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zysi [14]
2 years ago
14

The sound waves travel along this​

Physics
1 answer:
Anna35 [415]2 years ago
4 0

Explanation:

sound waves go through almost every thing exept solid walls

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A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?
Marizza181 [45]
According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons
6 0
3 years ago
Fill in the blank question.
puteri [66]
Answer is on the image with the explanation. I hope that might help you with the answers

4 0
1 year ago
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
3 years ago
If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled
Alisiya [41]

Answer:F_2=16\times F_1

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

F=\frac{kq_1q_2}{r^2}

where q_1 and q_2 is the charges  of particle

r=distance between charge particle

=\frac{kq^2}{r^2}when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

F_2=\frac{k(2q)^2}{(0.5r)^2}

F_2=\frac{kq^2}{r^2}\times 4\times 4

F_2=16\times F_1

         

6 0
3 years ago
PHYSICS HW HELP PLS
lana [24]
Violet cannot , ultraviolet can
red can, infrared cannot
3 0
3 years ago
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