Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differential equation dv m mg bv dt   . Here, m is the mass of the object, g is the acceleration due to gravity, and b > 0 is a constant. Consider an object that has a mass of 100 kilograms and an initial velocity of 10 m/sec (that is, v(0) = 10). If we take g to be 9.8 m/sec2 and b to be 5 kg/sec, find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

Answer:

length of one side of the coil is 0.1083 m

Explanation:

given data

no of turn = 269

coil rotates = 113 rad/s

magnetic field = 0.222-T

peak output = 79.2 V

solution

we will apply here formula for maximum emf induced in a coil that is

εo = N × A × B × ω    ........................1

and here Area A = l²

so

l = [tex]\sqrt{A}[/tex]

put here value of A from equation 1

I = [tex]\sqrt{\frac{\epsilon _o}{N\times B\times \omega} }[/tex]  

put here value and we will get

l =  [tex]\sqrt{\frac{79.2}{269\times 0.222\times 113}}[/tex]

l = 0.1083 m

Answer:

Explanation:

number of turns, n = 269

angular velocity, ω = 113 rad/s

magnetic field, B = 0.222 T

Voltage, V = 79.2 V

Let the area of the coil is A and the side of the square is s.

The maximum emf generated by the coil is

V = n x B x A x ω

79.2 = 269 x 0.222 x A x 113

A = 0.01174 m²

so, s x s = 0.01174

s = 0.108 m

Thus, the side of the square coil is 0.108 m.

The above is not complete. Couldn't find the it online but i have found a similar question which would help you solve the above  

Answer:

Explanation:

From the question we are given that

Mass = 70kg

Angle of elevation = 40°

Length of  stretch = 3.00 m

Height of performer above floor = Height of the net into the into which he is shot

  Time traveled = 2.14 s

Distance traveled = 26.8 m

First obtain the horizontal velocity = [tex]\frac{Distance}{time} =\frac{26.8}{2.14} = 12.523 \ m/s[/tex]

To obtain the initial we would divide the final velocity by the cosine of the angle of elevation

     Initial velocity = [tex]\frac{12.523}{cos\ 40} = 16.348m/s[/tex]

Next is to obtain the initial kinetic energy

This is equal to = [tex]\frac{1}{2} mv^2= \frac{1}{2} *70*16.348^2 = 9354 \ Joules[/tex]

Looking at the diagram in the second uploaded the cannon raised the performer before releasing him is [tex]30sin(40) = 19284 \ m[/tex]

So the potential energy given by the cannon is = mgh [tex]=70 *9.80*1.9284 = 1323 \ Joules[/tex]

 Hence the total energy the band gives the performer i.e the total energy stored in the band is  = 9354 +1323 = 10677 Joules

        To obtain the Spring constant we would use the stored energy formula

       i.e  Stored Energy [tex]=\frac{1}{2}kx^2[/tex]

And we have calculated the stored energy as 10677

      Substituting

               [tex]10677 = \frac{1}{2} *k * 3.00^2[/tex]

                     [tex]k = 2373 \ N/m[/tex]

The circus performer's stunt includes aspects of elastic potential energy and projectile motion. The performer is launched due to the elastic potential energy in the cannon bands and then follows the laws of projectile motion. The performer's mass does not influence the time of flight.

This problem demonstrates the principles of projectile motion and elastic potential energy applied to a real-world situation. The 70-kg circus performer is propelled out of the cannon because of the elastic potential energy stored in the stretched bands. Once launched, the performer's motion is an example of projectile motion.

In projectile motion, the horizontal and vertical motions are independent of each other. Therefore, it's possible to find the horizontal and vertical components of the performer's velocity separately and use them to construct the overall motion. The angle of 37° is used to calculate these components.

While the performer is in flight, the vertical motion is under constant acceleration due to gravity, while the horizontal motion has a constant velocity because air resistance is negligible. This allows us to calculate various aspects of the motion, such as the time it takes for the performer to land in the net.

Lastly, the performer's mass does not influence the time of flight in a vacuum. This illustrates an important principle of physics known as the independence of motion.

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The distance between the orbits of two satellites is 7.97 m.

Explanation:

Johannes Kepler was the first to propose three laws for the planetary motion. According to him, the orbits in which planets are rotating are elliptical in nature and Sun is at the focus of the ellipse. Also the area of sweeping is same.

So based on these three assumptions, Kepler postulated three laws. One among them is Kepler's third law of planetary motion. According to the third law, the square of the time taken by a planet to cover a specified region is directly proportional to the cube of the major elliptical axis or the radius of the ellipse.

So, [tex]T^{2} = r^{3}[/tex]

Thus, for the geosynchornous satellite, as the time taken is 24 hours, then the radius or the major axis of this satellite is

[tex](24)^{2}= r^{3} \\(2*2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{2*2*2*3} =(2)^{2} * (6)^{\frac{2}{3} } =13.21 m[/tex]

Similarly, for the another satellite orbiting in time period of 12 hours, the major axis of this satellite is

[tex](12)^{2}= r^{3} \\(2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{12} =5.24 m[/tex]

So, the difference between the two radius will give the distance between the two orbits, 13.21-5.24 = 7.97 m.

So the distance between the orbits of two satellites is 7.97 m.

Answer:

The speed of woman is 20.39 mi/h at and angle of 11.3 degrees wrt the surface of the water.

Explanation:

Given that,

Speed of women due west, [tex]v_w=4\ mi/h[/tex]

Speed of women due north, [tex]v_n=20\ mi/h[/tex]

We need to find the speed and direction of the woman relative to the surface of the water. The resultant speed is given by :

[tex]v=\sqrt{v_w^2+v_n^2}[/tex]

[tex]v=\sqrt{4^2+20^2}[/tex]

[tex]v=20.39\ mi/h[/tex]

Let [tex]\theta[/tex] is the direction of speed. It is given by :

[tex]\tan\theta=\dfrac{4}{20}[/tex]

[tex]\theta=11.3^{\circ}[/tex]

So, the speed of woman is 20.39 mi/h at and angle of 11.3 degrees wrt the surface of the water.

The speed of the woman relative to the surface of the water is approximately 20.4 mi/h, and she is moving approximately 11.3° south of west.

To find the speed and direction of the woman relative to the surface of the water, we can use vector addition. Since the woman is walking due west, her velocity vector is 4 mi/h west. The ship's velocity vector is 20 mi/h north. By adding these two vectors using vector addition, we can find the resultant velocity vector. The magnitude of the resultant velocity represents the speed of the woman relative to the surface of the water, while the direction represents the direction she is moving relative to the surface of the water.

Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:

Resultant Velocity: √(4^2 + 20^2) = √(16 + 400) = √416 ≈ 20.4 mi/h

Using trigonometry, we can find the direction of the resultant velocity:

Direction: tan^(-1)(4/20) ≈ 11.3° south of west

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Note: There is no image to take as a reference, so I'm assuming F2 directed to the right and F1 to the left, and F2=2F1

Answer:

[tex]\displaystyle a=\frac{F}{3M}[/tex]

to the right

Explanation:

Net Force

When several forces are applied to a particle or a system of particles, the net force is the sum of them all, considering each force as a vector. As for the second Newton's law, the total force equals the product of the mass by the acceleration of the system:

[tex]\vec F_n=m\cdot \vec a[/tex]

If the net force is zero, then the system of particles keeps at rest or at a constant velocity.

The system of particles described in the question consists of two objects of masses m1=M and m2, where

[tex]m_2=2m_1=2M[/tex]

Two forces F1=F and F2 act individually on each object in opposite directions and

[tex]F_2=2F_1=2F[/tex]

We don't get to see any image to know where the forces are applied to, so we'll assume F2 to the right and F1 to the left.

The net force of the system of particles is

[tex]F_n=2F-F=F[/tex]

The mass of the system is

[tex]m_t=m_1+m_2=3M[/tex]

Thus, the acceleration of the center of mass of the system is

[tex]\displaystyle a=\frac{F}{3M}[/tex]

Since F2 is greater than F1, the direction of the acceleration is to the right.

Note: If the forces were opposite than assumed, the acceleration would be to the left

Answer:

-26.1111

Explanation:

Hello!

The formula for converting Fahrenheit to Celsius can be written as follows:

C = [tex]\frac{(F-32)(5)}{9}[/tex]

Now let’s insert the value provided for Fahrenheit:

C = [tex]\frac{((-15)-32)(5)}{9}[/tex]

Now simplify the numerator:

C = [tex]\frac{(-235)}{9}[/tex]

Now divide:

C = -26.111

We have now proven that -15 degrees Fahrenheit is equal to -26.111 degrees Celsius.

I hope this helps!

Explanation:

Lets consider

Circumference of orbit = T

as it is mentioned in the question that a satellite is in orbit that is very close to the surface of planet. so

circumference of orbit = circumference of planet

Time period = T

radius of planet = R

orbital velocity = V

gravitational constant = G

mass of planet = m

Solution:

Time period for a uniform circular motion of orbit is,

T = [tex]\frac{2\pi R}{V}[/tex]

[tex]T = \frac{2\pi R }{\sqrt{\frac{GM}{R} } }[/tex]

[tex]T= 2\pi \sqrt{(\frac{R^3}{GM} )}[/tex]

[tex]M = \frac{4}{3} \pi R^{3}p[/tex]

where p = density

[tex]T = 2\pi \sqrt{\frac{R^{3} }{G\frac{4}{3}\pi R^{3} p } }[/tex]

[tex]T = \sqrt{\frac{3\pi }{Gp} }[/tex]

T = 2.17 hours = 7812 sec

(7812)² = [( 3×3.14)/6.67×[tex]10^{-11}[/tex]×ρ)]

ρ = 6.28/6.67×[tex]10^{-11}[/tex]×6.10×[tex]10^{-7}[/tex]

ρ = 6.28/40.687×[tex]10^{-18}[/tex]

ρ = 0.1543×[tex]10^{18}[/tex]kg/m³

ρ = 15.43×[tex]10^{16}[/tex]kg/m³

Final answer:

Calculating a planet's density based on a satellite's orbital period involves physics concepts like gravitational laws and Kepler's third law, applying them to given data about the satellite's orbit. Direct calculation requires additional information that is not provided in the question.

Explanation:

One can use Kepler's third law in combination with the formula for gravitational force to find the planet's density based on the period of a satellite in a circular orbit close to the planet's surface.

The key relationship to use is T^2 = (4π^2/GM)r^3, where T is the period of the satellite, G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit, which is approximately the radius of the planet if the satellite is very close to the surface. However, to find the density (ρ) of the planet, we use the formula ρ = M/(4/3πr^3).

Given that we do not have the mass (M) or radius (r) of the planet explicitly, we cannot directly calculate the density without additional information such as the gravitational constant (G) and the exact radius of the planet's orbit.

The solution involves using the period (T) given to manipulate these equations to express mass in terms of the period and then apply it to the density formula. This complex physics problem requires an understanding of orbital mechanics and gravitational laws.

Answer:

False

Explanation:

The potential difference between two points exists solely because of a source charge and depends on the source charge distribution. For a potential energy to exist, there must be a system of two or more charges. The potential energy belongs to the system and changes only if a charge is moved relatively to the rest of the system.

The tip of one prong of a tuning fork undergoes SHM of frequency 1000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the

(a) maximum acceleration,

(b) maximum velocity,

(c) acceleration at tip displacement 0.20 mm, and

(d) velocity at tip displacement 0.20 mm?

(a) 15795.5m/s²

(b) 2.5m/s

(c) 7897.7 m/s²

(d) 2.2m/s

The displacement, y, of a body undergoing simple harmonic motion (SHM) is given by

y = A sin (ωt + φ)             ------------------(i)

Where;

A = maximum displacement or amplitude of the body

ω = angular frequency of the body

t = time taken for the displacement

φ = phase constant

The velocity, v, of the body can be found by differentiating equation (i) as follows;

v = Aω cos (ωt + φ)          ------------------(ii)

Where;

Aω = maximum velocity or amplitude of the velocity of the body

Also, the acceleration of the body can be found by differentiating equation (ii) as follows;

a = -Aω² sin(ωt + φ)                  --------------------(iii)

Where;

-Aω² = maximum acceleration or amplitude of the acceleration of the body

(a) From equation (iii), the magnitude of the maximum acceleration [tex]a_{max}[/tex] is given by;

[tex]a_{max}[/tex] = Aω²           ----------------(iv)

Where;

A = amplitude = 0.40mm = 0.00040m

ω = 2 π f            [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]

=> ω = 2 x 3.142 x 1000 = 6284 rad/s

Substitute these values into equation (iv) as follows;

[tex]a_{max}[/tex] = 0.00040 x 6284² = 15795.5m/s²

Therefore, the magnitude of the maximum acceleration is 15795.5m/s²

========================================================

(b) From equation (ii), the magnitude of the maximum velocity [tex]v_{max}[/tex], is given by;

[tex]v_{max}[/tex] = Aω          ----------------(v)

Where;

A = amplitude = 0.40mm = 0.00040m

ω = 2 π f            [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]

=> ω = 2 x 3.142 x 1000 = 6284 rad/s

Substitute these values into equation (v) as follows;

[tex]v_{max}[/tex] = 0.00040 x 6284 = 15795.5m/s²

Therefore, the magnitude of the maximum velocity is 2.5m/s

========================================================

(c) Comparing equations (i) and (iii),  equation (iii) can be written as;

a = -ω² y            -------------------(vi)

Therefore, to get the acceleration at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m and ω = 6284rad/s into equation (vi) as follows;

a = - 6284² x 0.00020

a = - 7897.7 m/s²

Therefore, the magnitude of the acceleration at the tip displacement is 7897.7 m/s²

========================================================

(d) Recall that;

sin²θ + cos²θ = 1

=> cos²θ = 1 - sin²θ

=> cosθ = √(1 - sin²θ )

=> cos (ωt + φ)  = √(1 - sin² (ωt + φ))

Substitute this value into equation (ii) as follows;

v = Aω √(1 - sin² (ωt + φ))

v = ω√(A² - A²sin² (ωt + φ))              

Now, comparing the equation above and equation (i), the equation above can be written as;

v = ω√(A² - y²)           -------------(vii)

Therefore, to get the velocity at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m, ω = 6284rad/s and A = 0.00040m into equation (vii) as follows;

v = 6284√(0.00040² - 0.00020²)

v = 6284√(0.00000060)

v = 2.2m/s

Therefore, the magnitude of the velocity at the tip displacement is 2.2 m/s

Answer:

λ= 5.24 × 10 ⁻² nC/cm

Explanation:

Given:

distance r = 4.10 cm = 0.041 m

Electric field intensity E = 2300 N/C

K = 9 x 10 ⁹ Nm²/C

To find λ = linear charge density = ?

Sol:

we know that E= 2Kλ / r

⇒ λ = -E r/2K         (-ve sign show the direction toward the wire)

λ = (- 2300 N/C × 0.041 m) / 2 ×  9 x 10 ⁹ Nm²/C

λ = 5.24 × 10 ⁻⁹ C/m

λ = 5.24 nC/m = 5.24 nC/100 cm

λ= 5.24 × 10 ⁻² nC/cm

Answer:

(a) 11.3 L

(b) 10 M

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

Luminosity of Star =  11.3 L

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

Mass of Star = 10 M

Answer:

1. 11

2. 10

Explanation:

1. Using the formula L=M^3.5, and the fact that we were given the Mass,

L=2^3.5

L=11.3137085

L=11

Thus, the luminosity of the star would be 11 solar lumen

2. If we invert the formula, M=L^(2/7)

M=3160^(2/7)

M=9.99794159

M=10

Thus, the mass of the starwould be 10 solar masses.

The time the projectile (or 'turd') takes to hit the ground can be calculated using the formula (T = 2u*sin(θ)/g). This formula uses the initial velocity, the projection angle, and gravitational acceleration.

This question relates to the physics concept of projectile motion. Given the launch speed, in this case, 72 m/s, and the angle of projection, 12 degrees, you can calculate the time of flight, i.e., the time taken by the projectile (in this humorous context, a 'turd') to hit the ground.

The time of flight (T) for a projectile can be calculated using the formula: T = 2u*sin(θ)/g, where u is the initial velocity, θ is the projection angle, and g is the gravitational acceleration (approximately 9.81 m/s² on Earth). Substituting the given values, we get T = 2*72*sin(12)/9.81, which gives the time after which the ground should be covered by a hypothetical 'portal' to intercept the turd just before it hits the ground.

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Answer:

The sphere returns back to the neutral.

Explanation:

When a positively charged rod is brought near the far end of the rod A a redistribution of the charges occurs in the rod A. The free electrons in the rod A are attracted towards the positively charged rod. This causes positive charges to be shifted towards the other end of the rod A. The sphere B is in contact with the positively charged end of rod A. So it also acquires a positive charge.

When the positively charged rod is removed from the faf end of rod A, the electrons return to their original state and the charges redistributes themselves and a neutral state is re-established .

Answer:

[tex]5.3\times 10^3 N/C[/tex]

Explanation:

We are given that

Distance between plates=d=2.2 cm=[tex]2.2\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

[tex]\sigma=47nC/m^2=47\times 10^{-9}C/m^2[/tex]

Using [tex]1 nC=10^{-9} C[/tex]

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

[tex]E_1=\frac{\sigma}{2\epsilon_0}[/tex]

[tex]E_2=\frac{\sigma}{2\epsilon_0}[/tex]

[tex]E=E_1+E_2[/tex]

[tex]E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}[/tex]

Where [tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]

Substitute the values

[tex]E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}[/tex]

[tex]E=5.3\times 10^3 N/C[/tex]

Hence, the magnitude of E in the region between the plates=[tex]5.3\times 10^3 N/C[/tex]

The magnitude of the electric field E in the region between two parallel conducting plates carrying opposite charges of equal magnitude can be calculated using the formula E = σ / ε0. Given the surface charge density σ of 47.0 nC/m2 and the permittivity of free space ε0 as 8.85 × 10-12 F/m, the electric field E results as 5.31 × 103 N/C.

The system mentioned in the question is known as a parallel-plate capacitor. It is important to understand that the electric field between two large parallel conducting plates carrying opposite charges of equal magnitude is perpendicular to the plates and is constant in both magnitude and direction. The electric field E between the plates of a capacitor can be calculated by the formula E = σ / ε0, where σ is the surface charge density on one plate, and ε0 is the permittivity of free space, which is approximately 8.85 × 10-12 F/m.

Given in the question, the surface charge density σ is 47.0 nC/m2. To convert nanocoulombs to coulombs, multiply by 10-9, so σ = 47.0 × 10-9 C/m2. Applying the values to the formula, we get E = (47.0 × 10-9 C/m2) / (8.85 × 10-12 F/m) = 5.31 × 103 N/C. Therefore, the electric field E in the region between the plates is 5.31 × 103 N/C.

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The emf induced in the loop is 0.38 V.

Explanation:

The emf induced in any coil is directly proportional to the rate of change in flux linked with the coil as stated by Faraday's law.

Flux can be obtained by the product of magnetic lines of force with cross sectional area. As flux is the measure of magnetic lines of force crossing a certain cross sectional area.

As here the region is considered as circular loop, so the area will be area of the circle.

Then, [tex]emf=\frac{d(phi)}{dt}[/tex]

phi = Magnetic field * Area

[tex]emf = \frac{d(BA)}{dt} =B*2\pi*r\frac{dr }{dt}[/tex]

[tex]emf = 0.6*2*3.14*0.15*0.68=0.38 V[/tex]

So, the emf induced in the loop is 0.38 V.

Answer:

The final velocity is 8 m/s and its direction is along the positive x-axis

Explanation:

Given :

Mass, m₁ = 2000 g = 2 kg

Mass, m₂ = 4000 g = 4 kg

Initial velocity of mass m₁, v₁ = 24i m/s

Initial velocity of mass m₂, v₂ = 0

According to the problem, after collision the two masses are stick together and moving with same velocity, that is, [tex]v_{1f}[/tex].

Applying conservation of momentum,

Momentum before collision = Momentum after collision

[tex]m_{1} v_{1} +m_{2} v_{2} =(m_{1}+m_{2}) v_{1f}[/tex]

Substitute the suitable values in the above equation.

[tex]2\times24 +4\times0 =(2+4}) v_{1f}[/tex]

[tex]v_{1f}=8i\ m/s[/tex]

Answer:

67.5

Explanation:

[tex]F= BILsin\theta[/tex] where F is the magnitude of the force, B is the magnetic field, I is current, [tex]\theta[/tex] is the angle of inclination, L is the length of the wire.

Substituting 0.9 T for magnetic field B, 10 m for length, 15 A for current I and 30 degrees for [tex]\theta[/tex] then

[tex]F=0.9\times 15\times 10\times sin 30^{\circ}=67.5[/tex]

Answer:

The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

[tex]E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]

[tex]E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]

Put the value into the formula

[tex]E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}[/tex]

[tex]E_{1}=1.0183\times10^{3}\ N/C[/tex]

Using formula of electric field again

[tex]E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]

Put the value into the formula

[tex]E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}[/tex]

[tex]E_{2}=-1.064\times10^{3}\ N/C[/tex]

We need to calculate the resultant electric field

Using formula of electric field

[tex]E=E_{1}+E_{2}[/tex]

Put the value into the formula

[tex]E=1.0183\times10^{3}-1.064\times10^{3}[/tex]

[tex]E=-0.045\times10^{3}\ N/C[/tex]

We need to calculate the force exerted on an electron

Using formula of electric field

[tex]E = \dfrac{F}{q}[/tex]

[tex]F=E\times q[/tex]

Put the value into the formula

[tex]F=-0.045\times10^{3}\times(-1.6\times10^{-19})[/tex]

[tex]F=7.2\times10^{-18}\ N[/tex]

Hence, The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]

Answer:

(A) Fo = 72 Hz

(B) The pipe is open at both ends

(C) The length of the pipe is 2.38m

This problem involves the application of the knowledge of standing waves in pipes.

Explanation:

The full solution can be found in the attachment below.

For pipes open at both ends the frequency of the pipe is given by

F = nFo = nv/2L where n = 1, 2, 3, 4.....

For pipes closed at one end the frequency of the pipe is given by

F = nFo = nv/4L where n = 1, 3, 5, 7...

The full solution can be found in the attachment below.

Answer:

  B = 9.16 10⁻²  T

Explanation:

The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel

       q v B = q E

       v = E / B

       B = E / v

Let's calculate

      B = 4.4 10⁵ / 4.8 10⁶

      B = 9.16 10⁻²  T

To select a speed of 4.8 x 10^6 m/s in a velocity selector, a magnetic field strength of 9.17 x 10^-2 T is needed.

A velocity selector in a mass spectrometer uses a combination of electric field and magnetic field to select charged particles based on their speed. The electric field and magnetic field exert forces on the particles in opposite directions, allowing only particles with a specific speed to pass through unaffected. To determine the magnetic field strength needed to select a speed of 4.8 x 106 m/s, we can use the equation qvB = qE, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and E is the electric field strength.

Given the electric field strength of 4.4 x 105 V/m, we can rearrange the equation to solve for B:

B = E/v = (4.4 x 105 V/m) / (4.8 x 106 m/s) = 9.17 x 10-2 T

Therefore, the magnetic field strength needed to select a speed of 4.8 x 106 m/s is 9.17 x 10-2 Tesla.

Answer:

(a) the current in the coil is 12.03 A

(b) the maximum magnitude of the torque is 0.0854 N.m

Explanation:

Given;

number of turns, N = 185

radius of the coil, r = 1.6 cm = 0.016 m

magnetic dipole moment, μ = 1.79 A·m²

Part (a) current in the coil

μ = NIA

Where;

I is the current in the coil

A is the of the coil = πr² = π(0.016)² = 0.000804 m²

I = μ / (NA)

I = 1.79 / (185 x 0.000804)

I = 1.79 / 0.14874

I = 12.03 A

Part (b) the maximum magnitude of the torque

τ = μB

Where;

τ is the maximum magnitude of the torque

B is the magnetic field strength = 47.7 mT

τ = 1.79 x 0.0477 = 0.0854 N.m

Given Information:

Number of turns = N = 185 turns

Radius of circular coil = r = 1.60 cm = 0.0160 m

Magnetic field = B =  47.7 mT

Magnetic dipole moment = µ =1.79 A.m

Required Information:

(a) Current = I = ?

(b) Maximum Torque = τ = ?

Answer:

(a) Current = 12.03 A

(b) Maximum Torque = 0.0853 N.m

Explanation:

(a) The magnetic dipole moment µ is given by

µ = NIAsin(θ)

I = µ/NAsin(θ)

Where µ is the magnetic dipole moment, N is the number of turns, I is the current flowing through the circular loop, A is the area of circular loop and is given by

A = πr²

A = π(0.0160)²

A = 0.000804 m²

I = 1.79/185*0.000804*sin(90)

I = 12.03 A

(b) The toque τ is given by

τ = NIABsin(θ)

The maximum torque occurs at θ = 90°

τ = 185*12.03*0.00804*0.0477*sin(90°)

τ = 0.0853 N.m

Full Question:

Consider an automobile with a mass of 5510 lbm braking to a stop from a speed of 60.0 mph.

How much energy (Btu) is dissipated as heat by the friction of the braking process?

______Entry field with incorrect answer Btu

Suppose that throughout the United States, 350.0 x 10^6 such braking processes occur in the course of a given day.   Calculate the average rate (megawatts) at which energy is being dissipated by the resulting friction.

______Entry field with incorrect answer MW

Answer:

Energy dissipated as heat = 852.10 Btu

Average rate of energy dissipation = 3.64 MW

Explanation:

1 lb = 0.453592 Kg

Mass = 5510 lb = 0.453592 * 5510 Kg = 2499.29 Kg

1 mph = 0.44704 m/s

Velocity = 60 mph = 60 * 0.44704 m/s = 26.82 m/s

Let E be equal to energy acquired

[tex]E = 1/2 MV^{2}[/tex]

[tex]E = 0.5 * 2499.29 * 26.88^{2} \\E = 899.01 * 10^{3} Joules[/tex]

Energy dissipated as heat:

[tex]E= 8.99*10^{3} = 0.0009478 * 8.99*10^{3}\\E = 852.10 tu[/tex]

E = 852.10 Btu

Energy dissipated throughout the United States

[tex]E = 350 * 10^{6} * 899.01 *10^{3} \\E=3.1466 * 10^1^4 Joules\\[/tex]

Average power rate in MW, P

[tex]P = (3.1466*10^{14} )/(24*60*60)\\P=3.64*10^{9} \\P=3.64MW[/tex]

Answer:

0.208 N

Explanation:

Parameters given:

Current, I = 2A

Angle, A = 60°

Magnetic field strength, B = 0.2 T

Length, L = 0.6 m

Magnetic force is given as:

F = I * L * B * sinA

F = 2 * 0.6 * 0.2 * sin60

F = 0.24 * sin60.

F = 0.208 N

Answer:

0.20784 N

Explanation:

The force on conductor carrying current in a magnetic Field is given as,

F = BILsinθ..................... Equation 1

Where F = force on wire carrying current, B = magnetic Field, I = current, L = Length of the wire, θ = angle between the wire and the magnetic Field.

Given: B = 0.2 T, I = 2 A, L = 0.6 m, θ = 60°

Substitute into equation 1

F = 0.2(2)(0.6)sin60°

F = 0.24×0.8660

F = 0.20784 N

Answer:

The torque in the coil is  4.9 × 10⁻⁵ N.m  

Explanation:

T = NIABsinθ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2[/tex]

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁵ N.m

Answer:

Gain in kinetic energy is [tex]1.6 \times 10^{-11} J[/tex]

or [tex]10^{8} eV[/tex]

Explanation:

The potential difference between the cloud and ground  is 100 MV (million volts)

Charge on the electron q = [tex]1.6 \times 10^{-19} C[/tex]

Kinetic energy of an electron accelerated through this potential difference is the work done to move the electron.

hence kinetic energy gained is

[tex]KE= \Delta V \times q\\KE= 100 \times 10^6 \times 1.6 \times 10^{-19}\\KE=1.6 \times 10^{-11} J[/tex]

In terms of electron volts, the conversion factor is 1 electron volt (eV) =

[tex]1.6 \times 10^{-19} J[/tex]

So,

[tex]\frac{1.6 \times 10^{-11} J}{1.6 \times 10^{-19} J} \\= 10^8 eV[/tex]

Answer:

The clip will experience no force from the charged rod.

Explanation:

Final answer:

When a charged rod is brought near a suspended paper clip inside a cup with aluminum foil, polarization of the paper clip occurs, resulting in its attraction to the charged rod due to induced charges.

Explanation:

When a charged rod is brought near a paper clip suspended from a string inside a Styrofoam cup surrounded by aluminum foil, the concept of polarization is observed. As has been demonstrated with aluminum foil, the presence of the charged rod will induce a separation of charges within the materials of the paper clip and aluminum foil. This is due to electrons being free to move within the metals. If the rod is negatively charged, it will repel the electrons in the metal towards the far end, leaving the near end with a net positive charge, which is attracted towards the rod. Conversely, if the rod is positively charged, it will attract electrons towards itself, leaving the far end positively charged. The force of attraction to the near end will be greater than the force of repulsion from the far end because the near end is closer to the rod. Consequently, the net force on the paper clip will be towards the rod, resulting in the attraction of the paper clip towards the charged rod.

This concept can be applied generally to conductive objects. A very small amount of charge, on the order of a pC (picocoulomb, 1×10⁻¹² coulombs), is sufficient to significantly affect the electrical potential of such objects. Hence, the paper clip, which is a conductive object, is likely to become polarized and attracted to the rod based on these principles.

Answer:

The answers to the questions are;

(a) The velocity of the truck right after the collision is 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J

(c) The change in mechanical energy is due to energy consumed by the collision process.

Explanation:

(a) From the principle of conservation of linear momentum, we have

m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄

Where:

m₁ = Mass of the car = 1225.0 kg

m₂ = Mass of the truck = 9700.0 kg

v₁ = Initial velocity of the car = 25.000 m/s

v₂ = Initial velocity of the truck = 20.000 m/s

v₃ = Final velocity of the car right after collision = 18.000 m/s

v₄ = Final velocity of the truck right after collision

Therefore

1225.0 kg × 25.000 m/s  +  9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s  + 9700.0 kg × v₄

That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄

Making v₄ the subject of the formula yields

v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s

The velocity of the truck right after the collision to five significant figures = 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision can be found by

The change in kinetic energy of the car truck system

Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy

That is ΔK.E. = ∑ Final K.E -∑ Initial K.E.

ΔK.E. = [tex](\frac{1}{2} m_1v_3^{2}+\frac{1}{2} m_2v_4^{2}) - (\frac{1}{2} m_1v_1^{2} +\frac{1}{2} m_2v_2^{2} )[/tex]

         = ([tex]\frac{1}{2}[/tex]·1225·18²+ [tex]\frac{1}{2}[/tex]·9700·20.884²) - ([tex]\frac{1}{2}[/tex]·1225·25²+[tex]\frac{1}{2}[/tex]·9700·20²)

         = 2313736.0616 kg·m²/s² - 2322812.5 kg·m²/s² =  -9076.4384 kg·m²/s²

1 kg·m²/s² = 1 J ∴ -9076.4384 kg·m²/s² = -9076.4384 J

(c) The energy given off by way of the 9076.4384 J is energy transformed into other forms including

1) Frictional resistance between the tires and the road for the truck and car

2) Frictional resistance in the transmission system of the truck to increase its velocity

3) Sound energy, loud sound heard during the collision

4) Energy absorbed when the car and the truck outer frames are crushed

5) Heat energy in the form of raised temperatures at the collision points of the car and the truck.

6) Energy required to change the velocity of the car over a short distance.

(a) The velocity of the truck right after the collision is 20.95258 m/s east. (b)The change in mechanical energy of the car-truck system in the collision is -2729.87 J. (c) This reduction in mechanical energy is due to energy conversion into heat, sound, and deformation.

Let's analyze the collision using the principles of conservation of momentum and mechanical energy.

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

Before the Collision:

After the Collision:

Total momentum after the collision: 22050.0 kg·m/s + (9700.0 kg)(vt)

Setting initial and final momenta equal:

224625.0 kg·m/s = 22050.0 kg·m/s + 9700.0 kg * vt

Solving for vt:

vt = (224625.0 kg·m/s - 22050.0 kg·m/s) / 9700.0 kg = 20.952577 m/s

Therefore, the velocity of the truck right after the collision is 20.95258 m/s (east).

Kinetic energy is calculated using the formula KE = (1/2)mv².

Before the Collision:

After the Collision:

Change in mechanical energy: 2320082.5 J - 2322812.5 J = -2729.87 J

The mechanical energy of the system decreases by 2729.87 J.

The decrease in mechanical energy is due to some energy being converted into other forms such as heat, sound, and deformation of the vehicles during the collision.

Answer:

(A) V = 9.89m/s

(B) U = -2.50m/s

(C) ΔK.E = –377047J

(D) ΔK.E = –257750J

Explanation:

The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.

This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.

See attachment below for full solution.

Answer:

6.21 m/s

Explanation:

Using work energy equation then

[tex]U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})[/tex]

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, [tex]v_a[/tex] as 0, 9.81 for g then

[tex]58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s[/tex]

Answer:

525.94N/m

Explanation:

According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.

[tex]F=ke..................(1)[/tex]

where  F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.

In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.

Given;

F = 223N;

e = 0.424m

k = ?

We make substitutions into equation (1) and then solve for k.

[tex]223=k*0.424\\k=\frac{223N}{0.424m}\\k = 525.94N/m[/tex]