a distance d up the ladder.

Weight force Mg pushing down a distance L/2 up the ladder.

Reaction force R pushing left at the top of the ladder.

Sum of forces in the x direction:

∑F = ma

Nμ − R = 0

Sum of forces in the y direction:

∑F = ma

N − mg − Mg = 0

Sum of moments about the base of the ladder:

∑τ = Iα

mg (d cos θ) + Mg (L/2 cos θ) − R (L sin θ) = 0

Use the first equation to substitute for R:

mg (d cos θ) + Mg (L/2 cos θ) − Nμ (L sin θ) = 0

Use the second equation to substitute for N:

mg (d cos θ) + Mg (L/2 cos θ) − (mg + Mg) μ (L sin θ) = 0

Simplify and solve for d:

m (d cos θ) + M (L/2 cos θ) − (m + M) μ (L sin θ) = 0

m (d cos θ) = (m + M) μ (L sin θ) − M (L/2 cos θ)

d = [ (m + M) μ (L sin θ) − M (L/2 cos θ) ] / (m cos θ)

Plug in values and solve:

d = [ (67 kg + 12 kg) (0.39) (5 m sin 43°) − (12 kg) (2.5 m cos 43°) ] / (67 kg cos 43°)

d = 1.70 m

Rounded to two significant figures, the maximum distance is 1.7 m.

The maximal distance,  [tex]d_{max[/tex], the person will reach before the ladder slips will be 3.41 m.

Here, we need to consider the forces acting and torque equilibrium. Let's start by noting down the given data:

The forces acting on the ladder are:

Using Newton’s second law for horizontal and vertical equilibrium and setting torques about the base of the ladder:

1. Horizontal forces:

2. Vertical forces:

For torque equilibrium about the base:

Substitute: [tex]\[N = (m_l + m_p)g\][/tex]

And frictional force: f = μN

We solve for d:

Simplify:

Substituting values:

Answer:

The balloon falls to the ground before the professor gets there. The student is DEFINITELY in for some TROUBLE!

Explanation:

The balloon picks up speed due to gravity and we can calculate the time taken for it to fall to the ground as follows:

Gravity (g) = 9.81 m/s^2

Height or distance (s) = 11 meters

Initial Speed (u) = 0 m/s

[tex]s = u*t + 0.5 * (a*t^2)[/tex]

[tex]11 = 0*t + 0.5 (9.81*t^2)[/tex]

[tex]t= 1.4975 s[/tex]

So we can see that the balloon takes 1.4975 seconds to fall to the ground, and since the professor takes 2 seconds to get to that place, the balloon hits the ground right before the professor gets there.