Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?

Answers

Answer 1

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

Answer 2

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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Related Questions

wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength

Answers

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]

[tex]E_{left} = 9210.5 \ N/C[/tex]

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

[tex]E_{right} = -9210.5 \ N/C[/tex]

The electric field strength at the midpoint;

[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]

A ceiling fan is spinning at 45 revolutions per minute when it is switched to a higher speed. It accelerates uniformly, and 2.0 seconds later it is spinning at 110 revolutions per minute. Through how many radians did it rotate during the transition of speeds

Answers

Answer:

θ = 16.2 rad

Explanation:

First we find the angular acceleration by using first equation of motion in angular form:

ωf = ωi + αt

where,

ωf =final angular speed = (110 rev/min)(2π rad/1 rev)(1 min/60 s) = 11.5 rad/s

ωi =initial angular speed = (45 rev/min)(2π rad/1 rev)(1 min/60 s) = 4.7 rad/s

α = angular acceleration = ?

t = time = 2 s

Therefore,

11.5 rad/s = 4.7 rad/s + α(2 s)

α = (6.8 rad/s)/(2 s)

α = 3.4 rad/s²

Now, we use 2nd equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = rotation = ?

Therefore,

θ = (4.7 rad/s)(2 s) + (1/2)(3.4 rad/s²)(2 s)²

θ = 9.4 rad + 6.8 rad

θ = 16.2 rad

A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film

Answers

Answer:

t(min) = 94nm

Explanation:

The wavelength of the light incident is 500 nm.

The refractive index of the film is 1.33.

The minimum thickness of the soap film required for constructive interference.

The thickness of the film for the constructive interference is given by:

2*t= (m + 1/2) λ′

Now, λ′ = λ/μ = 500/1.33 = 376nm

The minimum thickness of the film

′t′ will be at m=0 :

2*t(min) = (0 + 1/2) 376

t(min) = 94nm

Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.

Answers

Complete Question

An emf is induced in a conducting loop of wire 1.12m long as its shape is.

changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.

Answer:

The induced emf is  [tex]\epsilon = 0.0863 \ V[/tex]

Explanation:

From the question we are told that

      The  time taken is  [tex]\Delta t = 0.125 \ s[/tex]

       The magnitude of the magnetic field is  B =  0.504 T

        The length of the loop wire is  [tex]l = 1.12 \ m[/tex]

Generally the circumference of the wire when in circular form is  

          [tex]C = 2 \pi r[/tex]

=>        [tex]l = 2 \pi r[/tex]

=>         [tex]r =[/tex][tex]\frac{l}{2 \pi}[/tex]

=>          [tex]r =[/tex][tex]\frac{1.12}{2 * 3.142}[/tex]

=>        [tex]r =[/tex][tex]0.1782 \ m[/tex]

Now the area of the wire as a circle is

           [tex]A = \pi r^2[/tex]

    =>     [tex]A = 3.142 * (0.1782)^2[/tex]      

     =>    [tex]A = 0.0998 \ m^2[/tex]

The  length of one side of the square is

         [tex]b = \frac{l}{4}[/tex]

         [tex]b = \frac{1.12}{4}[/tex]

         [tex]b = 0.28 \ m[/tex]

Now the area of the wire as a square is

          [tex]A_s = b^2[/tex]

=>          [tex]A_s =(0.28 )^2[/tex]

             [tex]A_s = 0.0784 \ m^2[/tex]

Generally the induced emf is mathematically represented as

        [tex]\epsilon = \frac{B * [A - A_s ]}{\Delta t }[/tex]

=>      [tex]\epsilon = \frac{0.504 * [0.0998 - 0.0784 ]}{0.125 }[/tex]

=>      [tex]\epsilon = 0.0863 \ V[/tex]

   

Sunlight reflected from a smooth ice surface is completely polarized. Determine the angle of incidence. (nice = 1.31.)

Answers

Answer:

Explanation:

We shall apply Brewster's law to solve the problem . Let i be the angle of reflection required. According to this law ,

tan i = μ where i is angle of incidence , μ is refractive index of the medium .

Here i = ? , μ = 1.31

tani = 1.31

i = 53° approx .

At the angle of 53° incidence , the reflected ray will be completely polarised .

Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom

Answers

Answer:

They both have the same total K.E at the bottom

Explanation:

This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved

So

K1 + U1 = K2 + U2

But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference

thus

K1 = 0 , U2 = 0

ΔK = ΔU = m g. h

But if the two inclines have the same height, and both hoops have the same mass m,

So difference in kinetic energy, must be the same for both hoops.

An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?

Answers

Answer:

3.9E-8

Explanation:

We know that

Mv²/r = Bqv

So

r= mv/Bq

But E is 1/2mv² which is 5.53eV

m²v² =2m x 5.53eV

mv = √( 2 x 9.1E-31)

So

r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19

= 3.9x10-8cm

Explanation:

An expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?

Answers

Answer:

The  value is  [tex]N = 3.708*10^{7} \ \ atoms[/tex]

Explanation:

From the question we are told that

    The pressure is  [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]

    The  temperature is  [tex]T = 26 + 273 = 299 \ K[/tex]

     The volume is  1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]

Generally according to the ideal gas law we have that

      [tex]PV = NkT[/tex]

here  k is the Boltzmann constant with a value  [tex]k = 1.38 *10^{-23} \ J/K[/tex]

  =>  [tex]N = \frac{PV}{ k T}[/tex]

=>     [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]

=>    [tex]N = 3.708*10^{7} \ \ atoms[/tex]

     

"How many wavelengths wide must a single slit be if, at a point 8o from the central maximum, there is a 72 rad phase difference between the top and bottom rays?"

Answers

Answer

82.3 wavelengths

Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.

Answers

Answer:

se the principle of induction.

place the two metallic spheres together,  now we bring the positively charged bar closer to the first sphere.

The charge that was induced in the sphere is distributed as infirm as possible,

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive

Explanation:

For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.

Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.

Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.

The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.

Answers

Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Required:
a. Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

b. Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm^2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

Answers

Answer:

1. 6.99x 10^-6V/m

2. 18m

Explanation:

See attached file

Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.43 A. Required:a. Detrmine the resistance of one of the bulbs.b. What would the current be if the bulbs were connected in series?

Answers

Answer:

a)3000ohm

b)4.44mA

Explanation:

a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

But according to ohm'law which can be expressed below

V=IR

R=V/I

R(total)= 120/0.36

= 333.33ohm

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

1/R(total)=9/R

1/333.33= 9/R

R= 3000ohm

Therefore, the resistance is 3000ohm

b)the bulbs were connected in series here, then for series connection we use below expression

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

R(total)=9R

= 9*3000

=27000ohm

I=VR

I=V/R

I= 120/27000

= 4.44*10⁻³A

4.44mA

Therefore, the current is 4.4mA

A person looks horizontally at the edge of a swimming pool. If its length is 5 m, and the pool is filled to the surface, to what depth (in m) could the observer see

Answers

Answer:

The observer could see to a depth of 4.38 m

Explanation:

Please check attachment for diagram.

Mathematically, from Snell law;

n1sin theta = n2 sin theta

1 * sin 90 = n2 * sin θR

where n2 = 1.33

1/1.33 = sin θR

Sin θR = 0.7519

θR = arc sin 0.7519

θR = 48.76

Now to get the height, we use the triangle

Using trigonometric ratio;

Tan( 90- θR) = H/5

H = 5 Tan( 90 - θR)

H = 5 Tan( 90-48.76)

H = 5 Tan41.24

H = 4.38 m

A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is

Answers

Answer:

At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage

Explanation;

Because at series connection the battery and resistor have equal voltage

Calculate the range of wavelengths (in m) for AM radio given its frequency range is 540 to 1,600 kHz. smaller value m larger value m (b) Do the same for the ultraviolet frequency range of 760 to 30,000 THz. smaller value m larger value m

Answers

Answer:

a)  λ = 555.5 m,  λ = 187.5 nm

 

Explanation:

The velocity of a wave is given by the relation

           c = λ f

           λ = c /f

a) length of the radii AM

            λ = 3 10⁸/540 10⁻⁷

            λ= 5.555 102 m

            λ = 555.5 m

             f = 1600 kHz

            λ = 3 108/1600 103

             λ = 1.875 102 m

            lam  = 187.5 nm

b) light = 760 Thz = 760 10-12 Hz

               

         λ=  c/f

            λ = 3 108/760 10-12

            λ = 3.947 10-9

           λ= 30000 Thz

           λ= c/f

      λ = 3 10⁸/ 30000 10-12

       λ = 1  m

if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?

Answers

Answer:

0.35 m³/s

Explanation:

When the pipe's depth is 0.4 m, the area of the circular segment is:

A = ½ R² (θ − sin θ)

The depth of the water is:

h = R (1 − cos(θ/2))

Solving for θ:

0.4 = 0.5 (1 − cos(θ/2))

0.8 = 1 − cos(θ/2)

cos(θ/2) = 0.2

θ/2 = acos(0.2)

θ = 2 acos(0.2)

θ ≈ 2.74 rad

The area is therefore:

A = ½ (0.5 m)² (2.74 − sin 2.74)

A = 0.338 m²

The cross-sectional area when the pipe is full is:

A = π (0.5 m)²

A = 0.785 m²

The flow velocity is constant:

v = v

Q / A = Q / A

(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)

Q = 0.35 m³/s

When a ray of light passes from glass to water it is?

Answers

Answer:

[tex]\huge\boxed{Refracted}[/tex]

Explanation:

When a ray of light passes from glass to water, it

1) is Slightly refracted (bending of light)

2) moves away from the normal.

Whenever a light ray travels from a denser medium to a rarer medium, it bends away from the normal.

Answer:

refraction

Explanation:

What is the maximum speed (in units of m/s) with which a car can round a
flat horizontal curve of radius (r=60 m), if the coefficient of static friction between tires and
road is (0.4)?
A) 5
B) 15.5
€) 240
D) 25.1​

Answers

Answer:

B)  15.5 m/s

Explanation:

r = 60m

μs = 0.4

using the formula max V = √r*g*μs (flat roadway)

v = sqrt(60 * 10 * 0.4)

v = 15.5 m/s

What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.

Answers

Answer:

The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Explanation:

Given;

work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J

Apply Einstein Photo electric effect;

E = K.E + Ф

Where;

E is the energy of the incident light

K.E is the kinetic of electron

Ф is the work function of silver surface

For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.

E = Ф

hf = Ф

[tex]h\frac{c}{\lambda} = \phi[/tex]

where;

c is speed of light = 3 x 10⁸ m/s

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

λ is the wavelength of the incident light

[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]

Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

The equivalent resistance of two resistors connected in series is always greater than the equivalent resistance of the same two resistors connected in parallel. True False

Answers

Answer:

True

Explanation:

Because the resistors in series is the sum of the two resistors given as

R= R1+R2

While that of resistors in parallel is the sum of the reciprocal of the resistance given as

1/R = 1/ R1+ 1/R2

So that of series connection will be greater

Physical properties of a mineral are a result of the arrangement of the atoms in the mineral. Use this fact to explain the following:_________
A. One mineral has a density of 2.7 g/ml while another has a density of 5.1 g/ml
B. The mineral mica cleaves into thin flat sheets while olivine does not show cleavage

Answers

Explanations:

a) The physical properties of a mineral is as a result of the arrangement of the atoms in the minerals. The reason behind one mineral having a density of 2.1 g/ml which is lower than that of another mineral with density of 5.1 g/ml is the packing density of the minerals. Minerals with high density have their atoms more closely packed together, leaving less space between the atoms. This characteristics means that they have more atomic mass per unit volume for a given molecular space, when compared to another mineral with its atoms less closely packed.

b) The property of cleavage is due to the crystalline structure of a mineral species. Cleavage is used to describe the ease with which minerals cleaves. Cleavage is due to a weak bonding strength between the molecules of the mineral, or a due to a greater lattice spacing across the the cleavage plane than in other planes within the crystal. The greater the lattice spacing, the weaker the strength of the bond across a plane.

From these, we can clearly see that the property of cleavage is due to the physical properties of the crystalline structure of the mineral species.

What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?​

Answers

Answer:

If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.

The simple reason is the desirability for genetic variation using recessive genes.

In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.

Hope this helps

A diver shines an underwater searchlight at the surface of a pond (n = 1.33). At what angle (relative to the surface) will the light be totally reflected?

Answers

Answer:

41.2°

Explanation:

Total internal reflection is the reflection of the incident ray at the interface between two media in which one of the media has a lower refractive index than the other. It occurs when the angle of incidence in the denser medium exceeds the critical angle.

The critical angle is the angle of incidence in the denser medium when the angle of incidence in the less dense medium is 90°.

Since

n= 1/sin C

C= sin^-(1/n)

C= sin^-(1/1.33)

C= 48.8°

Hence angle of incidence= 90-48.8 = 41.2°

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m ​ 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.

Answers

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]

[tex]v_{i}[/tex] is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]

[tex]\Delta x = 30 - 2.36[/tex]

[tex]\Delta x[/tex] = - 42

Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.

A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.

Answers

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

within which type of system is the total mass conserved but not the total energy

Answers

In a closed system the mass is conserved, but the energy is not conserved.

To find the answer, we have to study about different systems in thermodynamics.

What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.

Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.

Learn more about Thermodynamic system here:

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When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?

Answers

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?

Answers

Answer:

Explanation:

According to Equations of Projectile motion :

[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

[tex]Horizontal Range = vcosx * t[/tex]

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.

Answers

Answer:

82.2 mL

Explanation:

The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;

C1V1=C2V2

Where;

C1= concentration of stock solution

V1= volume of stock solution

C2= concentration of dilute solution

V2= volume of dilute solution

V2= C1V1/C2

V2= 1.48 × 55.6/ 1.0

V2= 82.2 mL

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