What is the frequency of this wave?

can someone pls explain and answer ​

Heya!!

For calculate aceleration, let's applicate second law of Newton:

[tex]\boxed{F=ma}[/tex]

⇒ Being:

→ F = Force = 12 N

→ m = Mass = 3 kg

→ a = aceleration = ?

Lets replace according formula and leave the "a" alone:

[tex]12\ N = 3\ kg * \textbf{a}[/tex]

[tex]\textbf{a} = 12\ N / 3\ kg[/tex]

[tex]\textbf{a} = 4\ m/s^{2}[/tex]

Result:

The aceleration of the object is of 4 m/s²

Answer: the car that is moving downhill

Explanation:

Answer:

The fundamental resonance frequency is 172 Hz.

Explanation:

Given;

velocity of sound, v = 344 m/s

total length of tube, Lt = 1 m = 100 cm

height of water, hw = 50 cm

length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm

For a tube open at the top (closed pipe), the fundamental wavelength is given as;

Node to anti-node (N ---- A) : L = λ / 4

λ = 4L

λ = 4 (50 cm)

λ = 200 cm = 2 m

The fundamental resonance frequency is given by;

[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]

Therefore, the fundamental resonance frequency is 172 Hz.

Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.

The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.

Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have

• net parallel force

∑ Force (//) = W (//) - F = m a

(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)

• net perpendicular force

∑ Force (⟂) = W (⟂) + N = 0

Notice that

W (//) = W sin(θ) … … … which is positive since it points down the plane

W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N

So the equations become

W sin(θ) - F = m a

-W cos(θ) + N = 0

Solving for a gives

a = (W sin(θ) - F ) / m

which is good enough if you know the magnitude of the friction force.

If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as

F = µ N

so that

a = (W sin(θ) - µ N ) / m

and the normal force itself has a magnitude of

N = W cos(θ)

so that

a = (W sin(θ) - µ W cos(θ) ) / m

The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so

a = (m g sin(θ) - µ m g cos(θ) ) / m

a = g (sin(θ) - µ cos(θ))

Answer:

Fx = 467.65N

Fy = 270N

Explanation:

Given

Force = 540N

angle of inclination = 40 degree

Horizontal component Fx = Fcos 30

Fx = 540cos30

Fx = 540(0.8660)

Fx = 467.65N

Hence the horizontal component is 467.65N

Vertical component Fy = Fsin 30

Fy = 540sin30

Fy = 540(0.5)

Fy = 270N

Hence the vertical component is 270N

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

Answer:

25

Explanation:

Answer:

it’s c not d

Explanation:

took the test

Answer: D!!!

Explanation: jus got it wrong from the other answer.

Answer:

What?

Explanation:

Answer:

t = 66.7 s

Explanation:

Given that,

Speed of a hare, v = 15 m/s

Speed of a turbocharged tortoise, v' = 3 m/s

The hare in a 250 meter race

Let the Hare takes time t. It can be calculated as follows :

[tex]t=\dfrac{250}{15}=16.67\ s[/tex]

Let a turbocharged tortoise takes t'. It can be calulated as follows :

[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]

To tie the race, required time is given by :

[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]

Hence, the correct option is (b) i.e. 66.7 seconds.

[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]

Answer:

The half-life is [tex] t_{1/2} = 1.005 h[/tex]

Explanation:

Using the decay equation we have:

[tex]A=A_{0}e^{-\lambda t}[/tex]

Where:

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]

[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]

[tex]0.5=e^{-\lambda*1 h}[/tex]

Taking the natural logarithm on each side we have:

[tex]ln(0.5)=-\lambda[/tex]

[tex]\lambda=0.69 h^{-1}[/tex]

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]

[tex] t_{1/2} = 1.005 h[/tex]

I hope it helps you!

Answer:

1.2 kg

__________________________________________________________

We are given:

Mass of the block = 2 kg

Coefficient of Static Friction = 0.6

__________________________________________________________

Friction Force on the Block:

Finding the Normal Force:

We know that the normal force will be equal and opposite to the weight of the 2 kg block

So, Normal Force = mg

replacing the variables with the given values

Normal Force = (2)(9.8)                    [Taking g = 9.8]

Normal Force = 19.6 N

Friction force on the Block:

We know that:

Coefficient of Static Friction =  Static Friction Force/Normal Force

replacing the variables

0.6 = Static Friction force / 19.6

Static Friction force = 0.6*19.6 N                 [Multiplying both sides by 19.6]

Static Friction force = 11.76 N

__________________________________________________________

Largest Mass that can Hang:

We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest

So, if the weight of the second block is less than the static friction force, it will hang

Weight of the second block ≤ 11.76

We know that weight = mg

mg ≤ 11.76

m(9.8) ≤ 11.76                                                   [since g = 9.8]

m ≤ 1.2 kg                                                        [dividing both sides by 9.8]

From this, we can say that the maximum mass of the second block is 1.2 Kg

Answer:

15N

Explanation:

F=ma so F=.3*50 therefore F=15N

The force of an egg that is thrown at a brick wall is equal to 15 N.

Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.

The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:

F = ma

Given, the mass of the egg, m = 0.3 Kg

The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²

The force of an egg that is thrown at a brick wall can be calculated as:

F = ma = 50 ×0.3 = 15 N

Learn more about force, here:

brainly.com/question/13191643

#SPJ2

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 6 × 4

We have the final answer as

Hope this helps you

Answer:

F = 11226.02 N

Explanation:

The roller coaster  car and passengers have a combined mass of 1620kg.

They descend the first hill at  an angle of 45.0 degrees to the horizontal.

We need to find force is the rollercoaster pulled  down the hill.

We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,

F = mgsinθ

= 1620 × 9.8 × sin(45)

= 11226.02 N

So, the roller coaster is pulled down the hill with a force of 11226.02 N.

Answer: 0.3m/s/s

(i'm really sorry if i'm wrong)

:(

Answer:

5.79

Explanation:

Vf=Vi+at

0=8.33+(-9.8)sin8.44t

t=8.33/(9.8sin8.44)=0.83/sin 8.44

=5.79

Answer:

The bell has a potential energy of 8550 [J]

Explanation:

Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.

[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]

Answer:

  about 2398 seconds

Explanation:

The relation between time, distance, and acceleration is ...

  d = (1/2)at²

  t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h

That is about ...

  (0.6660 h)(3600 s/h) ≈ 2397.7 s

The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

Answer:

RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]

Explanation:

We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.

The formula for RMS speed of a gas is given by :

[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]

Where

R is ideal gas constant, R = 8.314 J /mol K

T = 5300 K

m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]

Substituting all the values in above formula :

[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]

So, the RMS speed Helium atoms 5748.75 m/s.

Answer:10000000

Explanation:

Answer:

18.9s

Explanation:

Using the formula;

ω = v/r

Where;

ω = angular velocity (rad/s)

v = linear velocity (m/s)

r = radius of the circular track (m)

According to the given information, v = 30m/s, r = 90m

ω = v/r

ω = 30/90

ω = 3/9

ω = 0.3333 radians/seconds.

Since ω = 2π/T

Where;

π = 3.142

T = period (s)

ω = angular velocity

0.333 = 2 × 3.142/T

T = 2 × 3.142/0.333

T = 6.284/0.333

T = 18.87s

T = 18.9s

Answer:

the forces are the same

Explanation:

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is  [tex]N = 109 \ rev[/tex]      

Explanation:

From the question we are told that

    The speed of the car is  [tex]u = 28.4 \ m/s[/tex]

     The constant deceleration experienced is  [tex]a = 1.92 \ m/s^2[/tex]

      The radius of the tire is  [tex]r = 0.307 \ m[/tex]

Generally from kinematic equation we have that

      [tex]v^2 = u^2 + 2as[/tex]

Here  v is the final velocity which is  0 m/s

   So

         [tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]

=>      [tex]s = 210.04 \ m[/tex]

Generally the circumference of the tire is mathematically represented as

         [tex]C = 2 \pi r[/tex]

=>      [tex]C = 2 * 3.142 * 0.307[/tex]    

=>      [tex]C = 1.929 \ m[/tex]

Generally the number of revolution is mathematically represented as

         [tex]N = \frac{ s}{C}[/tex]    

=>     [tex]N = \frac{210.04}{1.929}[/tex]

=>     [tex]N = 109 \ rev[/tex]      

Answer:

The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.

Explanation:

Hope this helps Plz mark brainliest

Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v

where v is the velocity of the 2.22 kg block after collision. Solve for v :

6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v

(2.22 kg) v = 2.70 kg•m/s

v = (2.70 kg•m/s) / (2.22 kg)

v ≈ 1.22 m/s

i.e. a velocity of about 1.22 m/s to the right.