All categories

2 years ago

What is the correct definition of a compression?

Physics

2 answers:

olga_2 [115]2 years ago

7 0

Answer:

Compression - "Compression is a process in which the actual space covered by any sort of entity is reduced and the values of the given parameters are lowered."

Explanation:

Definition with respect to Stress and strain- "The change in actual length,L to a new length,ΔL due to applied stress is called as compression".

Definition with respect to longitudinal wave- "The region of the wave where the particles present are more close to each other is called as compression".

ASHA 777 [7]2 years ago

5 0

Answer:

The answer that I found is.....

<h3>Compression. A compression is a region in a longitudinal wave where the particles are closest together. Rarefaction. A rarefaction is a region in a longitudinal wave where the particles are furthest apart.</h3>

You might be interested in

The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro

qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So, Q = ∫∫∫ρdV

Q = ∫∫∫ρr²sinθdθdrdΦ

Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q = ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ

Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q = ∫∫0.2r⁴[-cosθ]drdΦ

Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q = ∫∫0.2r⁴-[- 2]drdΦ

Q = ∫∫0.2r⁴(2)drdΦ

Q = ∫∫0.4r⁴drdΦ

Q = ∫0.4r⁴dr∫dΦ

Q = ∫0.4r⁴dr[Φ]

Q = ∫0.4r⁴dr[2π - 0]

Q = ∫0.4r⁴dr[2π]

Q = ∫0.8πr⁴dr

Q = 0.8π∫r⁴dr

Q = 0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0

2 years ago

A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro

ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let $u_1, u_2 , v_f$ be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

$m=2.50\times 10^{4}$

Conserving momentum

$mu_1+3mu_2=4mv_f$

$4m+6m=4mv_f$

$v_f=2.5 m/s$

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

$Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J$

$Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J$

$Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ$

4 0

3 years ago

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the

Taya2010 [7]

Answer:

5.5 N

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} , mass = density x volume

buoyant force = density x volume x acceleration

where density is the density of air for the buoyant force

buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

force by helium = density x volume x acceleration

force by helium = 0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

force of its weight = mass of rubber x acceleration

force of its weight = 0.0125 x 9.8 = 0.1225 N

Tension = buoyant force + force by helium + force of weight of rubber

the force of weight of rubber and of helium act downwards, so they

carry a negative sign.

Tension = 6.54 - 0.917 - 0.1225 = 5.5 N

8 0

2 years ago

Failures to arrive at an answer in a scientific investigation can still be useful because they show us what doesn't work. Answer

kipiarov [429]

True
From what Ive learned its like how you create your own hypothesis for a science experiment it may not work completely that way but you learn what didnt work and what does work

4 0

3 years ago

The gun, mount, and train car of a railway had a total mass of 1.22 x 10^6 kg. The gun fired a projectile that was 80 cm in diam

kakasveta [241]

1) According to the law of conservation of momentum ..
Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh) 

1.22^6kg x 5.0 m/s = 7502kg x Vh 
Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s 

Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s 

2) Using the standard range equation .. R = u² sin2θ /g 

R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)

5 0

2 years ago