The liquid nitrogen temperature is 63 K. Convert to ºC.

Answer:

x = 3.5 cm

Explanation:

When spring is used as a lift tool

so the mass suspended on it will have spring force on it

So as per the force equation of mass we can say

[tex]F_{net} = ma[/tex]

now net force on the mass is

[tex]F_{net} = kx - mg[/tex]

[tex]F_{net} = kx - mg = ma[/tex]

here we have

[tex]kx = mg + ma[/tex]

now we have

[tex]x = \frac{mg + ma}{k}[/tex]

[tex]x = \frac{2.2(9.8) + 2.2(3.25)}{8.2}[/tex]

[tex]x = 3.5 cm[/tex]

Answer:

Fc =  5.41 N

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Newton's second law for the set of the three blocks

F =  13 N

m=3.0 kg+ 4.0 kg+5.0 kg = 12 kg

F = m*a

13 = 12*a

a = 13 / 12

a = 1.083 m/s² : acceleration of the set of the three blocks

Newton's second law for the  5.0 kg block

m= 5.0 kg

a = 1.083 m/s²

Fc: Contact force of the 4 kg block on the 5 kg block

Fc =  5.0 kg * 1.083 m/s²

Fc =  5.41 N

The 4.0 kg block exerts a force of 5.4 N on the 5.0 kg block when a 13 N force is applied to the entire system of blocks and accelerates the system at 1.08 m/s² on a frictionless surface.

We are dealing with a problem related to Newton's Third Law of Motion where a 13 N force is applied to a system of three blocks in contact with each other on a frictionless surface.

To find the force that the 4.0 kg block exerts on the 5.0 kg block, we need to calculate the acceleration of the system first and then use Newton's second law for just the last two blocks.

The total mass of the system is the sum of the masses of the three blocks: 3.0 kg + 4.0 kg + 5.0 kg = 12.0 kg. The total force provided by the 13 N force gives us an acceleration of the system:

a = F_{total} / m_{total} = 13 N / 12.0 kg = 1.08 m/s² (rounded to two decimal places)

Now, using Newton's second law, F = ma, for the 5.0 kg block, which is being accelerated by the force from the 4.0 kg block (F_{45}), we find:

F_{45} = m_{5.0kg} × a = 5.0 kg × 1.08 m/s²= 5.4 N

Therefore, the 4.0 kg block exerts a force of 5.4 N on the 5.0 kg block.

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Answer: 3MeV electron

Explanation:

m_e={9.1\times 10^{-31}      m_α=4\times m_e    m_a={9.1\times 10^{-31}

m_p=1.67\times 10^{-27}

(a)  K.E. Energy of electron =[tex]\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}[/tex]=3MeV

[tex]v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }[/tex]=1.05\times10^{18}

[tex]v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}[/tex]

(b) K.E. Energy of alpha particle =[tex]\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}[/tex]=10MeV

[tex]v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6}  }{4\times9.1\times 10^{-31} }[/tex]=0.88\times10^{18}

[tex]v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}[/tex]

(c) K.E. Energy of auger particle =[tex]\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}[/tex]=0.1MeV

[tex]v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }[/tex]=0.035\times10^{18}

[tex]v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}[/tex]

(d)  K.E. Energy of proton particle =[tex]\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}[/tex]=400keV

[tex]v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3}  }{1.67\times 10^{-27} }[/tex]=0.766\times10^{14}

[tex]v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}[/tex]

from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.

Answer:

2.93 x 10⁻⁴ volts

4.88 x 10⁻⁵ A

Explanation:

N = Number of turns in cylindrical coil = 400

r = radius of the coil = 2 cm = 0.02 m

Area of the coil is given as

A = πr²

A = (3.14)(0.02)²

A = 12.56 x 10⁻⁴ m²

ΔB = Change in magnetic field = 0.13 - 0.06 = 0.07 T

t = time interval = 2 min = 2 x 60 sec = 120 sec

Induced emf is given as

[tex]E=\frac{NA\Delta B}{t}[/tex]

[tex]E=\frac{(400)(12.56\times 10^{-4})(0.07)}{120}[/tex]

E = 2.93 x 10⁻⁴ volts

R = resistance of the resistor = 6 ohm

i = induced current

Using ohm's law

E = i R

2.93 x 10⁻⁴ = i (6)

i = 4.88 x 10⁻⁵ A

Answer:

5.6 mH

Explanation:

i1 = 3.20 A, i2 = 1.90 A, e = 14 mV = 0.014 V,

Let L be the coil's inductance.

[tex]e = -L\times \frac{\Delta i}{\Delta t}[/tex]

[tex]0.014 = -L\times \frac{1.90 - 3.20}{0.52}[/tex]

L = 0.0056 H

L = 5.6 mH

Answer:

a) [tex]t=3.199 seconds[/tex]

b) [tex]h = 11.97 m[/tex]

Explanation:

Since this problem belongs to the concept of projectile motion

a) we know,

[tex]Vcos\theta=\frac{R}{t}[/tex]

Where,

V = initial speed

Θ = angle with the horizontal

R = horizontal range

t = Time taken to cover the range 'R'

Given:

V = 27m/s

R = 60m

Θ = 46°

thus,

the equation becomes

[tex]27\times cos46^o=\frac{60}{t}[/tex]

or

[tex]t=\frac{60}{27\times cos46^o}[/tex]

[tex]t=3.199 seconds[/tex]

b)The formula for height is given as:

[tex]h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\[/tex]

where,

g = acceleration due to gravity = 9.8m/s²

substituting the values in the above equation we get

[tex]h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\[/tex]

or

[tex]h = 62.124-50.14[/tex]

or

[tex]h = 11.97 m[/tex]

It takes 2.22 seconds for the T-shirt to reach the fan. The fan is located at a height of 24.57 meters from the ground.

To find the time it takes for the T-shirt to reach the fan, we need to solve for the time in the horizontal motion. The horizontal distance from the gun to the fan is given as 60 m. Since the horizontal motion is constant velocity, we can use the equation d = vt and solve for t. Plugging in the values, we get t = 60 m / 27 m/s = 2.22 s.

To find the height of the fan from the ground, we need to solve for the vertical motion. The equation for vertical motion is y = yt + (1/2)gt^2, where y is the vertical displacement, yt is the initial vertical velocity, and g is the acceleration due to gravity. In this case, the vertical displacement is unknown, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values and solving for y, we get y = (1/2)(9.8 m/s^2)(2.22 s)^2 = 24.57 m.

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Answer:

Acceleration= 1,59 (meters/(second^2))

Direction= NE; 65,22°  above the east direction.

Explanation:

Resulting force= ( ((180N)^2) + ((390N)^2) ) ^ (1/2) = 429,53 N

Angle obove the east direction= ((cos) ^ (-1)) (180N / 429,53 N) = 65,22°

Acceleration= Resulting force / mass = (429,53 N) / (270 kg) =

= (429,53 kg × (meters/(second^2))) / (270kg) = 1,59 (meters/(second^2))

Answer:

(a i - 2a j) / 2

Explanation:

a1 = a i

a2 = a j

a3 = - (i + j) (2a) = - 2a i - 2a j

a4 = (i - j) (3a) = 3a i - 3a j

Let the mass of each particle is m.

Use the formula of acceleration of centre of mass

a cm = (m1 a1 + m2 a2 + m3 a3 + m4 a4) / (m1 + m2 + m3 + m4)

a cm = (a i + aj - 2a i - 2a j + 3a i - 3a j) / 4

a cm = ( 2a i - 4a j) / 4

a cm = (a i - 2a j) / 2

Final answer:

The acceleration of the center of mass for a system of 4 equally massive particles, each with different accelerations specified, averages out to zero. This conclusion is based on the principle that the center of mass's acceleration is the vector sum of individual accelerations divided by the total number of particles.

Explanation:

The question asks about the acceleration of the center of mass of a system of 4 particles in space, each with the same mass but different accelerations. To find this, we will use the principle that the acceleration of the center of mass is given by the vector sum of the accelerations of all particles divided by the number of particles, assuming equal mass for simplicity. The accelerations provided are: a1 = iâ, a2 = jâ×2, a3 = –(iâ + jâ)(2a), a4 = (iâ - jâ)(3a), where a > 0 is a constant.

To calculate the center of mass acceleration, we sum up the given accelerations:

a1 = iâ

a2 = jâ×2

a3 = –(iâ + jâ)(2a) = –(2iâ + 2jâ)a

a4 = (iâ - jâ)(3a) = 3iâa - 3jâa

Summing these accelerations: (iâ + 2jâ - 2iâa - 2jâa + 3iâa - 3jâa), simplifies to (2iâ + 2jâ) - (2iâ + 2jâ)a. Given a = 1 for equal mass, the acceleration of each term cancels out, resulting in zero acceleration for the center of mass. Therefore, the acceleration of the center of mass of the system of 4 particles is zero.

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]

[tex]v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}[/tex]

[tex]v = 16.52 m/s[/tex]

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]

[tex]v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}[/tex]

[tex]v = 7.47 m/s[/tex]

Answer:

525.2

Explanation:

w = 154 rad/s, e0 = 16.5 V, A = 1 x 3 = 3 cm^2 = 3 x 10^-4 m^2, B = 0.68 T

Let the number of turns be N.

By use of law of electromagnetic induction

e0 = N B A w

16.5 = N x 0.68 x 3 x 10^-4 x 154

N = 525.2

Answer:

N = 525 Turns

Explanation:

Peak Voltage of a coil is given by the formula

[tex]V_{peak} = NBA\omega[/tex]

here we know that

[tex]V_{peak} = 16.5 Volts[/tex]

B = 0.68 T

[tex]\omega = 154 rad/s[/tex]

[tex]Area = (0.01 m)\times (0.03 m)[/tex]

[tex]Area = 3 \times 10^{-4} m^2[/tex]

now we have

[tex]16.5 = N(0.68)(3 \times 10^{-4})(154)[/tex]

[tex]N = 525[/tex]

Answer:

I = [tex]R^{2}[/tex](K+5)

Explanation:

Given :

J = k+5

Now selecting a thin ring in the wire of radius "r" and thickness dr.

Current through the thin ring is

dI = J X 2πrdr

dI = (K+5) x 2πrdr

Now integrating we get

I = [tex]\int_{0}^{R} = (K+5).2\pi rdr[/tex]

I = (K+5) 2π[tex]\int_{0}^{R} rdr[/tex]

I = (K+5) 2π [tex]\frac{R^{2}}{2}[/tex]

I = [tex]R^{2}[/tex](K+5)

-- Water cannot be boled.

-- There is no such thing as a coffe aker.

-- There is no such thing as an immension-type heating element.

Be that as it may, and it very likely still is, as it were . . .

-- The latent heat of vaporization of water at 100°C is about 2250 kilo-joules per kg.

-- To evaporate half of the kg of water in the coffee maker requires 1125 kJ of heat energy.

-- To supply that amount of heat energy over a period of 10 minutes (600 seconds), it must be supplied at a rate of

(1,125,000 Joules / 600 seconds) = 1,875 joules/second.

-- That's 1,875 watts, or 1.875 kilowatts.

-- Choice-a is the choice when the solution is rounded.

Answer:

wavelength

Explanation:

An electromagnetic waves is produced due to interaction of oscillating electric and magnetic field when they interacts perpendicular to each other. For example, Gamma rays, X rays , ultraviolet rays, visible radiations, infrared rays, micro waves, radio waves.

All the electromagnetic waves have velocity equal to velocity of light.

If the frequency of electromagnetic wave is unknown then we find the wavelength of wave. the formula used is

Wave speed = wavelength x frequency

Answer:

change in momentum of the stone is 1.635 kg.m/s

Explanation:

let m = 0.500kg ball and M  be the mass of the stone, v be the velocity of the ball and V be the velocity of the stone

the  initial kinetic energy of the ball is 1/2(0.500)(20^2) = 100 J

the kinetic energy of the ball after rebounding is 70/100(100) = 70 J

Kb = 1/2mv^2

 v =  \sqrt{2k/m} = \sqrt{2(70)/0.500} = 16.73 m/s  

from the conservation of linear momentum, we know that:

mvi + MVi = mvf + MVf

MVf - MVi = mvi - mvf

MVf - MVi = (0.500)(20) - (0.500)(16.73)

                 = 1.635 kg.m/s  

therefore, the change is momentum of the stone is 1.635 kg.m/s

Answer:

Explanation:

There are two types of collision.

(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.

In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The kinetic energy of the system before collision = the kinetic energy after the collision

(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.

In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The total mechanical energy of the system before collision = total mechanical of the system after the collision

Answer:

Part a)

[tex]m_2 = 4.9 \times 10^7 kg[/tex]

Part b)

[tex]q_1 = q_2 = 5312.6 C[/tex]

Explanation:

Part a)

As we know that both charge particles will exert equal and opposite force on each other

so here the force on both the charges will be equal in magnitude

so we will have

[tex]F = m_1a_1 = m_2a_2[/tex]

here we have

[tex]6.3 \times 10^7(7) = m_2(9)[/tex]

now we have

[tex]m_2 = 4.9 \times 10^7 kg[/tex]

Part b)

Now for the force between two charges we can say

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now we have

[tex](6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}[/tex]

now we have

[tex]q_1 = q_2 = 5312.6 C[/tex]

To calculate the ratio of rotational energy to translational kinetic energy for a pitched baseball, use the formulas for rotational kinetic energy and translational kinetic energy. The moment of inertia is found by using the radius of the ball, and the rotational and translational kinetic energies can be determined using the mass, radius, and velocity of the ball. Finally, calculate the ratio of the two energies to get the answer.

To calculate the ratio of the rotational energy to the translational kinetic energy, we need to find the rotational kinetic energy and the translational kinetic energy of the pitched baseball.

The rotational kinetic energy can be calculated using the formula:

Rotational Kinetic Energy = (1/2) * I * ω²

Where I is the moment of inertia and ω is the angular speed.

The translational kinetic energy can be calculated using the formula:

Translational Kinetic Energy = (1/2) * m * v²

Where m is the mass of the ball and v is the velocity.

We know that the radius of the ball is 3.87 cm and it moves at a velocity of 36.6 m/s. The angular speed is 112 rad/s. Since the ball is treated as a uniform sphere, its moment of inertia can be calculated using the formula:

Moment of Inertia = (2/5) * m * r²

Substitute the given values into the formulas and calculate the rotational and translational kinetic energies. Then, find the ratio of the rotational energy to the translational energy.

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Answer:

The kinetic energy of the wagon is 967.0 J

Explanation:

Given that,

Force = 120 N

Mass = 55 kg

Height = 8 m

We need to calculate the kinetic energy of the wagon

Using newtons law

[tex]F = ma[/tex]

[tex]\dfrac{120}{55}=a[/tex]

[tex]a =2.2\ m/s^2[/tex]

Using equation of motion

[tex]v^2 =u^2+2as[/tex]

Where,

v = final velocity

u = initial velocity

s = height

Put the value in the equation

[tex]v^2=0+2\times2.2\times8[/tex]

[tex]v=5.93\ m/s[/tex]

Now, The kinetic energy is

[tex]K.E=\dfrac{1}{2}mv^2[/tex]

[tex]K.E=\dfrac{1}{2}\times55\times(5.93)^2[/tex]

[tex]K.E=967.0\ J[/tex]

Hence, The kinetic energy of the wagon is 967.0 J

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

Answer:

output work is not possible to have more than 294 kJ value so this is not reasonable claim

Explanation:

As we know that the efficiency of heat engine is given as

[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]

now we will have

[tex]T_2 = 290 K[/tex]

[tex]T_1 = 500 K [tex]

[tex]\eta = 1 - \frac{290}{500}[/tex]

[tex]\eta = 0.42[/tex]

now we know that efficiency is defined as

[tex]\eta = \frac{Work}{Heat}[/tex]

[tex]0.42 = \frac{W}{700}[/tex]

[tex]W = 294 kJ[/tex]

So output work is not possible to have more than 294 kJ value

To determine if the inventor's claim of developing a heat engine is reasonable, we can calculate the maximum theoretical efficiency of the engine and compare it to the claimed net work output and heat input.

A heat engine operates between a hot reservoir and a cold reservoir, absorbing heat from the hot reservoir and converting some of it into work, while rejecting the remaining heat to the cold reservoir. The efficiency of a heat engine is determined by the temperature of the reservoirs. In this case, the inventor claims that the heat engine receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting waste heat to a sink at 290 K. To determine if this claim is reasonable, we can calculate the theoretical maximum efficiency of the heat engine using the Carnot efficiency formula.

The Carnot efficiency formula is given by:

Efficiency = 1 - (Tc / Th)

where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this case, Tc = 290 K and Th = 500 K.

Plugging in these values into the formula, we have:

Efficiency = 1 - (290 K / 500 K) = 1 - 0.58 = 0.42 or 42%

Therefore, the maximum theoretical efficiency for a heat engine operating between these temperatures is 42%. Since the claimed net work output of the heat engine is 300 kJ, we can calculate the maximum heat input by dividing the net work output by the efficiency:

Maximum heat input = Net work output / Efficiency = 300 kJ / 0.42 = 714.3 kJ

Since the claimed heat input is 700 kJ, which is slightly less than the maximum calculated heat input, it is reasonable to say that the inventor's claim is possible.

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Answer:

104 m/s

Explanation:

L = length of the wire stretched between he two points = 80 cm = 0.80 m

f = fundamental frequency of vibration of the wire = 65.0 Hz

v = speed of propagation of transverse waves in the wire

Speed of propagation is given as

v = 2fL

inserting the values

v = 2 (65.0) (0.80)

v = (130) (0.80)

v = 104 m/s

Answer:

[tex]F_{net} = -0.7 \hat i + 4.77 \hat j[/tex]

Explanation:

Force due to q2 on q1 is along - X direction due to repulsion between them

so we have

[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]

[tex]F_{12} = \frac{(9\times 10^9)(86 \mu C)(32 \mu C)}{2.5^2}[/tex]

[tex]F_{12} = 3.96 N (-\hat i)[/tex]

Now force between q1 and q3 is given as

[tex]F_{13} = \frac{kq_1q_3}{r^2} \frac{(1.5\hat i + 2.2 \hat j)}{\sqrt{1.5^2 + 2.2^2}}[/tex]

[tex]F_{13} = \frac{(9\times 10^9)(86 \mu C)(53 \mu C)}{(1.5^2 + 2.2^2} \frac{(1.5\hat i + 2.2 \hat j)}{\sqrt{1.5^2 + 2.2^2}}[/tex]

[tex]F_{13} = (5.78)\frac{(1.5\hat i + 2.2 \hat j)}{2.66}[/tex]

[tex]F_{13} = (2.17)(1.5\hat i + 2.2 \hat j)[/tex]

Now net force on q1 is given as

[tex]F_{net} = F_{12} + F_{13}[/tex]

[tex]F_{net} = 3.96(-\hat i ) + (3.26 \hat i + 4.77 \hat j)[/tex]

[tex]F_{net} = -0.7 \hat i + 4.77 \hat j[/tex]

Answer:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, [tex]\dfrac{dB}{dt}=3\ T/s[/tex]

Current, I = 6 A

Resistance of the solenoid, R = 17 ohms

We need to find the radius of the solenoid (r). We know that emf is given by :

[tex]E=N\dfrac{d\phi}{dt}[/tex]

[tex]E=N\dfrac{d(BA)}{dt}[/tex]

Since, E = IR

[tex]IR=NA\dfrac{dB}{dt}[/tex]

[tex]A=\dfrac{IR}{N.\dfrac{dB}{dt}}[/tex]

[tex]A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}[/tex]

[tex]A=0.188\ m^2[/tex]

or

[tex]A=0.19\ m^2[/tex]

Area of circular cross section is, [tex]A=\pi r^2[/tex]

[tex]r=\sqrt{\dfrac{A}{\pi}}[/tex]

[tex]r=\sqrt{\dfrac{0.19}{\pi}}[/tex]

r = 0.24 m

So, the  radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.

Answer:

[tex]u = 0.057 J/m^3[/tex]

Explanation:

Energy density near the surface of the sphere is given by the formula

[tex]u = \frac{1}{2}\epsilon_0 E^2[/tex]

also for sphere surface we know that

[tex]E = \frac{V}{R}[/tex]

R = radius of sphere

V = potential of the surface

now we have

[tex]u = \frac{1}{2}\epsilon_0 (\frac{V^2}{R^2})[/tex]

now from the above formula we have

[tex]u = \frac{1}{2}(8.85 \times 10^{-12})(\frac{8500^2}{0.075^2})[/tex]

[tex]u = 0.057 J/m^3[/tex]

Answer:

a) [tex]E_photon =0.306 eV[/tex]

b) [tex]E_photon =0.166 eV[/tex]

Explanation:

The energy of the photon (E) for [tex]n^th[/tex] orbit of the hydrogen atom is given as:

[tex]E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})[/tex]

where,

[tex]E_o[/tex] = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

[tex]E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})[/tex]

[tex]E_photon =0.306 eV[/tex]

b) Now for the transition from n = 5 to n = 6

[tex]E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})[/tex]

[tex]E_photon =0.166 eV[/tex]

The energy of the photon for the n=4 to n=5 transition in hydrogen is 0.306 eV, and for the n=5 to n=6 transition, it is 0.166 eV, calculated using the Rydberg formula for hydrogen energy levels.

The question asks about the energy of photons associated with specific electronic transitions in a hydrogen atom. We can use the Rydberg formula to calculate these energies: Ephoton = |Ef - Ei|, where Ef and Ei are the energies of the final and initial states, and the energy levels of hydrogen are given by En = -13.6 eV / n2 for an electron in the nth energy level.

To calculate the energy of the photon that could cause an electronic transition from the n=4 state to the n=5 state of hydrogen (a), and the n=5 state to the n=6 state (b), we simply need to plug these values into the energy level formula and subtract:

E4 = -13.6 eV / 42 = -0.85 eV

E5 = -13.6 eV / 52 = -0.544 eV

Energy for transition (a): |E5 - E4| = |-0.544 eV + 0.85 eV| = 0.306 eV

E6 = -13.6 eV / 62 = -0.378 eV

Energy for transition (b): |E6 - E5| = |-0.378 eV + 0.544 eV| = 0.166 eV

Therefore, the photons must have energies of 0.306 eV and 0.166 eV to cause transitions (a) and (b) respectively.

Answer:

6.93 volts

Explanation:

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

r = radius of the orbit = 2.08 x 10⁻¹⁰ m

V = Electric potential due to proton on electron

Electric potential due to proton on electron is given as

[tex]V = \frac{kq}{r}[/tex]

Inserting the values

[tex]V = \frac{(9\times 10^{9})(1.6\times 10^{-19})}{2.08\times 10^{-10}}[/tex]

V = 6.93 volts

Final answer:

The electric potential due to the proton on an electron in an orbit with a radius of 2.08 x 10^-10 m in a hydrogen atom model is approximately -13.6 volts.

Explanation:

In the context of the model of a hydrogen atom, the electric potential due to the proton on an electron in an orbit with radius 2.08 x 10-10 m can be calculated using the formula for the electric potential V due to a point charge, which is V = k * Q / r. Here, k is the Coulomb's constant (approximately 8.99 x 109 N m2/C2), Q is the charge of the proton (1.602 x 10-19 C), and r is the distance from the proton to the electron, which is the radius of the orbit.

To find the electric potential, simply plug these values into the formula:

V = (8.99 x 109 N m2/C2) * (1.602 x 10-19 C) / (2.08 x 10-10 m)

After calculating, the electric potential is found to be approximately -13.6 volts, with the negative sign indicating that the potential energy associated with the electron is negative, which is common for bound states such as electrons in an atom.

Answer:

750.25 cm

Explanation:

θ = 13.78°, h = 184 cm

Let the distance between rock and camera is d.

Tan θ = h / d

tan 13.78 = 184 / d

d = 750.25 cm

By utilizing trigonometry and the tangent of an angle in a right triangle, it is determined that the camera (at a height of 184.0 cm from the Mars surface with an angle of depression of 13.78° to a rock) is approximately 761.4 cm away from the rock.

This question involves trigonometry, specifically the tangent of an angle in a right triangle. The robot captures an image of a rock with an angle of depression of 13.78°. The camera is 184.0 cm above the Mars surface. We can create a right triangle where the angle at the camera is 13.78°, the opposite side is 184.0 cm (the height of the camera), and the adjacent side is the distance between the camera and the rock, which we will call x.

By definition, tan(angle) = opposite/adjacent. Here, the angle is 13.78°, the opposite side is 184.0 cm and the adjacent side is x (unknown). To find x, we can use the following formula: x = opposite/tan(angle).

Therefore, x = 184.0 cm / tan(13.78°). Using a calculator, x is approximately 761.4 cm (to the nearest tenth). So the camera is approximately 761.4 cm away from the rock.

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Answer:

(a) 5142.86 m

(b) 317.5 m/s

(c) 49.3 degree C

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 =  1/2 x 100 x v^2

v = 317.5 m/s

(c) The temperature of normal human body, T1 = 37 degree C

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C

I believe the answer is A: True

Answer:

v = 10.84 m/s

Explanation:

using the equation of motion:

v^2 = (v0)^2 + 2×a(r - r0)

due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.

v^2 = 2×g(r - r0)

v = \sqrt{2×(-9.8)×(4 - 10)}

  = 10.84 m/s

therefore, the velocity at r = 4 meters is 10.84 m/s