front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire

Answers

Answer 1

Answer:

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

Explanation:

This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.

Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.

Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.


Related Questions

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .

Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.

Answers

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.

Answers

Answer:

T = 3.14 hours

Explanation:

We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.

We know that the radius of Mars is 3,389.5 km.

So, r = 1,787 + 3,389.5 = 5176.5 km

Using Kepler's law,

[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]

M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]

So,

[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]

or

T = 3.14 hours

So, the orbital period is 3.14 hours

g A solenoid 63.5 cm long has 960 turns and a radius of 2.77 cm. If it carries a current of 2.28 A, find the magnetic field along the axis at its center.Find the magnetic field on the solenoidal axis at the end of the solenoid.

Answers

Answer:

The  value is  [tex]B = 0.0043 \ T[/tex]

Explanation:

From the question we are told that

   The  length of the solenoid is  [tex]l = 63.5 = 0.635 \ m[/tex]

    The number of turns is  [tex]N = 960 \ turns[/tex]

    The  current is  [tex]I = 2.28 \ A[/tex]

Generally the magnetic field is mathematically represented as

      [tex]B = \mu _o * n * I[/tex]

Where  n is the number of turn per unit length which is mathematically evaluated as

      [tex]n = \frac{N}{l}[/tex]

     [tex]n = \frac{960}{0.635}[/tex]

     [tex]n = 1512 \ turns /m[/tex]

and  [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

So  

    [tex]B = 4\pi * 10^{-7} * 1512 * 2.28[/tex]

    [tex]B = 0.0043 \ T[/tex]

     

At TTT = 14 ∘C∘C, how long must an open organ pipe be to have a fundamental frequency of 262 HzHz ? The speed of sound in air is v≈(331+0.60T)m/sv≈(331+0.60T)m/s, where TT is the temperature in ∘C∘C.

Answers

Answer:

Length of pipe organ(L) = 0647 m (Approx)

Explanation:

Given:

Temperature (T) = 14°C

Fundamental frequency (F) = 262 Hz

Speed of sound (v) = 331 + 0.60(T) m/s

Find:

Length of pipe organ(L)

Computation:

Speed of sound (v) = 331 + 0.60(14) m/s

Speed of sound (v) = 339.4

Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)

Length of pipe organ(L) = 339.4 / 2 (262)

Length of pipe organ(L) = 0647 m (Approx)

Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth

Answers

Answer:

Vrel= 0.75c

Explanation:

See attached file

A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?

Answers

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU  is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU  - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?

Answers

Explanation:

F = ma

85 N = m (15 m/s²)

m ≈ 5.7 kg

What is the reason for the increase and decrease size of the moon and write down in a paragraph.

Answers

Answer:

The reason for the increase or decrease of the moon is due to the angular perception of the moon.

Explanation:

Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.

The zenith is the highest part of the sky and the horizon the lowest.

When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.

But when looking up at the sky as there is no reference point there will be a failure in the perception of size.

An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. Answer the following question:

The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true?

a. for every object submerged partially or completely in a fluid
b. only for an object that is floating
c. only for an object that is fully submerged and is sinking.
d. for no object submerged in a fluid

Answers

Answer:

Answer:

A. for every object submerged partially or completely in a fluid

Explanation:

This is following Archimedes principle which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether FULLYor PARTIALLY submerged, is equal to the weight of the fluid that the body displaces.

Explanation:

A 25 cm diameter circular saw blade spins at 3500 rpm. How fast would you have to push a straight hand saw to have the teeth move through the wood at the same rate as the circular saw teeth

Answers

Answer:

The answer is "45.79 m/s"

Explanation:

Given values:

diameter= 25 cm

w= 3500 rpm

Formula:

[tex]\boxed{v=w \times r} \ \ \ \ \ \ _{where} \ \ \ w = \frac{rad}{s} \ \ \ and \ \ \ r = meters[/tex]

Calculating r:

[tex]r= \frac{diameter}{2}[/tex]

  [tex]=\frac{25}{2}\\\\=12.5 \ cm[/tex]

converting value into meters: [tex]12.5 \times 10^{-2} \ \ meter[/tex]

calculating w:

[tex]w= diameter \times \frac{2\pi}{60}\\[/tex]

   [tex]= 3500 \times \frac{2\times 3.14}{60}\\\\= 3500 \times \frac{2\times 314}{6000}\\\\= 35 \times \frac{314}{30}\\\\= 35 \times \frac{314}{30}\\\\=\frac{10990}{30}\\\\=\frac{1099}{3}\\\\=366.33[/tex]

w= 366.33 [tex]\ \ \frac{rad}{s}[/tex]

Calculating v:

[tex]v= w\times r\\[/tex]

  [tex]= 366.33 \times 12.5 \times 10^{-2}\\\\= 366.33 \times 12.5 \times 10^{-2}\\\\= 4579.125 \times 10^{-2}\\\\\boxed{=45.79 \ \ \frac{m}{s}}[/tex]

An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room

Answers

Answer:

28.3°C

Explanation:

Using

T(t) = (T(0) - 20)*(e^(-k*t)) + 20

for some positive number k, and some initial temperature T(0).

Boundary conditions:

T(4) = T(0) - 5 _______ (i)

T(8) = T(0) - 7 _______ (ii)

==> solving for T(0) and k :

(i):

(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>

(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5

(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5

5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))

5 = (T(0) - 20) * ( 1 - e^(-k*4) )

(ii):

(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7

(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7

7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))

7 = (T(0) - 20) * (1 - e^(-k*8))

In both results, subsitute x = e^(-4k) and C = (T(0) - 20)

(i): 5 = C * (1 - x)

(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)

Substitute C*(1-x) from (i) into (ii):

(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5

back into (i):

(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3

C = T(0) - 20 ==>

T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3

= 28.3°C

In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit

Answers

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

Suppose a certain laser can provide 82 TW of power in 1.1 ns pulses at a wavelength of 0.24 μm. How much energy is contained in a single pulse?

Answers

Answer:

The energy contained in a single pulse is 90,200 J.

Explanation:

Given;

power of the laser, P = 82 TW = 82 x 10¹² W

time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s

the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m

The energy contained in a single pulse is calculated as;

E = Pt

where;

P is the power of each laser

t is the time to generate the power

E = (82 x 10¹²)(1.1 x 10⁻⁹)

E = 90,200 J

Therefore, the energy contained in a single pulse is 90,200 J

A radar pulse returns 3.0 x 10-4 seconds after it is sent out, having been reflected by an object. What is the distance between the radar antenna and the object

Answers

Answer:

The distance is  [tex]D = 45000 \ m[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 3.0 *10^{-4 } \ s[/tex]

   

Generally the speed of the radar is equal to the speed of light and this has a value  

      [tex]c = 3.0*10^{8} \ m /s[/tex]

Now the distance covered by the to and fro movement of the radar is mathematically evaluated as

      [tex]d = c * t[/tex]

=>    [tex]d = 3.0*10^{8} * 3.0*10^{-4}[/tex]

=>  [tex]d = 90000 \ m[/tex]

Therefore the distance between the radar antenna and the object is  

      [tex]D = \frac{d}{2}[/tex]

       [tex]D = \frac{ 90000}{2}[/tex]

      [tex]D = 45000 \ m[/tex]

The distance between the radar antenna and the object will be 45000 m.

What is a radar antenna?

A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.

The given data in the problem is;

t is the time=  3.0 x 10⁻⁴

d is the distance between the radar antenna and the object=?

c is the peed of light=3×10⁸ m/sec

The radar's speed is usually equal to the speed of light, and this has a value. The distance covered by the radars to and fro movement is now calculated mathematically as

[tex]\rm d= c \times t \\\\ \rm d= 3.0 \times 10^8 \times 3.0 \times 10^{-4} \\\\ d=90000 \ m[/tex]

As a result, the radar antenna's distance from the target is

[tex]\rm D=\frac{d}{2} \\\\ \rm D=\frac{90000}{2} \\\\ \rm D=\ 45000 \ m[/tex]

Hence the distance between the radar antenna and the object will be 45000 m.

To learn more about the radar antena refer to the link;

https://brainly.com/question/24067190

The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?

Answers

Answer:

N₁ = 1800 turns

So, the side of the transformer that plugs into the outlet has 1800 turns.

Explanation:

The transformer turns ratio is given by the following equation:

V₁/V₂ = N₁/N₂

where,

V₁ = Voltage of outlet = 120 V

V₂ = Device Voltage = 3 V

N₁ = No. of turns on outlet side = ?

N₂ = No. of turns on side of device = 45

Therefore,

120 V/3 V = N₁/45

N₁ = (40)(45)

N₁ = 1800 turns

So, the side of the transformer that plugs into the outlet has 1800 turns.

In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?

Answers

Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

What is the displacement of the object after 3 seconds?

Answers

Answer:

3 meters

Explanation:

A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with the horizontal. Determine the time taken by the projectile to hit the ground below.

Answers

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.

Answers

Answer:

Explanation:

magnetic field due to an infinite current carrying conductor

B = k x 2I / r where k = 10⁻⁷  , I is current in conductor and r is distance from wire

putting the given data

B = 10⁻⁷ x 2 x 100 / 8

= 25 x 10⁻⁷ T .

B )

earth's magnetic field = .5 gauss

= .5 x 10⁻⁴ T

= 5 x 10⁻⁵ T

percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100

= 5 %

C )

ii.  No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.

A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 540N . What are the frequencies of the fundamental and first two overtones? Express your answers using two significant figures. Enter your answers in ascending order separated by commas.

Answers

Answer:

[tex]v_1 = 301 Hz[/tex]

[tex]v_2 = 601 \ \ Hz[/tex]

[tex]v_3 = 901 \ Hz[/tex]

Explanation:

From the question we are told that

     The  length of the string is  [tex]l = 90 \ cm = 0.9 \ m[/tex]

     The mass of the string is  [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]

     The  distance  from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]

    The tension is [tex]T = 540 \ N[/tex]

Generally the frequency is mathematically represented as

        [tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]

Where n is and integer that defines that overtones

i.e  n =   1 is for fundamental frequency

      n =  2   first overtone

       n =3   second overtone

Also  [tex]\mu[/tex] is the linear density of the string which is mathematically represented as

           [tex]\mu = \frac{m_s}{l}[/tex]

=>        [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]

=>       [tex]\mu = 0.003889 \ kg/m[/tex]

So for   n = 1

     [tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v_1 = 301 \ Hz[/tex]

So for  n =  2

     [tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v_2 = 601 \ Hz[/tex]

So for  n =  3

     [tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v =901 \ Hz[/tex]

     

       

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.

Required:
What is the charge on the dust particle?

Answers

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

   

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

IMPORTANT ANSWER ALL 3 PLEASE!

Answers

Answer:

4. Liters

5. Celsius

6. Grams

What is 3/4 of 12 and 24

Answers

Answer:

3/4 of 12 = 16

3/4 of 24 = 32

Those are the answers based on how your question sounded

Explanation:

Answer:

27

Explanation:

3/4 of (12 and 24)

of means ×

and means +

therefore,3/4× (12+24)

3/4×(36)

3×9

27

How could a country benefit from making it into space?

Answers

Answer:

space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.

A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A What is the focal length of the lens

Answers

Answer:

The focal length of the lens is 2.5 cm

Explanation:

Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;

[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]

Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:

[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]

then, solving for the focal distance knowing that the object distance is 1.5 cm:

[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]

A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium. a) What is the total mechanical energy of the motion

Answers

Answer:

Explanation:i think this would help u

A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached

Answers

Answer:

The  displacement is  [tex]A_r = 6.071 \ cm[/tex]

Explanation:

From the question we are told that

   The initial displacement is [tex]A_o = 10 \ cm[/tex]

     The displacement at the end of first oscillation is  [tex]A_d = 9.05 \ cm[/tex]

     

Generally the damping constant of this damped oscillator is mathematically represented as  

           [tex]\eta = \frac{A_d}{A_o}[/tex]

substituting values

           [tex]\eta = \frac{9.05}{10}[/tex]

        [tex]\eta = 0.905[/tex]

The displacement after 4 more oscillation is mathematically represented as

       [tex]A_r = \eta^4 * A_d[/tex]

substituting values

      [tex]A_r = (0.905)^4 * (9.05)[/tex]

      [tex]A_r = 6.071 \ cm[/tex]

Answer:

Displacement reached is 6.0708 cm

Explanation:

Formula for damping Constant "C"

[tex]C^n=\frac{A_2}{A_1}[/tex]                  where n=1,2,3,........n

Where:

[tex]A_2[/tex] is the displacement after first oscillation    

[tex]A_1\\[/tex] is the initial Displacement

[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]

In our case, n=1.

[tex]C=\frac{9.05}{10}\\C=0.905[/tex]

After 4 more oscillation, n=4:

[tex]C^4=\frac{A_6}{A_2}[/tex]                                        

Where:

[tex]A_6[/tex] is the final Displacement after 4 more oscillations.

[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]

Displacement reached is 6.0708 cm

an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate

Answers

Answer:

The electric field  can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.

Explanation:

Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.

the charge density in an insulateed solid sphere of radius find the electric field at a distance of from the center of the solid

Answers

Answer:

Assuming the charged density in the insulated solid sphere of radius 3.1m is 8.8e-9, the electric field at 5.2 meters is 73.1256 [tex]i[/tex].

Explanation:

The electric charge linear density is equal to 8.8 x[tex]10^{-9}[/tex]

the radius of the sphere is 3.1m

The magnitude of the electric field at the radius of the sphere equal to 5.2 meters can be calculated with the formula ;

- E = λ / 4πε₀ [ r  / α ( α + r ) ] [tex]i[/tex]

Solution:

E =  8.8 x[tex]10^{-9}[/tex] / 4πε₀ [ 3.1/ 5.2( 5.2 + 3.1) ] [tex]i[/tex]

= 1018.0995 [0.07183] [tex]i[/tex]

=  73.1256 [tex]i[/tex]

HELP!!! 35 point question. answer at least 3 correctly. please include equations and how you did it

Answers

Answer:

9. (B) ¼ Mv²

10. (A) √(3gL)

11. 20 N

12. 5 m/s²

Explanation:

9. The rotational kinetic energy is:

RE = ½ Iω²

RE = ½ (½ MR²) (v/R)²

RE = ¼ Mv²

10. Energy is conserved.

Initial potential energy = rotational energy

mgh = ½ Iω²

Mg(L/2) = ½ (⅓ ML²) ω²

g(L/2) = ½ (⅓ L²) ω²

gL = ⅓ L² ω²

g = ⅓ L ω²

ω² = 3g / L

ω = √(3g / L)

The velocity of the top end is:

v = ωL

v = √(3gL)

11. Sum of torques about the hinge:

∑τ = Iα

-(Mg) (L/2) + (T) (r) = 0

T = MgL / (2r)

T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)

T = 20 N

12. Sum of forces on the block in the -y direction:

∑F = ma

mg − T = ma

Sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a / R)

T = ½ Ma

Substitute:

mg − ½ Ma = ma

mg = (m + ½ M) a

a = mg / (m + ½ M)

Plug in values:

a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))

a = 5 m/s²

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