if you make a sound by tapping on a glass of water what is the order of vibration
A. glass - water - air
C. air - water - glass
B. water - glass - air

Answer:

Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2

Explanation:

See the attached files.

To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.

Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.

The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.

Answer:

Covalent bonds can form between similar atoms.

Explanation:

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

Answer:

 v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]

Explanation:

For this exercise we must use the projectile launch ratios, let's start by finding the time it takes to reach the bottom of the cliff, the initial vertical velocity is zero

          y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

at the bottom of the cliff y = 0 and as the body is thrown horizontally the initial vertical velocity is zero

          0 = y₀ + 0 - ½ g t²

          t = [tex]\sqrt{2y_o/g}[/tex]

this time is the same as the horizontal movement.

Let's use Newton's second law to find the acceleration on this x-axis due to the force of the air

           F = m aₓ

they tell us that force is equal to the weight of the body

           -mg = maₓ

           aₓ = -g

the sign indicates that the acceleration is to the left

we write the kinematics equation

          x = x₀ + v₀ₓ t + ½ aₓ t²

They indicate that the final position is the foot of the cliff (x = 0), when it leaves the top it is at x₀ = 0 and has a velocity v₀ₓ = v

we substitute

          0 = 0 + v t + ½ (-g) t²

          v = ½ g t

we use the drop time

          v = ½ g [tex]\sqrt{\frac{2yo}{g} }[/tex]

          v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]

Answer:

eeeeeeeeeeeeeeeeeeeee

Answer:

31.68 m/s

Explanation:

The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.

As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:

[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]

Now we just add the time accelerating:

[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]

Answer:

mechanical wave

Explanation:

The sound vibrations cause waves of pressure that travel through a medium, such as air, water, wood or metal. Sound energy is a form of mechanical energy

Answer:

A mechanical waves

Explanation:

in nature include water waves, sound waves, seismic waves and internal water waves

Answer:

c, its axis is tilted

maybe

As it works its way around the sun, its tilted axis exposes different parts of earth.

C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis

Answer:

θ = 90º

Explanation:

The velocity is given by

          v = [tex]\frac{dr}{dt}[/tex]

calculate

          v = 3 i ^ + √2 j ^ + 2t k ^

acceleration is defined by

         a = dv / dt

         a = 2 k ^

one way to find the angle is with the dot product

         v. a = | v | | a | cos θ

         cos θ= v.a / | v | | a |

Let's look for the value of each term

       v. a = 4 t

        | v | = [tex]\sqrt{3^2 + 2 + (2t)^2 }[/tex] = [tex]\sqrt{ 11 + 4t^2}[/tex]

        | a | = 2

they ask us for the angle for time t = 0

         v. a = 0

        | v | = √11 = 3.317

we substitute

        cos θ = 0 /√11

        cos θ = 0

therefore the angles must be θ = 90º

Answer:

Impulse of force = 300Ns

Explanation:

Given the following data;

Mass = 25kg

Change in velocity = 12m/s

To find the impulse;

Impulse is given by the formula;

[tex] Impulse \; of \; force = mass * change \; in \; velocity [/tex]

Substituting into the equation, we have;

[tex] Impulse \; of \; force = 25 * 12 [/tex]

Impulse of force = 300Ns

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ([tex]a_c[/tex]) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ([tex]a_c[/tex]) is given by:

[tex]a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s[/tex]

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

Answer:

0.705 m/s²

Explanation:

a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.

Using newton's law of motion:

v² = u² + 2as

v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h,  s = distance = 67 m

9.72² = 0² + 2a(67)

134a = 94.484

a = 0.705 m/s²

b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:

v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m

v² = u² + 2as

9.72² = 9.72² + 2a(88)

176a = 9.72² - 9.72²

a = 0

c) During the last distance, the speed slows down from 35 km/h to 32 km/h.

u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m

v² = u² + 2as

8.89² = 9.72² + 2a(45)

90a = 8.89² - 9.72²

90a = -15.4463

a = -0.1716 m/s²

The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.

Answer:

a) The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.

b) the net electric flux is  -821.09 N.m²/c

c) the net charge inside the cylinder is -7.2666 nC

Explanation:

Given the data in the question;

a) (a) What type of charge distribution is inside the surface?

Based on the Image of the question below;

The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.

b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface?

we know that; Electric flux is;

∅ = -2 × E × Area

we substitute

∅ = -2 × 300 × π ( 0.66 m )²

= -821.09 N.m²/c

Therefore; the net electric flux is  -821.09 N.m²/c

c) What is the net charge inside the cylinder? nC

from Gause's law;

∅ = q/∈₀

q = ∅ ∈₀

we know permittivity ∈₀ = 8.85 × 10⁻¹²)

we substitute

q = -821.09 N.m²/c  × (8.85 × 10⁻¹²)

q =  -7.2666 × 10⁻⁹ C

q = -7.2666 nC

Therefore, the net charge inside the cylinder is -7.2666 nC

Answer:

I think A

Explanation:

because I don't think a unknowned number can be a division problem

Answer:

- the Thevenin equivalent circuit for the battery is uploaded below

- the battery delivered 117 watts of power to the starter motor

Explanation:

Given the data in the equation

diagram of the Thevenin equivalent circuit for the battery is uploaded below.

Current I = 10 A

Voltage 1 = 12 V

voltage 2 = 11.7 v

R = (V1 - V2) / I

R = (12-11.7)/10

R = 0.3 / 10

R = 0.03Ω

Thevenin equivalent circuit

[tex]R_{L}[/tex] = V2 / I = 11.7 / 10

[tex]R_{L}[/tex] = 1.17Ω

so, power delivered to the starter motor will be;

p = (V2)² / [tex]R_{L}[/tex]

P = ( 11.7 V )² / 1.17Ω

p =  136.89 / 1.17

p =  117 watts

Therefore, the battery delivered 117 watts of power to the starter motor

Answer:

has a charge

Explanation:

Answer:

gain electrons

Explanation:

I looked it up and that was what I found

side note: ions do have a charge

Answer:

250N same direction

Explanation:

100+150 = 250N

same direction

Answer:

C. The star is receding from you with the speed of 3000 km/s

Explanation:

To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.

We solve for V

V = 3x10⁶m/s

V = 3000km/s

We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.

Thank you!

Answer:

13.4436

Explanation:

Answer:

22

Explanation:

before:

2m

What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?

(Change in momentum - momentum after-

after:

2m

momentum before) p - mv

Answer:

A

Explanation:

In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.

Hope this helps!

Answer:

a

Explanation:

AAAA

Answer:

Explanation:

Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.

Answer:

Explanation:

We shall first calculate electric potential at the centre of square due to four charges of Q .

potential = k Q / r , k = 9 x 10⁹ , Q is charge which is placed at distance r from the point where potential is measured .

Distance of corner of square to centre = √ 2 a / 2 = a / √2 = .707 a

potential due to one charge

= k x Q / .707 a

= 9 x 10⁹ x Q / .707a

= 12.73 x 10⁹ Q/a

Potential due to 4  charges

= 4 x 12.73 x 10⁹ Q/a

= 50.92 x 10⁹ Q/a .

Potential at infinity = 0

Work done in carrying charge q from infinity to centre of square

= potential difference x charge

= ( 50.92 x 10⁹ Q/a - 0 ) x q

= 50.92 x 10⁹ x Qq / a  J .

Answer:

1.875 x 10⁶ m /s .

Explanation:

Force on electron = E e where E is electric field and e is charge on electron

acceleration generated = Ee / m where m is mass of the electron .

Putting the values

acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹

= .879 x 10¹² m /s²

v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m

v² = 0 + 2 x .879 x 10¹² x 2

v = 1.875 x 10⁶ m /s .

Answer:

Magnitude of it's displacement = 1,000 km

Explanation:

Given:

Distance towards east = 800 km

Distance towards north = 600 km

Find:

Magnitude of it's displacement

Computation:

Magnitude of it's displacement = √800² + 600²

Magnitude of it's displacement = √640000 + 360000

Magnitude of it's displacement = √10000000

Magnitude of it's displacement = 1,000 km

Answer:

a)   #_total = 4 10²³ photons / h,  b) # _photon_area = 3  10²² photons

Explanation:

a) Let's start by calculating the energy of an emitted photon

         E₀ = h f

         c = λ f

substituting

          E₀ = h c /λ

          E₀ = [tex]\frac{6.62 \ 10^{-34} \ 3 \ 10^{8} }{550 \ 10^{9} }[/tex]

          E₀ = 3.6 10⁻¹⁹ J

Let's use a direct proportion rule (res rule) if a photon has Eo, how many photons are there in 55W

          #_foton = 55 / E₀

          # _photon = 55 / 3.6 10⁻¹⁹ = 15.27 10¹⁹ photons

This version is with 100% if the conversion is 75%, how many footnes are there

         #_foton_real = # _foton 75/100

         #real_photon = 15.27 10¹⁹ 0.75

         # _real_photon = 11.45 10¹⁹ photons

this is the broadcast in a second

         #_total = # _real_photon   t

         #_total = 11.45 10¹⁹ 3600

         #_total = 4.1 10²³ photons / h

         #_total = 4 10²³ photons / h

b) This number of photons is constant, so after being emitted they are distributed on the surface of a sphere, in this case of radius r = 1.0 m

the volume of a sphere is

          A = 4π r²

          A = 4π 1²

          A = 12,566 m²

the area of ​​the plate is

           A₁ = l₁ l₂

           A₁ = 0.10 0.10

           A₁ = 1 10⁻² m

Let's use a direct proportion rule, if there are 4.1 10²³ photons in an area A, how much are there in an area A₁

          # _photon_area = #_total   A₁ / A

          # _photon_area = [tex]4.1 \ 10^{23} \ \frac{1 \ 10^{-2} }{12.566}[/tex]

          # _photon_area = 3.26  10²²

as the number of photons must be a whole number

            # _photon_area = 3  10²² photons

Answer:

Current through them, voltage across them.

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

[tex] V = IR[/tex]

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

A resistor can be defined as an electronic component that opposes the free flow of current from one point to another in an electrical circuit.

In a series circuit of two resistors, the resistors have the same current through them; in a parallel circuit of two resistors, the resistors have the same voltage across them.

Answer:

the banking angle of the road is 24.2⁰

Explanation:

Given;

speed of the vehicles considered, v = 75 mi/h

Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s

                         75 mi/h --------> ?

=   75 x 0.44704 m/s = 33.528 m/s

radius of the curve, r = 255 m

The banking angle of the road is calculated as;

[tex]\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0[/tex]

Therefore, the banking angle of the road is 24.2⁰

The angle of banking is 24 degrees.

As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.

The angle of banking depends on the speed of the vehicle and the radius of the curve.

θ = v^2/rg

speed = 75.0 mi/h or 33.5 m/s

r = 255 m

g = 9.8 ms-1

θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1

θ =   tan-1(1122.3/2499)

θ =  24 degrees

Learn more about banking: https://brainly.com/question/12134554