Which of the following are properties of a conductor in electrostatic equilibrium? The force on a charge near the conductor is zero. Electric field lines just outside the conductor are perpendicular to the surface. All net charge in the conductor is on the surface. The electric field inside the conductor is zero.

Answer:

12.9

Explanation:

1.6x15=24

0.4x4.5=1.8

24+1.8=25.8

1.6+0.4=2

25.8/2=12.9

Answer:

C. Its velocity is perpendicular to its acceleration

Explanation:

Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude

Answer:

455.9 Hz, Flat

Explanation:

The beat frequency is basically the difference between the frequency of the B4 note and the frequency of the A4 tuning fork. This means the B4 note is 15.9 Hz off of 440Hz, and 67.4 Hz off of 525.3 Hz. As B4 is between the two notes, it would make sense to find its frequency by adding 15.9 to 440 and subtracting 67.4 from 523.3, both if which give us a frequency 455.9 Hz for the B4 key. This is because the note doesn't change for the different turning forks so both differences should result in the same frequency. Because the note should be 493.9 frequency but instead has a frequency of 455.9 Hz, it is flat because the frequency is lower than it is supposed to be.

Hope this helped!

This question deals with the phenomenon of "Beat Frequency".

The frequency being played by the B4 key is "455.9 Hz", which is "lower (flat) than it should be".

When two waves with a slightly different frequency interfere with each other, they produce a pattern known as beat frequency. The beat frequency is the difference between the frequencies of the interfering wave. Since the frequency of B4 key produces beat frequency with both the C5 tuning fork (which has a frequency of 523.3 Hz) and the A4 (which has a frequency of 440 Hz). Hence, the frequency of the B4 key must be between the range of these frequencies, that is, 440 Hz to 523.3 Hz.

ANALYZING THE BEAT FREQUENCY WITH A4 TUNING FORK:

Beat Frequency with A4 tuning fork = Frequency of B4 - Frequency of A4

Frequency of B4 = Frequency of A4 + Beat Frequency with A4 tuning fork

Frequency of B4 = 440 Hz + 15.9 Hz

Frequency of B4 = 455.9 Hz

ANALYZING THE BEAT FREQUENCY WITH C5 TUNING FORK:

Beat Frequency with C5 tuning fork = Frequency of C5 - Frequency of C5

Frequency of B4 = Frequency of C5 - Beat Frequency with C5 tuning fork

Frequency of B4 = 440 Hz - 67.4 Hz

Frequency of B4 = 455.9 Hz

Both the answers validate each other.

Since the expected frequency of the B4 key is 493.9 Hz, which is greater than the actual frequency produced by the B4 key. Hence, we can say that "the frequency of the key lower than it should be ("flat")"

The attached picture shows the general formula of the beat frequency.

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Answer:

a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.

Explanation:

Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.

a) How long does it take for the police to catch the speeding car:

Kinematic equation of each vehicle's position are described:

Speeder

[tex]s_{A} = s_{A,o}+v_{A}\cdot t[/tex]

Police Car

[tex]s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]

If [tex]s_{A} = s_{B}[/tex], [tex]s_{A,o} = s_{B,o}[/tex], [tex]v_{A} = 18.056\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the resulting expression is done:

[tex]v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]

[tex]\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0[/tex]

[tex]t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0[/tex]

[tex]t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}[/tex]

[tex]t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }[/tex]

[tex]t = 18.056\,s[/tex]

The police will take 18.056 seconds to catch the speedy car.

b) How far did the police car travel before police caught up with the speeder?

The distance travelled by the police is: ([tex]s_{B,o} = 0\,m[/tex],  [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and  [tex]t = 18.056\,s[/tex])

[tex]s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}[/tex]

[tex]s_{B} = 326.019\,m[/tex]

The police will travel 326.019 meters before catching the speedy car.

c) What is the speed of the police car when catches up with the speeder?

The speed of the police car is represented by the following formula:

[tex]v_{B} = v_{B,o} + a_{B}\cdot t[/tex]

Where [tex]v_{B}[/tex] is the speed of the police car, measured in meters per second.

Given that [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], [tex]t = 18.056\,s[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the final speed of the police car when catches up with the speeder is:

[tex]v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)[/tex]

[tex]v_{B} = 36.112\,\frac{m}{s}[/tex]

The speed of the police car when catches up with the speeder is 36.112 meters per second.

Answer:

Explanation:

The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as

Input = Output + Accumulation

The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.

Answer:

The acceleration of the particle is 2 m/s²

The position of the particle is 9m

Explanation:

Given;

particles velocity, v = 2t m/s

The acceleration of the particle is given by;

a = dv/dt

a = d(2t) /dt

a = 2 m/s²

The acceleration of the particle is 2 m/s²

The position of the particle is given by;

s = ut + ¹/₂at²

Since the particle was released from rest, u = 0

s = 0(3) + ¹/₂(2)(3)²

s = 0 + 9 m

The position of the particle when t = 3 s is 9m

You haven't described whether they're connected in series or in parallel.  Actually, there are eight (8) different ways they can be arranged, and each way has a different equivalent resistance.

==> All 3 resistors in series.  Equivalent resistance = (2+3+4) = 9.000 ohms

==> All 3 resistors in parallel.

Equivalent resistance = 1 / (1/2 + 1/3 + 1/4) = (12/13) ohm or 0.923 ohm

==> The 2 ohm resistor in series with (the 3 and the 4 in parallel)

Equivalent resistance = 2 + 1/(1/3 + 1/4) =  3.714 ohms

==> The 3 ohm resistor in series with (the 2 and the 4 in parallel)

Equivalent resistance = 3 + 1/(1/2 + 1/4) =  4.333 ohms  

==> The 4 ohm resistor in series with (the 2 and the 3 in parallel)

Equivalent resistance = 4 + 1/(1/2 + 1/3) =  5.200 ohms

==> The 2 ohm resistor in parallel with (the 3 and the 4 in series)

Equivalent resistance = 1.556 ohms

==> The 3 ohm resistor in parallel with (the 2 and the 4 in series)

Equivalent resistance = 2.000 ohms

==> The 4 ohm resistor in parallel with (the 2 and the 3 in series)

Equivalent resistance = 2.222 ohms

Speed=distance/time.

7.74mins=464.4secs.

72.1=distance/464.4.

distance=464.4×7.21.

=3348.324m.

Answer:

Explanation:

According to newtons first law of motion;

Force = mass * acceleration

F = ma

where a = v-u/t

v is the final velocity

u is the initial velocity

t is the time taken

F = m(v-u/t)

Ft = m(v-u)

Given parameters;

F = 910,000N

t = 120m = 120 *60 = 7200secs

m = 1400kg

u = 0m/s

Required parameter

Final velocity v

Substituting the given parameters into the formula to get v;

Ft = m(v-u)

910,000 * 72000 = 1400(v-0)

910,000 * 72000  = 1400v

v = (910,000 * 72000)/1400

v = 910,000 * 720/14

v = 910,000 * 60

v = 468,000,000 m/s

Hence the final velocity of the drag racer is 468,000,000 m/s

Answer: his distance from the pole is 90.16 ft

Explanation:

based on the diagram of the question i will upload along this answer;

45 - d is equal to part of the pole below the horizontal line

d/x = tan(16) ; x = d/tan(16)............equ1

(45-d) / x = tan(12); x = (45-d) / tan(12)-----------equ2

∴ d/tan(16) = (45-d) / tan(12)

d.tan(12) = (45-d).tan(16)  

d.tan12 = (45×tan16) - d.tan16

0.2125d = 12.9034 - 0.2867d

0.2125d + 0.2867d = 12.9034

0.4992d = 12.9034

d = 12.9034/0.4992

d = 25.85 ft

now substitute value of d into any of our previous equation, lets take equation 1

x = 25.85 / tan(16)

x = 25.85 / 0.2867

x = 90.16 ft

Therefore his distance from the pole is 90.16 ft

Answer:

85.12 μAmp

Explanation:

The battery power output = 17.15 Amp-hr

If the battery is to last 23 years, we have to calculate how many hours there are in 23 years

in one year there are 24 hours x 365 day = 8760 hrs

in 23 years there are 23 x 8760 = 201480 hours

maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = 85.12 μA

The momentum of the truck is "56250 kgm/s".

According to the question,

Mass,

Velocity,

As we know,

→ [tex]p = mv[/tex]

By putting the values, we get

→    [tex]= 2250\times 25[/tex]

→    [tex]= 56250 \ kgm/s[/tex]

Thus the answer is right.

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Answer:

0.00004666N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

So lets say that the power absorbed

Will be = Intensity x Area

Which Is = 1.4 x 10^3 x(10)

So

14000 Watt = 14 kWatt

However we know that radiation pressure is equal to

time-averaged intensity all over the speed of light in free space

So

P = (1.4 x 1000)/c

But

F= P x A

So

((1.4 x 1000)/(3 x1 0^8)) x 10

Which is

=0.000046666N

Explanation:

Answer:

q = -2.19 x 10⁻⁵ C

Explanation:

Given;

mass of the particle, m = 1.5 g = 0.0015 kg

magnitude of electric field, E = 670 N/C

Electric field is given by;

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

where;

q is the magnitude of the

f is the force of the charge

f = mg

[tex]E = \frac{F}{q}\\\\E = \frac{mg}{q}\\\\q = \frac{mg}{E}\\\\q = \frac{0.0015*9.8}{670} \\\\q = 2.19*10^{-5} \ C[/tex]

Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative

q = -2.19 x 10⁻⁵ C

Answer:

6.705555556 m/s

Explanation:

60.35 divide by 9

Answer:

The  value  is  [tex]t = 10 \  s[/tex]

Explanation:

From the question we are told that

  The  height is  [tex]h = 500 \ m[/tex]

  The acceleration is  [tex]g =  10 \  m/s[/tex]

Generally from the second equation of motion we have that

    [tex]h  =  ut  +  \frac{1}{2} a t^2[/tex]

Here  the initial velocity u =  0  m/s  given that the stone started from rest

       So

            [tex]500 =  0 (t) +0.5 *  (10) t^2[/tex]

=>         [tex]t = 10 \  s[/tex]

Answer:

1.78 m/s

Explanation:

Distance to putt = 8 m

Distance the ball stops to the putt = 1.5 m

therefore distance traveled by the ball = 8 - 1.5 = 6.5 m

The ball stops at this point 1.5 m from the putt, therefore its final velocity at this point = 0 m/s

the ball was struck with an initial velocity of 1.6 m/s

Using the equation

[tex]v^2[/tex] = [tex]u^2[/tex] + 2as

where v is the final velocity of the ball = 0 m/s

u is the initial speed of the ball = 1.6 m/s

a is the acceleration of the grass

s is the distance the ball travels = 6.5 m

substituting values, we have

[tex]0^2[/tex] = [tex]1.6^2[/tex] + 2(a x 6.5)

0 = 2.56 + 13a

13a = -2.56

a = -2.56/13 = -0.197 m/s^2

If this acceleration of the grass is assumed to be constant, then to the initial speed needed to make the putt will be calculated from

[tex]v^2[/tex] = [tex]u^2[/tex] + 2as

where

v is the final speed at the putt = 0 m/s

u is the initial speed with which the ball is struck = ?

a is the acceleration of the grass = -0.197 m/s^2

s is the distance to the putt = 8 m

substituting values, we have

[tex]0^2[/tex] = [tex]u^2[/tex] + 2(-0.197 x 8)

0 = [tex]u^2[/tex] - 3.152

[tex]u^2[/tex] = 3.152

u = [tex]\sqrt{3.152}[/tex] = 1.78 m/s

Answer:

[tex] \huge{ \boxed{ \sf{30000 \: kg \: {m/s}^{2}}}}[/tex]

Explanation:

[tex] \text{ \underline{ Given}} : [/tex]

[tex] \star[/tex] Mass of a truck ( m ) = 3000 kg

[tex] \star[/tex] Velocity of a truck ( v ) = 10 m / s

Finding the momentum :

[tex] \boxed{ \sf{momentum \: ( \: p \: ) \: = \: mass \: ∗ \: \: velocity}}[/tex]

[tex] \hookrightarrow{ \sf{momentum \: ( \: p \: ) \: = 3000 \: ∗ \: 10}}[/tex]

[tex] \hookrightarrow{ \sf{ \: momentum \: ( \: p \: ) \: = 30000 \: kg \: m/ {s}^{2} }}[/tex]

Hope I helped!

Best wishes ! :D

~[tex] \sf{TheAnimeGirl}[/tex]

Given:-

To calculate: Momentum of the body.

We know,

p = mv

where,

Thus,

p = (3000 kg)(10 m/s)

→ p = 30000 kg m/s (Ans.)

Answer:

lamp oil.

Explanation: the ping pong ball doesnt sink through the lamp oil.

Answer:

less precise than

Explanation:

Answer:

less precise than

Explanation:

its right on edge2020

Answer:

The answer is false

Explanation:

Though the mostly used SI unit of measurement or the most popular units are the

Length,

Time and

Mass

i.e meter (m), seconds (s), kilogram (kg)

Aside all the above stated units for measurements there are other four basic units which are itemized  bellow.

they are

1. Amount of substance - mole (mole)

2. Electric current - ampere (A)

3. Temperature - kelvin (K)

4. Luminous intensity - candela (cd)

Complete Question

A boy throws a ball on a spring scales which oscillates about the equilibrium position with a period of T = 0.5 sec. The amplitude of the vibration is A = 2 cm. Assume the ball does not bounce from the scales’s surface afterwards. Assume the vibration of the scale is expressed mathematically as x(t) = Acos(t + ). Find:

a) frequency

b) the maximum acceleration

c) the maximum velocity

Answer:

a

[tex]f =  2  \ Hz[/tex]

b

[tex]a_{max} =  3.2 \ m/s^2[/tex]

c

[tex]v_{max} =  0.25 \  m/s[/tex]

Explanation:

From the question we are told that

   The period is  [tex]T  =  0.5 \ sec[/tex]

    The  amplitude is  [tex]A =  2 \ cm  =  0.02 \ m[/tex]

    The  vibration of the scale is  [tex]Acos(wt +  \phi )[/tex]

Generally the frequency is mathematically represented as

   [tex]f =  \frac{1}{T}[/tex]

=>   [tex]f =  \frac{1}{0.5}[/tex]

=>    [tex]f =  2  \ Hz[/tex]

The  maximum acceleration is mathematically represented as

    [tex]a_{max} =  A *(2 \pi f)^2[/tex]

=>    [tex]0.02 *  (2 * 3.142  *  2)^2[/tex]

=>  [tex]a_{max} =  3.2 \ m/s^2[/tex]

The  maximum velocity is mathematically represented as  

     [tex]v_{max} =  A *  (2 \pi f)[/tex]

=>    [tex]v_{max} =  0.02 *  (2 * 3.142 *  2)[/tex]

=>    [tex]v_{max} =  0.25 \  m/s[/tex]

Answer:

The change in momentum is:  [tex]13.5\,\frac{kg\,m}{s}[/tex]

Explanation:

Let's define that vectors pointing up are positive, and vectors pointing down negative.

Then we express the initial momentum of the ball as the product of its mass times the velocity, and include the negative sign since this momentum is pointing down (velocity vector is pointing down):

[tex]P_i=-0.5\,(15) \frac{kg\,m}{s} =-7.5\, \frac{kg\,m}{s}[/tex]

the final momentum is positive (pointing up) and given by the product:

[tex]P_f=0.5\,(12) \frac{kg\,m}{s} =6\, \frac{kg\,m}{s}[/tex]

Therefore, since the change in momentum is defined as the difference between the final momentum minus the initial one, we get:

[tex]P_f-P_i=6-(-7.5)\,\frac{kg\,m}{s} =13.5\,\frac{kg\,m}{s}[/tex]

Answer:

b. disorder in the universe is increasing with the passage of time

Explanation:

The second law of thermodynamics states that the total entropy of an isolated system  increases over time . The entropy of a system measures the randomness or disorder  in the system . So the we can state,  in terms of disorder , the second  law of thermodynamics as follows .

The disorder in the universe is increasing with the passage of time .

The second law of thermodynamics leads us to conclude that disorder in the universe is increasing with the passage of time. So, here, option b is the correct option.

Entropy is a measure of the disorder or randomness in a system. The second law of thermodynamics states that in any spontaneous process, the total entropy of an isolated system (including both the system and its surroundings) tends to increase or, at best, remain constant. This means that over time, natural processes tend to lead to an increase in the overall disorder or randomness of the system. It's important to note that while the increase in entropy is a statistical tendency on a macroscopic scale, there can still be localized decreases in entropy within a system. 

Learn more about the law of thermodynamics here.

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Answer:

False

Explanation:

If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges

Answer:

  Q_total = 2 10⁻⁶ C

Explanation:

Let's apply the conditions of static equilibrium to this case, in the adjoint we can see a diagram of the forces.

X axis

    Tₓ - Fe = 0

    Tₓ = Fe

Y axis  

    [tex]T_{y}[/tex] - W = 0

    T_{y} = W

Let's use trigonometry to find the stress components, the angle is measured with respect to the vertical

    sin 10 = Tₓ / T

    cos 10 = T_{y} / T

    Tₓ = T sin 10

    T_{y} = T cos 10

we substitute

    T sin 10 = Fe

    T cos 10 = W

     T = mg / cos10

     T = 0.120 9.8 / cos 10

     T = 1,194 N

     Fe = 1,194 sin 10

     Fe = 0.2073 N

the electric force is

     F = k q₁q₂ / r²

in this case, as the cans touch, they have the same charge and the distance r is searched for by trigonometry

     sin 10 = y / L

     y = L sin 10

     y = 0.30 sin 10

     y = 0.052 m

this is the distance from the vertical to one can the distance between the two cans is

     y_total = 2y

     y_totlal = 2  0.052 = 0.104 m

     Fe = F = k q² / r²

     q = √ (F r² / k)

       q = √ (0.2073 0.104²/9 10⁹) = √ (24.913 10⁻¹⁴)

       q = 4.99 10⁻⁷ C

This is the charge of a can, as the transfer is carried out by contact, the flannel transfers half of its charge to the cans and these when separating face one keeps half of the transferred charge, therefore the total charge of the flannel is

     Q_total = 4 q

     Q_total = 19.97 10⁻⁷ C

     Q_total = 2 10⁻⁶ C

Answer:

A. Using

I= P/ A

So = P/ 2π R²

So=777E3 watt/ 2*πx (²5.5E3m)

4.3*10^ -3W/m²

So pressure= 2I/c

=2* 4.3*10^ -3W/m²/ 3*10^ 8

2.7*10^-11p.a

B.

Using Emax= √2I/Eoc

So

√ 2*4.3*10^-3/8.8E-12*3*10^8

= 5.48V/m

So

Bmax= Emax/c

5.58/3E8

= 1.83*10-9T

D.average density is given by

= Eo(Emax/√2)²

= 8.85E-12(5.48/√2)²

1.3*10^10W/m²

Answer:

it is B 1:50

Explanation:

just did it on apex