PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium

Answer:

Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel [tex]N_1=2\ rev/s[/tex]

angular speed [tex]\omega_1=2\pi N_1=4\pi\ rad/s[/tex]

mass of wheel [tex]m_1=4.5\ kg[/tex]

diameter of wheel [tex]d_1=0.30\ m=30\ cm[/tex]

radius of wheel [tex]r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm[/tex]

mass of clay [tex]m_2=2.8\ kg[/tex]

the radius of the chunk of clay [tex]r_2=8\ cm[/tex]

Moment of inertia of Wheel

[tex]I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2[/tex]

Combined moment of inertia of wheel and clay chunk

[tex]I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2[/tex]

Conserving angular momentum

[tex]\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi[/tex]

Common frequency of wheel and chunk of clay is

[tex]\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s[/tex]

Answer:

Rainy

Explanation:

It is the only one that has nothing to do with the case

Answer: Hello, Mark me as Brainliest! :)

If soldiers march across the bridge with a cadence equal to the bridge's natural frequency and impart $$1.00 × 10^4 J$$ of energy each second, how long does it take for the bridge's oscillations to go from 0.100 m to 0.500 m amplitude. $ 5 \times 10^7 \text{J} $ . \\ b) $ 12 \times 10^4 \text{s}$ .

Your Welcome!

Explanation:

Answer:

biology

Explanation:

Animals and plants

Answer:

1) 204.6663938

2) 19.51219512

3) 368

4) 135.81

5) 43932240

6) 63830.2

7) 2470.8

8) 1.718213058

9) 1005550.526

10) 586.365

Most of the questions you asked were in repeating decimal form.

Explanation:

Answer:

height

weight of ball

time of ball falling

Answer:

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Explanation:

We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.

Formula for resistance is;

R = V/I

R = 12.6/2.07

R = 6.087 ohms

Since R1 and R2 are connected in series.

Thus; R1 + R2 = 6.087 ohms

R1 = 6.087 - R2

We are also told that when they are connected in parallel, the current is 8.98 A.

Thus, R = 12/8.98

R = 1.403 ohms

Thus;

(1/R1) + (1/R2) = 1/1.403

Let's put 6.087 - R2 for R1;

(1/(6.087 - R2)) + (1/R2) = 1/1.403

Multiply through by 1.403R2(6.087 - R2) to get;

1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)

Expanding gives;

1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²

(R2)² - 6.087R2 + 8.54 = 0

Using quadratic formula, we have;

R2 = 2.193 ohms or 3.894 ohms

Thus,

R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894

R1 = 3.894 or 2.193

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Answer:

Gamma Rays have the highest frequencies

Explanation:

This is because Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies compared to the other light frequency. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation which means the answer has to be gamma rays. Brainly Please!!!! Here are screenshots that may help

Answer:

102.5N

Explanation:

Given that  a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W

The power = 4.12 × 746 = 3073.52 W

Using the formula

Power = force × velocity

3073.52 = force × 30

Force = 3073.52 / 30

Force = 102.5 N

Since most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road, therefore,

the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s is 102.5 N

Answer:

the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space

Explanation:

Answer:

look pkch

Explanation:

Answer: The daughter is named Susie.

Explanation: LIL SUSIE!!!

                      HUH? DIDN'T UNDERSTAND THE QUESTION!

                                        HAVE A GREAT DAY!!!!!

Answer:6/3 Li

Explanation:

I’m not sure what the person under me is talking about but yeah

Answer:

true

Explanation:

force or powerbecause he pushes a disk

Answer:

The answer should be D

Explanation:

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

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

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]

Therefore, the electric field strength at the mid-point between the two rings is zero.

Answer:

[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]

[tex]44.43^{\circ}[/tex], second order does not exist

Explanation:

n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]

[tex]\lambda[/tex] = Wavelength

m = Order

Distance between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]

[tex]\lambda=400\ \text{nm}[/tex]

m = 1

We have the relation

[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]

The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].

[tex]\lambda=700\ \text{nm}[/tex]

m = 1

[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]

Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.

The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].

Second order angle does not exist.

Answer:

a)  Wapp = 520 N

b)  ΔUf = 447 N

c) Wn = 0

d) Wg = 0

e) ΔK = 73 J

f) v = 1.96 m/s

Explanation:

a)

       [tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]

b)

       [tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]

       [tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]

c)

d)

e)

        [tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]

f)

       [tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]

a) The work done by the applied force   [tex]W_{AP}=520\ J[/tex]

b) The change in the internal energy [tex]\Delta U=447\ J[/tex]

c) Work done by normal force  [tex]W_n=0[/tex]

d) Work done by gravitation   [tex]W_g=0[/tex]  

e) The change in KE will be [tex]\Delta KE=73\ J[/tex]

f) The final speed v = 1.96 m/s

The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.

If the object achieve movement due to the work then the energy in the object will be kinetic energy.

If the object attains some height  against the gravity then the energy in the object will be potential energy.

Now it is given in the question that

The horizontal force   [tex]F_h=130\N[/tex]

mass of the object  m= 38 kg

Coefficient of friction [tex]\mu=0.3[/tex]

Displacement of the object [tex]\delta x=4\ m[/tex]

(a) The work done will be

[tex]W=F_h\tines \Delta x[/tex]

[tex]W=130\times 4=520\ J[/tex]

(b) The increase in the internal energy

The increase in the internal energy of the box is due to the energy generated by the force of friction so

[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]

here  [tex]F_f[/tex] is the frictional force and is given as

[tex]\mu=\dfrac{F_f}{R}[/tex]

Here R is the normal reaction and its value will be weight of the box in opposite direction.

[tex]\mu=\dfrac{F_f}{-mg}[/tex]

[tex]F_f=-mg\times \mu[/tex]

[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]

[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]

(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.

(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.

(e) The change in the KE of the box.

The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be

[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]

(f) The speed of the box

The KE of the box will be

[tex]KE=\dfrac{1}{2} mv^2[/tex]

[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]

[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]

Thus

a) The work done by the applied force   [tex]W_{AP}=520\ J[/tex]

b) The change in the internal energy [tex]\Delta U=447\ J[/tex]

c) Work done by normal force  [tex]W_n=0[/tex]

d) Work done by gravitation   [tex]W_g=0[/tex]  

e) The change in KE will be [tex]\Delta KE=73\ J[/tex]

f) The final speed v = 1.96 m/s

To know more about Work and energy follow

https://brainly.com/question/25959744

Answer:

The angular speed is 23.24 rad/s.

Explanation:

Given;

mass of the disk, m = 7 kg

radius of the disk, r = 0.2 m

applied force, F = 42 N

distance moved by disk, d = 0.9 m

The torque experienced by the disk is calculated as follows;

τ = F x d = I x α

where;

I is the moment of inertia of the disk = ¹/₂mr²

α is the angular acceleration

F x r = ¹/₂mr² x α

The angular acceleration is calculated as;

[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]

The angular speed is determined by applying the following kinematic equation;

[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]

initial angular speed, ωi = 0

angular distance, θ = d/r = 0.9/0.2 = 4.5 rad

[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]

Therefore, the angular speed is 23.24 rad/s.

Answer:

35 N

Explanation:

F = ma

centripetal force = 10(3.5) = 35 N

Answer:

F = 56 N

Explanation:

       [tex]F = \frac{W}{\Delta x} =\frac{560J}{10m} = 56 N (1)[/tex]

Answer:

static force

Explanation:

mark me brainliest

Answer:

B. kinetic energy

Explanation:

Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.

Answer:

1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa

2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]

Explanation:

1. From Boyles' law;

[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]

[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa

[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]

[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]

Thus,

1 x [tex]10^{5}[/tex] x 9.6 =  [tex]P_{2}[/tex] x 12

 [tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]

     = 80000

[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa

2. Pressure, P = ρhg

where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).

Thus,

P = 1020 x 15 x 9.8

  = 149940

P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]

Answer:

1)   x = x₀ + vot - ½ c t² - 1/6 bt³,    v = v₀ - ct - ½ b t²

2)   v₁ = 5.25 m/s,         v₂ = -8 m/s

Explanation:

1) For this exercise, the relationship of the body is not constant, so you must use the definition of speed and position to find them.

acceleration is

           a = c + bt

a) the relationship between velocity and acceleration

           a = [tex]\frac{dv}{dt}[/tex]

           dv = -a dt

The negative sign is because the acceleration is contrary to the speed to stop the vehicle.

we integrate

           ∫ dv = - ∫ a dt

           ∫ dv = -∫ (c + bt) dt

            v = -c t - ½ b t²

This must be valued from the lower limit, the velocity is vo, up to the upper limit, the velocity is v for time t

             v - v₀ = -c (t-0) - ½ b (t²-0)

             v = v₀ - ct - ½ b t²

b) the velocity of the body is

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

             dx = v dt

we replace and integrate

              ∫ dx = ∫ (v₀ - c t - ½ bt²) dt

              x-x₀ = v₀ t - ½ c t² - ½ b ⅓ t³

Evaluations from the lower limit the body is at x₀ for t = 0 and the upper limit the body is x = x for t = t

           x - x₀ = v₀ (t-0) - ½ c (t²-0) + [tex]\frac{1}{6}[/tex]  (t³ -0)

           x = x₀ + vot - ½ c t² - 1/6 bt³

2) The speed when you reach the traffic light

Let's write the data that indicates, the initial velocity is vo = 15 m / s, the initial position is xo = 315m, let's use the initial values ​​to find the constants.

       t = 25 s x = 20

we substitute

          20 = 315 + 15 25 - ½ c 25² - 1/6 b 25³

         0 = 295 + 375 - 312.5 c - 2604.16 b

         670 = 312.5 c + 2604.16 b

we simplify

         2.144 = c + 8.33 b

Now let's use the equation for velocity,

        v = v₀ - ct - ½ b t²

        v = 15 - c 25 - ½ b 25²

        v = 15 - 25 c - 312.5 b

let's write our two equations

        2.144 = c + 8.33 b

        v = 15 - 25 c - 312.5 b

Let's examine our equations, we have two equations and three unknowns (b, c, v) for which the system cannot be solved without another equation, in the statement it is not clear, but the most common condition is that if the semaphore does not change, it follows with this acceleration (constant) to a stop

               a = c + b 25

from the first equation

              c = 8.33 / 2.144 b

              C = 3.885 b

we substitute in the other two

            v = 15 - 25 (3.885 b) - 312.5 b

            v = 15 - 409.6 b

final acelearation

            a = 28.885 b

let's use the cinematic equation

               [tex]v_{f}^2[/tex]= v² - 2 a x

                0 = v² - 2a 20

               0 = v² - (28.885b) 40

               v² = 1155.4 b

we write the system of equations

               v = 15 - 409.6 b

               v² = 1155.4 b

resolve

              v²= 1155.4 ( [tex]\frac{15 -v }{409.6}[/tex] )

              v² = 2.8 ( 15 -v)

              v² + 2.8 v - 42.3 = 0

              v=  [ -2.8 ±[tex]\sqrt {2.8^2 + 4 \ 42.3) }[/tex] ]/2 = [-2.8 ± 13.3]/2

              v₁ = 5.25 m/s

              v₂ = -8 m/s