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2 years ago

5. What is the amplitude of the waves shown in the diagram below?

Physics

1 answer:

Nitella [24]2 years ago

3 0

Answer:

0.54 m

Explanation:

Example below.

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(15 points) (Asap!!) In what two ways can you increase the elastic potential energy of a spring?

adelina 88 [10]

Hello There

Answers: The elastic potential energy can be increased by: 

1) Getting a spring with a higher spring constant

2) Increasing the length at which the spring is compressed.

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris

5 0

2 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago

The ancient Greek recommendation of rest in the treatment of abnormal behavior soon gave rise to ______. A. Trephining B. The as

Lilit [14]

C. Is the correct answer

7 0

2 years ago

Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t

Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

$\tau = Fdsin \theta$

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

$\theta$ is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so $\theta=90^{\circ}, sin \theta=1$

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

4 0

2 years ago

A tank contains one mole of carbon dioxide gas at a pressure of 6.70 atm and a temperature of 23.0°C. The tank (which has a fixe

quester [9]

Answer: 888.45 K or 615.3 °c

Explanation:

According to Gay Lussacs law which states that at constant volume, pressure of an ideal gas is directly proportional to it's absolute temperature.

P/T = Constant

Therefore, P1/T1 = P2/T2

P1 = 6.7 atm

T1= 23°c = 273.15 + 23 = 296.15K

Since P2 is tripled, then,

P2 = 6.7 x 3= 20.1 atm

T2 = (20.1 x 296.15) ÷ 6.7

T2 = 888.45 K

Or in celcius 615.3°c

5 0

2 years ago