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Travka [436]
3 years ago
8

Help with these three

Physics
1 answer:
Elenna [48]3 years ago
8 0
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
g = 9.81 \frac{m}{s {}^{2} }
let's do it for this case:
f_{g}  = m \times g \\ f _{g}  = 65kg \times 9.81 \frac{m}{s {}^{2} }  = 637.65
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} }  \\ f = 325n
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
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A wave on a string is described by
liberstina [14]

Explanation:

A wave on a string is described is given by :

D(x,t)=2\ cm\ sin[(12.57\ rad/m)-(638\ rad/s)t]

The linear density of the string is 5 g/m.

Where

x is in meters and t is in seconds

The general equation of a wave is given by :

y=A\ sin(kx-\omega t)

(2) The speed of the wave in terms of tension is given by :

v=\sqrt{\dfrac{T}{\mu}}

Also, v=\dfrac{\omega}{k}

So, \dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}

T=\dfrac{\mu \omega^2}{k^2}

T=\dfrac{5\times 10^{-3}\times (638)^2}{(12.57)^2}

T = 12.88 N

(3) The maximum displacement of a point on the string is equal to the amplitude of the wave. So, the maximum displacement is 2 cm.

(4) The maximum speed of a point on the string is given by :

v=A\omega

v=0.02\times 638

v = 12.76 m/s

Hence, this is the required solution.

5 0
3 years ago
Particle motion in surface waves is __________ motion.
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Answer:

A combination of longitudinal & transverse

Explanation:

6 0
3 years ago
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Scuba divers use compressed oxygen gas to go diving. The oxygen's volume decreases as pressure is increased to fit more in an ox
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I feel like it’s Boyle’s Law, correct me if I’m wrong .
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Which of the following is/are the best example(s) of elastic collision(s)? Explain why you chose your answer. A) A collision bet
ivanzaharov [21]

Answer:

A

Explanation:

option A is correct

the best example of elastic collision among the following is collision between two billiard balls. In collision of two billiard balls the energy lost is very little approximately zero. hence we can say that the kinetic energy is conserved. since, the kinetic energy is conserved we can say that the collision is elastic

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3 years ago
The box resting on the inclined plane above has a mass of 20kg. The incline sits at a 30o angle. Find the friction force between
tekilochka [14]

The friction force between the box and the incline if the box does not slide down the incline will be 0.577

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.

Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle

We have to find the friction force between the box and the incline if the box does not slide down the incline

Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:

F₁ = F₂

μmgcosΘ = mgsinΘ

μ = (mgsinΘ)/(mgcosΘ)

μ = tanΘ

μ = 0.577

Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577

Learn more about friction force here:

brainly.com/question/24386803

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