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MaRussiya [10]
3 years ago
7

Examples of reaction force and action force hewlp​

Physics
2 answers:
mrs_skeptik [129]3 years ago
6 0

Answer:

Action-Reaction Force Examples in Everyday Life

Recoil of a Gun.

Swimming.

Pushing the Wall.

Diving off a Raft.

Space Shuttle.

Explanation:

hope this helps

Andreas93 [3]3 years ago
6 0
A swimmer swimming forward
A ball is thrown against a wall
A person pushes against a wall and the wall exerts an equal and opposite force against the person
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During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr
WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

E = \frac{\Delta V}{d}

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV  

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0
2 years ago
If a star is moving away from you at a constant speed, how do the wavelengths of the absorption lines change as the star gets fa
Minchanka [31]

Answer:

they stay shifted the same amount to the red

Explanation:

Redshift is given by

z=\dfrac{\lambda_o-\lambda_e}{\lambda_e}

Where,

\lambda_o = Wavelength observed

\lambda_e = Wavelength emitted

Also

Transverse redshift is given by

1+z=\dfrac{1}{\sqrt{1-v^2/c^2}}

v = Velocity of object

c = Speed of light = 3\times 10^8\ m/s

So, if the velocity is constant the redshift remains the same

8 0
3 years ago
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0
3 years ago
Read 2 more answers
You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Base
shepuryov [24]

Answer:

Explanation:

 For entry of light into tube of unknown refractive index

sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.

sin ( 90 - 25 ) / sin( 90 - C)  = μ

sin65 / cos C = μ

sinC = 1.33 / μ  , where 1.33 is the refractive index of body liquid.

From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

TanC = 1.4675

C= 56°

sinC = 1.33 / μ

μ = 1.33 / sinC

= 1.33 / sin56

= 1.33 / .829

μ = 1.6   Ans

3 0
3 years ago
An ideal gas is confined within a closed cylinder at atmospheric pressure (1.013 * 105 Pa) by a piston. The piston moves until t
likoan [24]

Answer:

911700\ \text{Pa}

Explanation:

P_1 = Initial pressure = 1.013\times 10^5\ \text{Pa}

V_1 = Initial volume

V_2= Final volume = \dfrac{V_1}{9}\\\Rightarrow \dfrac{V_1}{V_2}=9

Temperature is the same in the initial and final state

From the ideal gas law we have

P_1V_1=P_2V_2\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=P_1\times9\\\Rightarrow P_2=1.013\times 10^5\times 9\\\Rightarrow P_2=911700\ \text{Pa}

The final pressure of the system is 911700\ \text{Pa}.

5 0
3 years ago
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