A current density of 7.00 10-13 A/m2 exists in the atmosphere at a location where the electric field is 185 V/m. Calculate the electrical conductivity of the Earth's atmosphere in this region.

Answer:

Explanation:

We cannot walk on a surface which has no friction.

To move across the surface, take a stone and throw it in the opposite direction of motion so that you get a reaction in the direction of motion and then you move across the surface.

Answer:

[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]

Explanation:

Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.

By using the definition of refractive index

refractive index of the medium is given by

n = speed of light in vacuum / speed of light in medium

n = c / v  ..... (1)

Use Snell's law

n = Sin i / Sin r

Where, r be the angle of refraction

From equation (1)

c / v = Sin i / Sin r

Sin r = v Sin i / c

[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]

Answer:

Part a)

t = 1.05 s

Part b)

[tex]\theta = 8.78 rad[/tex]

Explanation:

Initial angular speed is given as

[tex]\omega_i = 40 rev/min = 0.66 rev/s[/tex]

[tex]\omega_i = 2\pi (0.66) = 4.19 rad/s[/tex]

angular acceleration is given as

[tex]\alpha = 8 rad/s^2[/tex]

now we have

part a)

final angular speed = 120 rev/min

[tex]\omega_f = 2\pi(\frac{120}{60} rev/s)[/tex]

[tex]\omega_f = 12.57 rad/s[/tex]

now by kinematics we have

[tex]\omega_f = \omega_i + \alpha t[/tex]

[tex]12.57 = 4.19 + 8 t[/tex]

[tex] t = 1.05 s[/tex]

Part b)

Angle turned by the blades is given by

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]

[tex]\theta = 4.19(1.05) + \frac{1}{2}(8)(1.05)^2[/tex]

[tex]\theta = 8.78 rad[/tex]

Answer:

a) V = 195.70 m/s

b) f=3.02 × 10⁻⁴ Hz

c) T = 3311.25 seconds

Explanation:

Given:

Wavelength, λ = 646 Km = 646000 m

Distance traveled = 3410 Km = 3410000 m

Time = 4.84 h = 4.84 × 3600 s = 17424 seconds

a) The speed (V) of the wave is given as

V = distance / time

V = 3410000 m/ 17424 seconds

or

V = 195.70 m/s

b) The frequency (f) of the wave is given as:

f = V / λ

f= 195.70 / 646000

f=3.02 × 10⁻⁴ Hz

c) The time period (T)  is given as:

T = 1/ f

T = 1/ (3.02 × 10⁻⁴) Hz

T = 3311.25 seconds

Answer:

Part A)

[tex]\alpha = 745 rad/s^2[/tex]

Part B)

[tex]\theta = 4563.1 rad[/tex]

Explanation:

Drill starts from rest so its initial angular speed will be

[tex]\omega_i = 0[/tex]

now after 3.50 s the final angular speed is given as

[tex]f = 2.49 \times 10^4 rev/min[/tex]

[tex]f = {2.49 \times 10^4}{60} = 415 rev/s[/tex]

so final angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega_f = 2607.5 rad/s[/tex]

now we have angular acceleration given as

[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]

[tex]\alpha = \frac{2607.5 - 0}{3.50}[/tex]

[tex]\alpha = 745 rad/s^2[/tex]

Part b)

The angle through which it is rotated is given by the formula

[tex]\theta = \frac{(\omega_f + \omega_i)}{2}\delta t[/tex]

now we have

[tex]\theta = \frac{(2607.5 + 0)}{2}(3.50)[/tex]

[tex]\theta = 4563.1 rad[/tex]

Answer:

The magnetic field strength is required [tex] 2.84\times10^{-14}\ T[/tex]

Explanation:

Given that,

Speed of proton[tex]v = 3.60\times10^{6}\ m/s[/tex]

Mass of proton[tex]m_{p}=1.67\times10^{-27}\ kg[/tex]

Charge[tex]q =1.60\times10^{-19}\ C[/tex]

When a proton moves horizontally, at a right angle to a magnetic field .

Then, the gravitational force balances the magnetic field

[tex]mg=Bqv\sin\theta[/tex]

[tex]B = \dfrac{mg}{qv}[/tex]

Here, [tex]\theta = 90^{\circ}[/tex]

Where, B = magnetic field

q = charge

v = speed

Put the value into the formula

[tex]B = \dfrac{1.67\times10^{-27}\times9.8}{1.60\times10^{-19}\times3.60\times10^{6}}[/tex]

[tex]B = 2.84\times10^{-14}\ T[/tex]

Hence, The magnetic field strength is required [tex] 2.84\times10^{-14}\ T[/tex]

Final answer:

To balance the gravitational force on a proton moving at a right angle through a magnetic field, the required field strength is found using the magnetic force formula, resulting in a necessary field strength of 2.86 x 10⁻²³ T.

Explanation:

The student is asking about the magnetic force required to counteract the gravitational force on a proton moving horizontally through a magnetic field. To solve this, we need to use the formula F = qvB, where F is the magnetic force, q is the charge of the proton, v is the velocity of the proton, and B is the magnetic field strength we wish to find.

First, we calculate the weight of the proton using W = mg, where m is the mass of the proton and g is the acceleration due to gravity (9.81 m/s²). This weight is the force we aim to balance with the magnetic force.

Now, let's calculate the weight of the proton: W = (1.67 × 10⁻²⁷ kg) × (9.81 m/s²) = 1.64 × 10⁻²⁶ N.

To keep the proton moving horizontally, the magnetic force needs to equal the proton's weight. So we set F to W and solve the equation for B:

B = W/(qv) = (1.64 × 10⁻²⁶ N) / ((1.60 × 10⁻²⁹ C) × (3.60 × 10⁶ m/s))

B = 2.86 × 10⁻² T

Therefore, a magnetic field strength of 2.86 × 10⁻² Tesla is required to just balance the weight of the proton and keep it moving horizontally.

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, [tex]A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2[/tex]

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, [tex]\rho=11\times 10^{-8}\ \Omega-m[/tex]

We need to find the resistance of the iron rod. It is given by :

[tex]R=\rho\dfrac{L}{A}[/tex]

[tex]R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}[/tex]

[tex]R=0.000077 \Omega[/tex]

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

Answer:

The mass of neon is 13.534 g.

Explanation:

Given that,

Volume = 3.8 L

Pressure = 6.8 atm

Temperature = 470 K

Atomic mass of neon =20.2 g/mol

Gas constant R = 8.314 = 0.082057 L atm/mol K

We need to calculate the mass of neon

Using equation of gas

[tex]PV=nRT[/tex]

[tex]6.8\times3.8=n\times0.082057\times470[/tex]

[tex]n=\dfrac{6.8\times3.8}{0.082057\times470}[/tex]

[tex]n= 0.670[/tex]

We know that,

[tex]n = \dfrac{m}{molar\ mass}[/tex]

[tex]m = n\times molar\ mass[/tex]

[tex]m=0.670\times20.2[/tex]

[tex]m=13.534\ g[/tex]

Hence, The mass of neon is 13.534 g.

Answer:

5.3 cm

Explanation:

Let the charges on the spheres be q and Q.

r = 9.20 cm

F = k Q q / (9.20)^2 ..... (1)

Let the new distance be r' and force F'

F' = 3 F

F' = k Q q / r'^2

3 F = k Q q / r'^2 ...... (2)

Divide equation (1) by (2)

F / 3 F = r'^2 / 84.64

1 / 3 = r'^2 / 84.64

r' = 5.3 cm

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let [tex]\rho_{m}[/tex] be the density of metal and [tex]\rho_{w}[/tex] be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

[tex]W =0.10400\ N[/tex]

[tex]mg=0.10400[/tex]

[tex]\rho V g=0.10400[/tex]

[tex]Vg=\dfrac{0.10400}{\rho_{m}}[/tex].....(I)

The weight of metal in water is

Using buoyancy force

[tex]F_{b}=0.10400-0.08400[/tex]

[tex]F_{b}=0.02\ N[/tex]

We know that,

[tex]F_{b}=\rho_{w} V g[/tex]....(I)

Put the value of [tex]F_{b}[/tex] in equation (I)

[tex]\rho_{w} Vg=0.02[/tex]

Put the value of Vg in equation (II)

[tex]\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02[/tex]

[tex]1000\times\dfrac{0.10400}{0.02}=\rho_{m}[/tex]

[tex]\rho_{m}=5200\ kg/m^3[/tex]

Hence, The density of the metal is 5200 kg/m³.

The density of the metal bracelet can be determined through the principles of Archimedes using the difference of its weight in air and water. After calculating the buoyant force, the volume of water displaced was determined, which was also the volume of the bracelet. Dividing its weight in the air by this volume gave us the density, which is approximately 5,434 kg/m³.

In order to find the density of the metal bracelet, we use Archimedes' principle, which states that the buoyant force (force exerted on a submerged object) is equal to the weight of the fluid displaced by the object.

The first step is to calculate the difference in weight in air and water, which gives the buoyant force, i.e., the weight of the water displaced. So, the difference is 0.10400 N - 0.08400 N = 0.02000 N.

Then, we can find the volume of the water displaced by using the formula for weight, w = mg. Here, w is the buoyant force, and g is the acceleration due to gravity (9.8 m/s² on Earth). So, m = w / g = 0.02000 N / 9.8 m/s² = 0.00204 kg. This is the mass of the water displaced, which is also the volume since 1 kg of water is 1 liter (or 1000 cm³).

Finally, the density of the metal bracelet can be calculated by dividing its weight in air by the volume of water displaced, and by gravity. That is, density = 0.10400 N / (0.00204 kg x 9.8 m/s²) = 5,434.4 kg/m³. So the density of the metal bracelet is approximately 5,434 kg/m³.

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Using the acceleration due to gravity on the planet, we calculate its mass to be 8.97 x 10²⁴ kg. For the mass of the star, using Kepler's third law, we find it to be approximately 1.99 x 10³⁰ kg, assuming the planet's orbit is about 1 AU.

To find the mass of the planet and the star, we can use Newton's form of Kepler's third law and Newton's universal law of gravitation. Let's denote the mass of the star as M and the mass of the planet as m.

Mass of the Planet

The acceleration due to gravity (g) on the surface of the planet is given by:

g = Gm / r²

Where:

G is the gravitational constant (approximately 6.674 x 10⁻¹¹ N m2 kg-²).

m is the mass of the planet.

r is the radius of the planet, which is half of its diameter.

Rearranging the formula to solve for m:

m = g r² / G

Substituting the given values (g = 59.7 m/s² and r = 9 x 106 m), we get:

m = 59.7 x (9 x 106)² / 6.674 x 10⁻¹¹
= 8.97 x 10²⁴ kg

Mass of the Star

Using Newton's version of Kepler's third law, we get:

M = 4π² a3 / G T²

Where: a is the semi-major axis of the planet's orbit, which is the average distance from the planet to the star.

T is the orbital period of the planet around the star.

Assuming that the planet's orbit is approximately 1 Astronomical Unit (AU) since the period is close to 1 Earth year, and converting 402 Earth days to seconds, we find M.

T = 402
days x 24 hours/day x 3600 seconds/hour = 3.47 x 10⁷ seconds

a = 1 AU = 1.496 x 10¹¹ m

Then the mass of the star M is:

M = 4π² (1.496 x 1011)3 / (6.674 x 10⁻¹¹ x (3.47 x 10⁷)²) = 1.99 x 10³⁰ kg

Answer:

Answer to the question: 91.44m

Explanation:

L=100 yds = 300 ft = 91.44 m

To convert 100 yards to meters for an American football field, you multiply 100 yards by 3 to get 300 feet, then convert feet to meters using the given conversion factor, resulting in approximately 91.4 meters.

To convert the length of an American football field from yards to meters, we first need to understand the conversion factors between these units. We know that 1 yard equals 3 feet, and we are given that 1 meter equals 3.281 feet. Starting with the length of the field in yards, which is 100 yards, we convert yards to feet and then convert feet to meters.

Step-by-Step Conversion

Therefore, the length of an American football field is roughly 91.4 meters, excluding the end zones.

Answer:

The magnetic field strength, B = 110 T

Explanation:

It is given that,

Cyclotron frequency, [tex]\nu=3\times 10^{12}\ Hz[/tex]

We need to find the magnetic field strength. The formula for cyclotron frequency is given by :

[tex]\nu=\dfrac{qB}{2\pi m}[/tex]

B is the magnetic field strength

q and m are the charge and mass of electron.

[tex]B=\dfrac{2\pi m\nu}{q}[/tex]

[tex]B=\dfrac{2\pi\times 9.1\times 10^{-31}\ kg\times 3\times 10^{12}\ Hz}{1.6\times 10^{-19}\ C}[/tex]          

B = 107.20 T

or

B = 110 T (approx)

So, the magnetic field strength of the electron is 110 T. Hence, this is the required solution.

Using the charge, mass, and cyclotron frequency of the electron, the magnetic field strength is calculated to be approximately 0.084 T.

The frequency of the electron in the magnetic field, called the cyclotron frequency, can be used to find the strength of the magnetic field. The equation for cyclotron frequency is ƒ = qB/2πm, wherein ƒ is the frequency, q is the charge, B is the magnetic field, and m is the mass of the electron. In this case, to solve for B (magnetic field), we can rearrange the equation to B = 2πmƒ/q.

Substituting the given values, ƒ = 3.0 x 10^12 Hz, q = 16 x 10^-19 C, and m = 9.1 x 10^-31 kg, into the formula, and performing the proper calculation, the magnetic field strength is found to be approximately 0.084 T, which corresponds to the option (D).

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Answer:

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

Explanation:

Let unknown velocity be v.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 6i + m x 0i = 6m i

Final momentum = m x (4.8cos 30 i + 4.8sin 30 j) + m x v = 4.16 m i + 2.4 m j + m v

Comparing

4.16 m i + 2.4 m j + m v = 6m i

v = 1.84 i - 2.4 j

Magnitude of velocity

       [tex]v=\sqrt{1.84^2+(-2.4)^2}=3.02m/s[/tex]

Direction,  

        [tex]\theta =tan^{-1}\left ( \frac{-2.4}{1.8}\right )=-53.13^0[/tex]     

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

To find the final velocity of the first puck, apply conservation of momentum and solve for v'. The final velocity of the first puck is 4.8 m/s.

To find the final velocity of the first puck, we can apply the principle of conservation of linear momentum. The initial momentum of the system is given by m1v1 + m2v2, where m1 is the mass of the first puck, m2 is the mass of the second puck, v1 is the initial velocity of the first puck, and v2 is the initial velocity of the second puck.

Since the collision is elastic, the total momentum before and after the collision is conserved. So, m1v1 + m2v2 = (m1 + m2)v', where v' is the final velocity of both pucks after the collision. We can plug in the given values to find the final velocity of the first puck.

Let's solve for v': m1v1 + m2v2 = (m1 + m2)v' => (m1)(6.0 m/s) + (m2)(0 m/s) = (m1 + m2)(4.8 m/s).

From the given information, we know that the two pucks are identical, so m1 = m2. Substituting this into the equation, we get (m)(6.0 m/s) = 2(m)(4.8 m/s), where m is the mass of each puck. Simplifying this equation, we find the mass cancels out, leaving 6.0 m/s = 2(4.8 m/s). Solving for v', we find v' = 4.8 m/s.

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Answer:

The top of the ladder go down by 0.15 m.

Explanation:

Here we have right angled triangle.

Hypotenuse = 4.6 m

Bottom angle = 66º

Length from ladder bottom to wall bottom = 4.6 cos66 = 1.87 m

Length from ladder top to wall bottom = 4.6 sin66 = 4.20 m

New length from ladder bottom to wall bottom =  1.87 + 0.31 = 2.18 m

By Pythagoras theorem

New length from ladder top to wall bottom is given by

          [tex]\sqrt{4.6^2-2.18^2}=4.05m[/tex]

Distance  the top of the ladder go down = 4.20 - 4.05 = 0.15 m

Given:

initial angular speed, [tex]\omega _{i}[/tex] = 21.5 rad/s

final angular speed, [tex]\omega _{f}[/tex] = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

[tex]\alpha = \frac{\omega_{f} - \omega _{i}}{t}[/tex]

Now, putting the given values in the above formula:

[tex]\alpha = \frac{28.0 - 21.5}{3.50}[/tex]

[tex]\alpha = 1.86 m/s^{2}[/tex]

Therefore, angular acceleration is:

[tex]\alpha = 1.86 m/s^{2}[/tex]

Answer:

a) Tmax=827.12N

b) Tmin = 626.22N

c) 776.895 N

Explanation:

Given:

Mass of wrecking ball M1=63.9 Kg

Mass of the chain M2=20.5 Kg

acceleration due to gravity, g=9.8m/s²

Now,

(a)The Maximum Tension generated in the chain,

    Tmax=(M1+M2)×g)

    Tmax=(M1+M2)×(9.8 m/s²)

    Tmax=(63.9+ 20.5)×(9.8 m/s²)  

    Tmax=827.12N

(b) The Minimum Tension Tmin will be due to the weight of the wrecking ball only

Mathematically,

Tmin=weight of the wrecking ball

Tmin = 63.9kg×9.8m/s²

Tmin = 626.22N

(c)Now. the tension at 3/4 from the bottom of the chain

In this part we have to use only 75% of the chain i.e the weight acting below the point of consideration

thus, the tension will be produced by the weight of the 3/4 part of the chain and the wrecking ball

Therefore, the weight of the 3/4 part of the chain = [tex]\frac{3}{4}\times 20.5\times 9.8 N[/tex]

= 150.675 N

Hence, the tension at a point 3/4 of the way up from the bottom of the chain will be  = 150.675 N + (63.9×9.8) N = 776.895 N

Answer:

The answers are: a)Fcp=0,23N b)As Fcp=0,93N, it increases 4 times when the speedis doubled

Explanation:

Let´s explain what´s the centripetal force about: It´s the force applied over an object moving on a curvilinear path. This force is directed to the rotation center.

This definition is described this way:

[tex]Fcp*r=m*V^2[/tex] where:

Fcp is the Centripetal Force

r is the horizontal circle radius

m is the ball mass

V is the tangencial speed, same as the rotation speed

w is the angular speed

Here we need to note that the information we have talks about 1cycle/0,639s (One cycle per 0,639s). We need to express this in terms of radians/seconds. To do it we define that 1cycle is equal to 2pi, so we can find the angular speed this way:

[tex]w=(1cycle/0,639s)*(2pi/1cycle)[/tex]

So the angular speed is [tex]w=9,83rad/s[/tex]

Now that we have this information, we can find the tangencial speed, which will be the relation between the angular speed and the circle radius, this way:

[tex]V=w*r[/tex] so the tangencial speed is:

[tex]V=(9,83rad/s)*(0,149m)[/tex]

[tex]V=1,5m/s[/tex]

Now we have all the information to find the Centripetal Force:

[tex]Fcp=(m*V^2)/(r)[/tex]

[tex]Fcp=((0,0154kg)*(1,5m/s)^2)/(0,149m)[/tex]

a) So the Centripetal Force is: [tex]Fcp=0,23N[/tex]  

b) If the tangencial speed is doubled, its new value will be 3m/s. replacing this information we will get the new Centripetal Force is:

[tex]Fcp=((0,0154kg)*(3m/s)^2)/(0,149m)[/tex]

The Centripetal Force is: [tex]Fcp=0,93N[/tex]

Here we can see that if the speed is doubled, the Centripetal Force will increase four times.

Answer:

The wavelength of x-ray is 0.272 nm.

Explanation:

Inter planer spacing, d = 0.399 nm = 3.99 × 10⁻¹⁰ m

First order constructive interference occurs when the beam makes an angle of 20° with the planes. We need to find the wavelength of the x-rays. The condition for constructive interference is given by :

[tex]n\lambda=2d\ sin\theta[/tex]

Here, n = 1

[tex]\lambda=2\times 3.99\times 10^{-10} m\ sin(20)[/tex]

[tex]\lambda=2.72\times 10^{-10}\ m[/tex]

[tex]\lambda=0.272\ nm[/tex]

So, the  wavelength of the x-rays is 0.272 nm. Hence, this is the required solution.

Answer:

Option d is the correct option Kinetic energy is minimum while as potential energy is maximum

Explanation:

At the top most point of the flight since it cannot reach any further up in the vertical direction thus the potential energy at this position shall be maximum. Now since the total energy of the projectile is conserved so the remaining kinetic energy shall be minimum at that point so as the sum of the kinetic and potential energies remain constant.

Answer:

The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]

Explanation:

Given that,

Mass of object = 6.22 kg

Distance = 1.02 m

We need to calculate the number of electron

Using formula of electric force

[tex]F_{e}=\dfrac{k(q)^2}{r^2}[/tex]....(I)

We know that,

[tex]q = Ne[/tex]

Put the value of q in equation (I)

[tex]F_{e}=\dfrac{k(Ne)^2}{r^2}[/tex].....(II)

Using gravitational force

[tex]F_{G}=\dfrac{Gm^2}{r^2}[/tex].....(III)

Equating equation (II) and (III)

[tex]F_{e}=F_{G}[/tex]

[tex]\dfrac{k(Ne)^2}{r^2}=\dfrac{Gm^2}{r^2}[/tex]

[tex]N=\sqrt{\dfrac{G}{k}}\times\dfrac{m}{e}[/tex]....(IV)

Put the value in the equation(IV)

[tex]N=\sqrt{\dfrac{6.67\times10^{-11}}{9\times10^{9}}}\times\dfrac{6.22}{1.6\times10^{-19}}[/tex]

[tex]N=3.35\times10^{9}[/tex]

Hence, The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]

Answer:The sled slides 16.875m before rest.

Explanation:

[tex]a=\frac{F}{m} =\frac{12N}{20kg}[/tex]

a=0.6 m/s²

[tex]Vf=0=Vi-a.t[/tex]

[tex]t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg[/tex]

[tex]d= Vi.t - \frac{a.t^{2}}{2}[/tex]

[tex]d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m[/tex]

Explanation:

The mass of freight car, m₁ = 30,000 kg

Velocity of freight car, u₁ = 0.85 m/s

Mass of hopper, m₂ = 110,000 kg

(a) Let v is the final velocity of the loaded freight car. Initial momentum of the car before the dump, [tex]p_i=30000\ kg\times 0.850\ m/s=25500\ kg-m/s[/tex]

Final momentum, [tex]p_f=(30000\ kg+110000\ kg)v=140000\ v[/tex]

According to the conservation of momentum,

initial momentum = final momentum

25500 kg-m/s = 140000 v

v = 0.182 m/s

So, the  final velocity of the loaded freight car is 0.182 m/s.

(b) Initial kinetic energy, [tex]k_i=\dfrac{1}{2}\times 30000\ kg\times (0.850\ m/s)^2=10837.5\ J[/tex]

Final kinetic energy, [tex]k_f=\dfrac{1}{2}(m_1+m_2)v^2[/tex]

[tex]k_f=\dfrac{1}{2}\times (30000\ kg+110000\ kg)\times (0.182\ m/s)^2=2318.68\ J[/tex]

So, loss in kinetic energy, [tex]\Delta k=k_f-k_i[/tex]

[tex]\Delta k=2318.68\ J-10837.5\ J=-8518.82\ J[/tex]

So, 8518.82 J of kinetic energy is lost after collision. Hence, this is the required solution.

Using the conservation of momentum, we find the final velocity of the loaded freight car to be 0.182 m/s. By comparing the initial and final kinetic energies, we find that 8,459.89 Joules of kinetic energy is lost during the process.

This problem is a classic example of conservation of momentum, and to a lesser extent, conservation of kinetic energy. Firstly, we calculate the initial momentum of the system. Momentum is mass multiplied by velocity, so the initial momentum of the freight car is 30,000 kg * 0.850 m/s = 25,500 kg*m/s. After the scrap metal is dumped into the freight car, the total mass of the system is now 30,000 kg + 110,000 kg = 140,000 kg. Because momentum must be conserved, we calculate the final velocity of the freight car as the total momentum divided by the total mass, 25,500 kg*m/s / 140,000 kg = 0.182 m/s.

Secondly, before the scrap metal is added, the initial kinetic energy of the freight car is 0.5 * 30,000 kg * (0.850 m/s)² = 10,781.25 Joules. After the metal is added, the final kinetic energy is 0.5 * 140,000 kg * (0.182 m/s)² = 2,321.36 Joules. The change in kinetic energy can then be obtained by subtracting the final kinetic energy from the initial one, which gives us 8,459.89 Joules as the amount of kinetic energy that is lost.

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Answer:

Total number of different selections  = 455

Explanation:

Number of selections from n objects if we select r things [tex]=^nC_r[/tex]            

Here we need to find number off selections of 3 apples from 15 apples.

Number off selections of 3 apples from 15 apples

                    [tex]=^{15}C_3=\frac{15\times 14\times 13}{1\times 2\times 3}=455[/tex]

Total number of different selections  = 455

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

[tex]H = 4.905 m[/tex]

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

[tex]a = g = 9.81 m/s^2[/tex]

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

[tex]v_f = 0[/tex]

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

[tex]\Delta y = 0 = v_y t + \frac{1}{2} at^2[/tex]

[tex]0 = v_y (2) - \frac{1}{2}(9.81)(2^2)[/tex]

[tex]v_y = 9.81 m/s[/tex]

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we have

[tex]0 - (9.81^2) = 2(-9.81)H[/tex]

[tex]H = 4.905 m[/tex]

The acceleration of the ball is -9.8 m/s², the velocity of the ball at max height is 0 m/s, the initial velocity is 9.8 m/s and the max height is 4.9 m.

The questions refer to the motion of a tennis ball under gravity. (a) The acceleration of the ball while it is in flight is -9.8 m/s², due to Earth's gravity. This will be the case throughout the ball's flight, regardless of whether it's moving up or down. (b) The velocity of the ball when it reaches its maximum height is 0 m/s, as the ball stops moving vertically for an instant before it starts coming down. (c) We can calculate the initial velocity using the equation v = u + at. Here, v is final velocity, u is initial velocity, a is acceleration and t is time. As v = 0 m/s at max height, a = -9.8 m/s² and t = 1 s (time taken to reach max height, half of total time), we find u to be 9.8 m/s. (d) The maximum height can be calculated using the equation h = ut + 0.5at². Here, h is height, u is initial velocity, t is time and a is acceleration. Subsituting u = 9.8 m/s, a = -9.8 m/s², t = 1 s, we get h to be 4.9 m.

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Answer:

focal length of eye lens is 3.28 cm

Explanation:

As we know that the magnification of the microscope is given by the formula

[tex]M = \frac{L}{f_o}\frac{D}{f_e}[/tex]

now we will have

[tex]L = 47.8 cm[/tex]

D = 25 cm

M = 2000

[tex]f_o = 1.82 mm = 0.182 cm[/tex]

now we need to solve above equation to find focal length of eye lens

so here we will have

[tex]f_e = \frac{L}{f_o}\frac{D}{M}[/tex]

[tex]f_e = \frac{47.8}{0.182}\frac{25}[2000}[/tex]

[tex]f_e = 3.28 cm[/tex]

Answer:1200

Explanation:

Given data

Upper Temprature[tex]\left ( T_H\right )=240^{\circ}\approx 513[/tex]

Lower Temprature [tex]\left ( T_L\right )=20^{\circ}\approx 293[/tex]

Engine power ouput[tex]\left ( W\right )=910 W[/tex]

Efficiency of carnot cycle is given by

[tex]\eta =1-\frac{T_L}{T_H}[/tex]

[tex]\eta =\frac{W_s}{Q_s}[/tex]

[tex]1-\frac{293}{513}=\frac{910}{Q_s}[/tex]

[tex]Q_s=2121.954 W[/tex]

[tex]Q_r=1211.954 W[/tex]

rounding off to two significant figures

[tex]Q_r=1200 W[/tex]

Answer:

1200

Explanation:

The amount of work required to pump all of the water over the side of the pool is 471,238.9 Newton Meters or Joules.

The work done to pump water out of a pool involves the concept of physics specifically related to potential energy, gravity, and volume. The work done to move a certain volume of water is given by the formula: Work = Weight x Height.

First, we need to find the volume of the water in the pool. The pool's shape resembles a cylinder, and the volume is given by the formula for a cylinder: Volume = pi × (diameter/2)²× height. Given a diameter of 8 meters and a height of 1.5 meters, the volume to be moved is pi * (8/2)² × 1.5 = 48pi cubic meters.

The weight of this water can be calculated by multiplying its volume by its density. The density of water is 1000 kg/m³. Therefore, the weight of the water is Volume x Density x g (acceleration due to gravity), which is 48pi × 1000 × 9.8 = 471,238.9 kg×m²/s² or Newton Meter (Nm) which is the unit of work.

So, the amount of work required to pump all the water over the side is 471,238.9 Nm or Joules (J).

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The work required to pump all the water over the side of a circular swimming pool of diameter 8 meters and water depth 1.5 meters is calculated using the formula Work = Force x Distance. The Force required is given by the weight of the water, which depends on its volume and density. The result is about 353,429.16 Joules.

To calculate the work required to pump water out of a swimming pool, we can use the formula for work: Work = Force x Distance. The force required is equal to the weight of the water which depends on the volume of the water and its density.

First, let's calculate the volume of the water in the pool. Given that it's a circular pool with a diameter of 8 meters, the radius is 4 meters. The depth of the water is 1.5 meters. So, volume (V) = πr²h = π×(4m)²×(1.5m) = 24π cubic meters.

The density of water is 1000 kg/m³. Therefore, the weight of water = Volume x Density x Gravity = 24π m³ ×1000 kg/m³ × 9.8 m/s² = 235619.44 kg×m/s², or Newtons. This is the force we need to overcome to lift the water.

The distance that we want to lift this mass is the depth of the pool, assuming we are pumping the water over the side of the pool which is 1.5 meters high. So, Work = Force x Distance = 235619.44 N×1.5m = 353429.16 Joules.

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Answer:

The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].

Explanation:

Given that,

Mass [tex]M=6.65\times10^{-27}\ kg[/tex]

Temperature = 20.0°C

We need to calculate the root-mean square speed

Using formula of root mean square speed

[tex]v_{rms}=\sqrt{\dfrac{3kTN_{A}}{M}}[/tex]

Where, N = Avogadro number

M = Molar mass

T = Temperature

k = Boltzmann constant

Put the value into the formula

[tex]v_{rms}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times293\times6.022\times10^{23}}{4\times10^{-3}}}[/tex]

[tex]v_{rms}=1351.37\ m/s[/tex]

We need to calculate the de Broglie wavelength

Using formula of de Broglie wavelength

[tex]P=\dfrac{h}{\lambda}[/tex]

[tex]mv=\dfrac{h}{\lambda}[/tex]

[tex]\lambda=\dfrac{6.626\times10^{-34}}{6.65\times10^{-27}\times1351.37}[/tex]

[tex]\lambda=7.373\times10^{-11}\ m[/tex]

Hence, The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].

Final answer:

The de Broglie wavelength of helium atoms moving at the root-mean-square speed is approximately 4.779 × 10^-10 meters.

Explanation:

To find the de Broglie wavelength of helium atoms moving at the root-mean-square speed, we can use the equation:

λ = h / (m * v)

where λ is the de Broglie wavelength, h is Planck's constant (6.626 × 10-34 J · s), m is the mass of the helium atom (6.65 × 10-27 kg), and v is the root-mean-square speed.

The root-mean-square speed of helium atoms at room temperature can be found using the equation:

v = √(3 * k * T / m)

where k is the Boltzmann constant (1.38 × 10-23 J/K) and T is the temperature in Kelvin (20.0 + 273 = 293 K).

Plugging the values into the equations and solving for λ:

λ = (6.626 × 10-34 J · s) / (6.65 × 10-27 kg * √(3 * 1.38 × 10-23 J/K * 293 K / 6.65 × 10-27 kg))

λ = 4.779 × 10-10 m

Therefore, the de Broglie wavelength of the helium atoms moving at the root-mean-square speed is approximately 4.779 × 10-10 meters.

Answer:

6.97 m/s, 344 degree

Explanation:

mass of car, m = 900 kg, uc = 15 m/s, vc = 5 m/s, θ = 40 degree

mass of truck, M = 1500 kg, uT = 0, vT = ?, Φ = ?

Here, vT be the velocity of truck after collision and Φ its direction above x axis.

Use conservation of momentum in X axis

900 x 15 + 1500 x 0 = 900 x 5 Cos 40 + 1500 x vT Cos Φ

13500 - 3447.2 = 1500 vT CosΦ

vT CosΦ = 6.7 .....(1)

Use conservation of momentum in y axis

0 + 0 = 900 x 5 Sin 40 + 1500 vT SinΦ

vT SinΦ = - 1.928 .....(2)

Squarring both the equations and then add

vT^2 = 6.7^2 + (-1.928)^2

vT = 6.97 m/s

Dividing equation 2 by 1

tan Φ = - 1.928 / 6.7

Φ = - 16 degree

Angle from + X axis = 360 - 16 = 344 degree