A child is riding a tricycle along a straight horizontal road. The pedals are attached directly to the front wheel, which has a radius of 15 cm. The rear wheels are smaller and have a radius of 8 cm. There is no slip between the wheels and the road surface. If the child is pedaling at 16 rpm (revolutions per minute), then the angular speed of the rear wheels is closest to:

Answer:

1199 miles

Explanation:

1 hour 30 minutes = 1 + 30/60 = 1.5 hours

2 hours 15 minutes = 2 + 15/60 = 2.25 hours

The distance she flew in the 1st segment is:

1.5*345 = 517.5 miles

The distance she flew in the 2nd segment is:

2.25 * 345 = 776.25 miles

Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown

776.25 * cos(45) = 549 miles in-line with the 1st segment and

776.25* sin(45) = 549 miles perpendicular to the 1st segment:

So the distance from the end to her starting position is

[tex]\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles[/tex]

Answer:

lol..WHAT IS THE QUESTION?!

Explanation:

Answer:

Frequencies are 0.86Hz

Explanation:

Given v=0.74m/s

d= 0.43m

Lambda/2=0.43m

Thus lambda= 0.86m

And F= v/lambda

0.74/0.86= 0.86Hz

Answer:

The obsrved  frquencies are [tex]f= n (0.86) Hz[/tex] n = 1,2,3,...,n.

Explanation:

   From the question we are told that

        The speed of the wave is v = 0.740 m/s

         The distance from the post is [tex]d = 43.0cm = \frac{43}{10} = 0.43m[/tex]

Generally frequency is mathematically represented as

           [tex]f = \frac{v}{\lambda }[/tex]

so for the first frequency which the the fundamental frequency(first harmonic frequency)  the clothesline(string) would form only one loop and hence the length between the  vertical post and throcky hands is mathematically represented as

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

=>    [tex]\lambda = 2L[/tex]

    So

             [tex]f_f =f_1= \frac{v}{2L}[/tex]

       Substituting values

              [tex]f_f = f_1 = \frac{0.740}{2* 0.43}[/tex]

                  [tex]= 0.860 Hz[/tex]

For the second  harmonic frequency  i.e is when the clothesline(string) forms two loops the length is mathematically represented as

        [tex]L = \lambda[/tex]

So the second harmonic frequency is            

           [tex]f_2 = \frac{v}{L}[/tex]

        [tex]f_2 = 2 * \frac{v}{2L}[/tex]

                [tex]f_2 =\frac{0.740}{0.430}[/tex]

                    [tex]= 1.72Hz[/tex]

For the third harmonic frequency i.e  when the clothesline(string) forms three loops the length  is mathematically represented as

                  [tex]L = 3 * \frac{\lambda }{2}[/tex]

So the wavelength is

               [tex]\lambda = \frac{2L}{3}[/tex]

And the third harmonic frequency is mathematically evaluated as

           [tex]f_3 = \frac{v}{\frac{2L}{3} }[/tex]

               [tex]= 3 * \frac{v}{2L}[/tex]

               [tex]f_3=2.58Hz[/tex]

Looking at the above calculation we can conclude that the harmonic frequency of a vibrating string(clothesline) can be mathematically represented as an integer multiple of the fundamental frequency

  i.e

          [tex]f = n f_f[/tex]

         [tex]f = n \frac{v}{2L}[/tex]

           [tex]f= n (0.86) Hz[/tex]

Where n denotes integer values  i.e  n = 1,2,3,....,n.

Note : The length that exist between two nodes which are successive is equivalent to half of the wavelength so when one loop is form the number of nodes would be 2 and on anti-node

Final answer:

The weight of the pendulum bob on Planet X remains 2.0 kg, the same as on Earth, because the mass of the pendulum bob does not affect the period of a simple pendulum. The difference in periods indicates a change in gravitational strength, not mass.

Explanation:

The question revolves around understanding how the period of oscillation of a simple pendulum changes with gravity on different planets. Since the mass of the pendulum bob does not affect the period of a simple pendulum, which depends only on the length of the pendulum and the gravitational acceleration (g), the weight of the pendulum bob on Planet X would still be 2.0 kg, the same as on Earth. However, the difference in periods between Earth and Planet X indicates a difference in gravitational acceleration, implying that g on Planet X is weaker than on Earth. The formula for the period (T) of a simple pendulum is T = 2π√(l/g), where l is the length of the pendulum and g is the gravitational acceleration. Since the mass of the pendulum bob does not factor into this equation, the weight of the pendulum bob on Planet X remains the same, but its apparent weight will change according to the planet's gravitational pull.

Answer:

No

Explanation:

The frequency of oscillation of spring mass system is independent of its amplitude.

The frequency of oscillation will not change because it is not a function of amplitude.

In a real spring-mass system, the spring has a non-negligible mass m. Since not all of the spring's length moves at the same velocity v as the suspended mass  M,  

To determine the frequency of oscillation, the effective mass of the spring is defined as the mass that needs to be added to  M to correctly predict the behaviour of the system.

For vertical springs, however, we need to remember that gravity stretches or compresses the spring beyond its natural length to the equilibrium position. After we find the displaced position

Thus the frequency of oscillation will not change because it is not a function of amplitude.

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Answer:

the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s

Explanation:

Given that :

Mass attached to the free end of the string, m = 423 g = 0.423 kg

Moment of inertia of pulley, I = 0.0352 kg m²

Radius of the pulley, r = 12.5 cm = 0.125 meters

Depth of fallen mass, h = 1.25 m

Acceleration due to gravity, g = 9.8 m/s²

Change in potential energy = mgh

= 0.423 × 9.8 × 1.25

=5.18175 J

From the question, we understand that the change in potential energy is used to raise and increase the kinetic energy of hanging mass and  the rotational kinetic energy of pulley.

As such;

[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2[/tex]

where;

[tex]\omega[/tex]  is the angular velocity of the pulley

v is the velocity of the mass after falling 1.25 m

where:

[tex]v = r \omega[/tex]

[tex]\omega = \frac{v}{r}[/tex]

replacing [tex]\omega = \frac{v}{r}[/tex]  into above equation; we have:

[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I( \frac{v}{r})^2[/tex]

[tex]5.18175= (\frac{1}{2} m + \frac{1}{2}* (\frac{I}{r^2})v^2 \\ \\ v^2 = \frac{5.18175}{0.5*0.423+0.5*\frac{0.0352}{0.1252}} \\ \\ v^2 = 3.873047 \\ \\ v = 1.968 \ m/s[/tex]

Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s

the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s

What all information we have?

Mass attached to the free end of the string, m = 423 g = 0.423 kg

Moment of inertia of pulley, I = 0.0352 kg m²

Radius of the pulley, r = 12.5 cm = 0.125 meters

Depth of fallen mass, h = 1.25 m

Acceleration due to gravity, g = 9.8 m/s²

Change in potential energy = mgh

Change in potential energy = 0.423 × 9.8 × 1.25

Change in potential energy =5.18175 J

As such:

5.18175 J= mv²+1w²

where;

w  is the angular velocity of the pulley

v is the velocity of the mass after falling 1.25 m

v=rw

w=v/r

Replacing into above equation;

5.18175 J=mv² + 1/2 (w/r²) ²

5.18175 = (m + /* (4) v²

v² = 3.873047

v² =1.968 m/s

Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s.

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Answer:

Similarities between magnetic fields and electric fields: Electric fields are produced by two kinds of charges, positive and negative. Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like poles repel; unlike poles attract.

they attract each other

Explanation:

(a) [tex]\(a_c \approx 39.8 \, \text{m/s}^2\)[/tex]

(b) [tex]\(r \approx 11.6 \, \text{m}\)[/tex]

(c) [tex]\(v_2 \approx 28.4 \, \text{m/s}\)[/tex]

(a) calculation of the centripetal acceleration [tex](\(a_c\))[/tex],

[tex]\[ a_c = \frac{g}{\sin(\theta_1)} \][/tex]

Given [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_1 = 15.0°\)[/tex], we find:

[tex]\[ a_c = \frac{9.8 \, \text{m/s}^2}{\sin(15.0°)} \][/tex]

Calculating this gives [tex]\(a_c \approx 39.8 \, \text{m/s}^2\).[/tex]

(b) The radius ((r)) of the curve can be found using the formula:

[tex]\[ r = \frac{v^2}{a_c} \][/tex]

Substituting [tex]\(v = 21.5 \, \text{m/s}\) and \(a_c \approx 39.8 \, \text{m/s}^2\),[/tex]we get:

[tex]\[ r = \frac{(21.5 \, \text{m/s})^2}{39.8 \, \text{m/s}^2} \][/tex]

This calculation yields [tex]\(r \approx 11.6 \, \text{m}\).[/tex]

(c) For a new deflection angle [tex]\(\theta_2 = 7.00°\)[/tex], we find the new centripetal acceleration [tex](\(a_{c2}\))[/tex] using:

[tex]\[ a_{c2} = \frac{g}{\sin(\theta_2)} \][/tex]

Substituting [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_2 = 7.00°\)[/tex], we get:

[tex]\[ a_{c2} = \frac{9.8 \, \text{m/s}^2}{\sin(7.00°)} \][/tex]

Calculating this gives [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\).[/tex]

Then, using [tex]\(a_{c2} = \frac{v_2^2}{r}\)[/tex], we find the new speed [tex](\(v_2\)):[/tex]

[tex]\[ v_2 = \sqrt{a_{c2} \cdot r} \][/tex]

Substituting [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\) and \(r \approx 11.6 \, \text{m}\), we get \(v_2 \approx 28.4 \, \text{m/s}\).[/tex]

Answer:

0.243 m/s

Explanation:

From law of conservation of motion,

mu+m'u' = V(m+m')................. Equation 1

Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.

make V the subject of the equation

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s

Substitute into equation 2

V = (260000×0.32+52500×(-0.14))/(260000+52500)

V = (83200-7350)/312500

V = 75850/312500

V = 0.243 m/s

Answer:

by sweating

Explanation:

Given Information:  

Current = I = 0.1 A

Resistance = R = 100 kΩ

Required Information:  

Voltage = V = ?

Answer:  

Voltage = V = 1000 V

Step-by-step explanation:  

We know that electrocution depends upon the amount of current flowing through the body and the voltage across the body.

V = IR

Where I is the current flowing through the body and R is the resistance of body.

If electrocution can be avoided when the current is below 0.1 A then

V = 0.1*10×10³

V = 1000 Volts

Therefore, 1000 V is the maximum voltage with which a person could come into contact while avoiding electrocution, any voltage more than 1000 V may result in fatal electrocution.

Also note that human body has very low resistance when the body is wet therefore, above calculated value would not be applicable in such case.

In a system where an A-36 steel pipe is snugly fit between two fuel tanks, the rise in fuel temperature will cause thermal stresses in the pipe due to the restraint from the fuel tanks. This is related to the thermal properties of the material and the rate of change due to temperature rising, from which the average normal stress can be calculated.

Determining the average normal stress developed in an A-36 steel pipe when fuel flows through it at varying temperatures requires knowledge of thermal expansion and associated stress in materials. This is related to thermal properties and the rate of change due to temperature rise.

When temperature rises along the pipe such as mentioned, from room temperature to 130∘C and 80∘C, the steel pipe expands. However, the pipe is restrained by the fuel tanks acting as springs, leading to development of stress within the pipe. The average normal stress can be calculated by dividing the force exerted by the expansion (or contraction) by the cross-sectional area of the pipe:

F/A = σ

Where, F is the force and A is the cross-sectional area of the pipe. The force can be obtained from Hooke's law for springs (F = k Δx), where k is the stiffness of the tank walls acting as springs, and Δx is the change in length due to thermal expansion.

The average normal stress is a measure of the extent to which the pipe is going through physical changes due to thermal variations.  

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Answer:

The direction is due south

Explanation:

From the question we are told that

     The energy of the electron is [tex]E = 19.0keV = 19.0 *10^3 eV[/tex]

      The earths magnetic field is [tex]B = 42.3 \muT = 42.3 *10^{-6} T[/tex]

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

          [tex]\= F = q (\= v * \=B)[/tex]

On the first uploaded image is an  illustration of the movement of the electron

    Looking at the diagram  we can see that in terms of direction  the magnetic force  is

             [tex]\= F =q(\=v * \= B)= -( -\r i * - \r k)[/tex]

                [tex]= -(- (\r i * \r k))[/tex]

generally  i cross k = -j

      so the equation above becomes

             [tex]\= F = -(-(- \r j))[/tex]

                [tex]= - \r j[/tex]

This show that the direction is towards the south  

Answer:

3.74 %

Explanation:

Given,

Mass of the fired projectile, m = 0.1 Kg

Mass of the stationary target, M = 2.57 Kg

Speed of the particle after collision

Using conservation of momentum

m v = (m + M) V

0.1 x v = (0.1 + 2.57) V

V = 0.0374 v

Initial KE = [tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}\times 0.1\times v^2[/tex]

Final KE = [tex]\dfrac{1}{2} (M + m) V^2 = \dfrac{1}{2}\times 2.67\times (0.0374 v)^2[/tex]

Now,

the percent of the projectile's KE carried out by target

  [tex]=\dfrac{\dfrac{1}{2}\times 2.67\times (0.0374 v)^2}{ \dfrac{1}{2}\times 0.1\times v^2}\times 100[/tex]

 [tex]= 3.74\ \%[/tex]

Answer:

Turbulence mixing will happen mostly in air

Explanation:

Answer:

Pls refer to attached file

Explanation:

Answer:

  t = 1.77 s

Explanation:

The equation of a traveling wave is

       y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

      v = λ f

Where the frequency and period are related

     f = 1 / T

we substitute

      v = λ / T

let's develop the initial equation

    y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

    y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

         2π /λ = 1.65

          λ = 2π / 1.65

          λ = 3.81 m

         2π / T = 4.64

          T = 2π / 4.64

          T = 1.35 s

we seek the speed of the wave

           v = 3.81 / 1.35

           v = 2.82 m / s

Since this speed is constant, we use the uniformly moving ratios

          v = d / t

           t = d / v

           t = 5 / 2.82

           t = 1.77 s

According to the information given, the Heisenberg uncertainty principle would be given by the relationship

[tex]\Delta x \Delta v \geq \frac{h}{4\pi m}[/tex]

Here,

h = Planck's constant

[tex]\Delta v[/tex] = Uncertainty in velocity of object

[tex]\Delta x[/tex] = Uncertainty in position of object

m = Mass of object

Rearranging to find the position

[tex]\Delta x \geq \frac{h}{4\pi m\Delta v}[/tex]

Replacing with our values we have,

[tex]\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}[/tex]

[tex]\Delta x \geq 5.79*10^{-9}m[/tex]

Therefore the uncertainty in position of electron is [tex]5.79*10^{-9}m[/tex]

Answer:

See attached handwritten document for answer

Explanation:

Answer:

Explanation:

a) You can compute the force by using the expression:

[tex]F=k\frac{q_1q_2}{[(x-x_o)^2+(y-y_o)^2+(z-z_o)^2]^{\frac{1}{2}}}[/tex]

where k=8.98*10^9Nm^2/C^2 and q1, q2 are the charges. By replacing for the forces you obtain:

[tex]F_T=k[\frac{(1*10^{-3}C)(10*10^{-9}C)}{[(3-0)^2+(2-3)^2+(-1-1)^2]}}][(3-0)\hat{i}+(2-3)\hat{j}+(-1-1)\hat{k}]\\\\ \ \ \ \ +k[\frac{(-2*10^{-3}C)(10*10^{-9}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}}][(-1-0)\hat{i}+(-1-3)\hat{j}+(4-1)\hat{k}]\\\\F_T=6.41*10^{-3}N[3i-j-2k]-6.9*10^{-3}N[-1i-4j+3k]\\\\=0.026N\hat{i}-0.034N\hat{j}-0.033\hat{k}[/tex]

b)

[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}]+k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}]\\\\E=641428.5N/C+690769.23N/C=1332197.73N/C[/tex]

[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}][3i-2j-2k]+\\\\k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}][-1i-4j+3k]\\\\E=641428.5N/C[3i-2j-2k]-690769.23N/C[-1i-4j+3k]=2615054.7i+1480219.92j-4973626.23k\\\\|E|=\sqrt{(E_x)^2+(E_y)^2+(E_z)^2}=5810896.56N/C[/tex]

where we you have used that E=kq/r^2

Answer:

D) 0.0372 N m

Explanation:

r = 45/2 cm = 22.5 cm = 0.225 m

As 1 revolution = 2π rad we can convert to radian unit

2.4 rev/s = 2.4 * 2π = 15.1 rad/s

18.2 rev = 18.2 * 2π = 114.35 rad

We can calculate the angular (de)acceleration using the following equation of motion

[tex]-\omega^2 = 2\alpha \theta [/tex]

[tex]- 15.1^2 = 2*\alpha * 114.35[/tex]

[tex]\alpha = \frac{-15.1^2}{2*114.35} = -0.994 rad/s^2[/tex]

The moment of inertia of the solid uniform sphere is

[tex]2mr^2/5 = 2*1.85*0.225^2/5 = 0.0375 kgm^2[/tex]

The net torque acting on this according to Newton's 2nd law is

[tex]T = I\alpha = 0.0375 * 0.994 = 0.0372 Nm[/tex]

Answer:

(D) The net torque acting on this sphere as it is slowing down is closest to  0.0372 N.m

Explanation:

Given;

mass of the solid sphere, m =  1.85 kg

radius of the sphere, r = ¹/₂ of diameter = 22.5 cm

initial angular velocity, ω = 2.40 rev/s = 15.08 rad/s

angular revolution, θ = 18.2 rev = 114.37 rad

Torque on the sphere, τ = Iα

Where;

I is moment of inertia

α is angular acceleration

Angular acceleration is calculated as;

[tex]\omega_f^2 = \omega_i^2 +2 \alpha \theta\\\\0 = 15.08^2 + (2*114.37)\alpha\\\\\alpha = \frac{-15.08^2}{(2*114.37)} = -0.994 \ rad/s^2\\\\\alpha = 0.994 \ rad/s^2 \ (in \ opposite \ direction)[/tex]

moment of inertia of solid sphere, I = ²/₅mr²

                                                           = ²/₅(1.85)(0.225)²

                                                           = 0.03746 kg.m²

Finally, the net torque on the sphere is calculated as;

τ = Iα

τ = 0.03746 x 0.994

τ = 0.0372 N.m

Therefore, the net torque acting on this sphere as it is slowing down is closest to  0.0372 N.m

Answer:

Explanation:

The picture attached shows the full explanation

Answer:

There is no induced current

Explanation:

[Find the attachment]

Answer:

Explanation:

find the answer below

Answer:

Explanation:

The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:  

Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.  

E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.

DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.  

The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.

Answer:

a) x = 0.144 m

b) W = 0.15 J

c) E = 0.14 J

d) The block will rise 0.07m after it is released

Explanation:

a) The elastic force equals the gravitational force

F = kx = mg

x = 0.07, m = 0.1 kg, g = 9.81 m/s²

0.07k = 0.1 * 9.8

k = (0.1*9.8)/0.07

k = 14 N/m

When the force, F = 3N

F = kx

3 = 14x

x = 3/14

x = 0.214 m

The position of the block = 0.214 - 0.07 = 0.144m

B) Determine the work you did stretching the spring.

Energy stored in the spring when x = 0.07

E = 0.5 kx²

E = 0.5 * 14 * 0.07²

E = 0.0343 J

Energy stored in the spring when x = 0.214

E = 0.5 kx²

E = 0.5 * 14 * 0.214²

E = 0.32 J

Potential energy lost due to gravity = mgh

PE = 0.1 * 9.81 * 0.144

PE = 0.141 J

So to calculate the work done:

0.0343 + W = 0.32 - 0.141

W = 0.15 J

c) Energy in the spring

E = 0.32 - 0.0343 - 0.15

E = 0.1357 = 0.14 J

d)

[tex]1/2 *k *0.214^{2} = 1/2 kx^{2} + mg(0.214+x)\\0.32 = 7x^{2} + 0.1*9.8(0.214+x)\\[/tex]

Solving for x, x = 0.07 m

The block will rise 0.07m after it is released

Answer:

a) 1.38°

b) 7.53*10^11 m/s/s

c) 6.52*10^-9m

Explanation:

a) to find the angle you can use the dot product between two vectors:

[tex]\vec{v}\cdot\vec{B}=vBcos\theta\\\\\theta=cos^{1}(\frac{\vec{v}\cdot\vec{B}}{vB})[/tex]

v: velocity of the electron

B: magnetic field

By calculating the norm of the vectors and the dot product and by replacing you obtain:

[tex]B=\sqrt{(20)^2+(50)^2+(30)^2}=61.64mT\\\\v=\sqrt{(40)^2+(30)^2+(50)^2}=70.71m/s\\\\\vec{v}\cdot\vec{B}=[(20)(40)+(50)(30)-(30)(50)]mTm/s=800mTm/s\\\\\theta=cos^{-1}(\frac{800*10^{-3}Tm/s}{(70.71m/s)(61.64*10^{-3}T)})=cos^{-1}(0.183)=1.38\°[/tex]

the angle between v and B vectors is 1.38°

b) the change in the speed of the electron can be calculated by the change in the momentum in the following way:

[tex]\frac{dp}{dt}=F_e=qvBsin\theta\\\\\frac{dp}{dt}=(1.6*10^{-19}C)(70.71m/s)(61.64*10^{-3}T)(sin(1.38\°))=6.85*10^{-19}N[/tex]

due to the mass of the electron is a constant you have:

[tex]\frac{dp}{dt}=\frac{mdv}{dt}=6.85*10^{-19}N\\\\\frac{dv}{dt}=\frac{6.85*10^{-19}N}{9.1*10^{-31}kg}=7.53*10^{11}(m/s)/s[/tex]

the change in the speed is 7.53*10^{11}m/s/s

c) the radius of the helical path is given by:

[tex]r=\frac{m_ev}{qB}=\frac{(9.1*10^{-31}kg)(70.71m/s)}{(1.6*10^{-19}C)(61.64*10^{-3}T)}=6.52*10^{-9}m[/tex]

the radius is 6.52*10^{-9}m

Answer:

(a) Current is 2831.93 A

(b) [tex]8.40A/m^2[/tex]

(c) [tex]\rho =15.52\times 10^{-9}ohm-m[/tex]

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

[tex]A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2[/tex]

Resistance [tex]R=11.9mohm=11.9\times 10^{-3}ohm[/tex]

Potential difference V = 33.7 volt

(A) current is equal to

[tex]i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A[/tex]

(B) Current density is equal to

[tex]J=\frac{i}{A}[/tex]

[tex]J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2[/tex]

(c) Resistance is equal to

[tex]R=\frac{\rho l}{A}[/tex]

[tex]11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}[/tex]

[tex]\rho =15.52\times 10^{-9}ohm-m[/tex]

The x component of the change in momentum of cart A is -0.20 kg·m/s, while for cart B, it is +0.20 kg·m/s. The sum of these changes in momentum is 0 kg·m/s, showing conservation of momentum in the system.

The student has provided information about a collision between two carts, which involves concepts of momentum and conservation of momentum from mechanics in physics.

Part A: Change in Momentum of Cart A

The x component of the change in momentum of cart A can be calculated using the formula Δp = m(v_final - v_initial), where m is mass and v is velocity. Here, Δp = 1.0 kg (0.50 m/s - 0.70 m/s) = -0.20 kg·m/s. The negative sign indicates that the momentum decreased.

Part B: Change in Momentum of Cart B

Similarly, for cart B, Δp = 0.10 kg (1.6 m/s - (-0.40 m/s)) = 0.20 kg·m/s. The positive sign signifies an increase in momentum.

Part C: Sum of Changes in Momentum

The total change in momentum for both carts is the sum of their individual momentum changes, which is -0.20 kg·m/s + 0.20 kg·m/s = 0 kg·m/s. This means that the total momentum of the system is conserved.

Answer:

75.5degrees

Explanation:

Magnitude of magnetic flux= BA

If rotated through an angle= BAcos theta

= (0.25)BA=BAcos theta

= costheta= 0.25

= theta= cos^-1 0.25

=75.5degrees

Answer:

[tex]62.1566632757\ ^{\circ}C[/tex]

[tex]15.9715157681\ ^{\circ}C[/tex]

Explanation:

[tex]\Delta T[/tex] = Change in termperature

[tex]\Delta L[/tex] = Change in length

We have the relation

[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{18.77-12.66}{t-0}\\\Rightarrow t=\dfrac{18.77-12.66}{0.0983}\\\Rightarrow t=62.1566632757\ ^{\circ}C[/tex]

The temperature is [tex]62.1566632757\ ^{\circ}C[/tex]

[tex]\dfrac{\Delta L}{\Delta T}=\dfrac{22.49-12.66}{100-0}=\dfrac{14.23-12.66}{t-0}\\\Rightarrow t=\dfrac{14.23-12.66}{0.0983}\\\Rightarrow t=15.9715157681\ ^{\circ}C[/tex]

The temperature is [tex]15.9715157681\ ^{\circ}C[/tex]

(A)

According to the question,

The change in length will be:

= [tex]l_2-l_1[/tex]

= [tex]22.49-12.66[/tex]

= [tex]9.83 \ cm[/tex]

The change per degree will be:

= [tex]\frac{Change \ in \ length}{Temperature}[/tex]

= [tex]\frac{9.83}{100}[/tex]

= [tex]0.0983 \ cm/deg[/tex]

Now,

The change in length,

= [tex]18.77-12.66[/tex]

= [tex]6.11 \ cm[/tex]

hence,

The temperature,

= [tex]\frac{6.11}{0.0983}[/tex]

= [tex]62.2^{\circ} C[/tex]

(B)

The change in length,

= [tex]14.23-12.66[/tex]

= [tex]1.57 \ cm[/tex]

hence,

The temperature will be:

= [tex]\frac{1.57}{0.0983}[/tex]

= [tex]15.98^{\circ} C[/tex]

Thus the above answers are correct.

Learn more about temperature here:

https://brainly.com/question/847609