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faust18 [17]
2 years ago
7

We have seen that when you generate a wave on a stretched spring, as long as the medium doesn’t change (that is, the tension and

the linear mass density remain the same), the speed of the wave on the spring is constant. Does this mean that the particles of the spring always have zero acceleration? Explain why or why not.
Physics
1 answer:
TiliK225 [7]2 years ago
6 0

Answer:

No, the acceleration is not always zero.

Explanation:

It does not mean that the acceleration of the particle is zero.

The velocity of wave is different from the velocity of particle.

The acceleration of wave is different from the acceleration of particle.

the acceleration of the particle is given by

a =- w^2 y

where, w is the angular frequency and y is the displacement from the mean position.

So, the acceleration is zero at mean position only and it varies as the position changes.  

You might be interested in
Researchers in the Antarctic measure the temperature to be -32°F. What is this temperature on the following scales?(a) the Celsi
Illusion [34]
Answer:

a) -35.6°C

b) 237.4 K

Explanation:

To convert temperature from degree celsius to degree fahrenheit, use the formula below:

T_c=\frac{5}{9}(T_f-32)

a) Therefore to convert -32°F to celsius, substitute it into the celsius

\begin{gathered} T_c=\frac{5}{9}(-32-32) \\  \\ T_c=\frac{5}{9}(-64) \\  \\ T_c=-35.6^0C \end{gathered}

b) To covert to the Kelvin scale, use the formula below

\begin{gathered} T_k=T_c+273 \\  \\ T_k=-35.6+273 \\  \\ T_k=237.4K \\  \end{gathered}

8 0
1 year ago
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert
Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, Vi and Ui are its initial vertical and horizontal components

R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi   = 10*sin(30) = 5 m/s\\Ui  = 10*cos(30) = 8.66 m/s

To find:

Max Height h achieved

Calculation:

1) Using the 3^{rd} equation of motion, we know

2*a*s = Vf^{2}  - Vi^{2}

2) In terms of gravity g height h and  the vertical component of Velocity Vf , Vi.

3) As Vf = 0 as at maximum height the vertical component of velocity is zero maximum height achieved

2*g*h = Vf^{2}  -Vi^{2}

putting values

4) h = 0.784 m/s

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0
3 years ago
Someone got this paper?
Snezhnost [94]
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.

Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms

Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A

Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.

Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A

Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
6 0
3 years ago
A person is lifting a heavy box using a lever. What is the purpose of the lever in this situation?
ruslelena [56]

Answer:

to reduce the <em>force</em> needed to lift the box and <em>change</em> the direction of the force

Explanation:

1. "A lever consists of a rigid bar that is able to pivot at one point. This point of rotation is known as the fulcrum. A force is applied at some point away from the fulcrum (typically called the effort)."

By this definition, we know that force is needed to lift an object using a lever.

2.<u> "When the input and output forces are on opposite sides of the fulcrum, </u><u>the lever changes the direction of the applied force.</u> This occurs only with first-class levers. When both the input and output forces are on the same side of the fulcrum, the direction of the applied force does not change"

For example, on a sew saw, if a force is applied on one end, you on the other side/end would go up, meaning <u>a change in direction</u>.

3. Lastly, we know <u><em>a lever is typically used to reduce work</em></u>, in other words, the force needed to move something.

Basically, if we were to put a lever into an equation:

reduced force + change in direction = lever

(<em>the expection</em>) <u>unless load and force are on the same side</u>, there will be <u>no change in direction. </u>

For example, if you and your friend sit on the same side of a sew saw, the sew saw would not go up or down, meaning no change in direction.

So if not stated otherwise you can assume the load and force are on opposite sides. The purpose of a lever in that situation would be to reduce the force needed to lift the box and change the direction of the force.

*While reading my explanation, it may be helpful to look up a diagram containing a lever, with a load, fulcrum, and applied force.

6 0
2 years ago
7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del
GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = T_{y} / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0
3 years ago
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