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2 years ago

What is the first step dealing with an injured relationship

Physics

2 answers:

RideAnS [48]2 years ago

7 0

Well there is two if u are badly injured go check it out if u are a bit hurt rest it while having ice on it

Alex_Xolod [135]2 years ago

6 0

Communicate your issues and agree on a solution

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The periodic table arranges elements by increasing _________.

Olegator [25]

Answer:

Atomic number

Explanation:

Hope it helps you in your learning process.

4 0

2 years ago

If a spring has a spring constant of 5 N/m and it is stretched 20 cm, what is the force of the Spring

Nastasia [14]

Answer:

1 N

Explanation:

First the equation is momentum = Force / distance

20 cm = 0.2 m

5 N/m = F / 0.2 m

F = 1 N

7 0

2 years ago

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object $v=1\ m/s$

radius of circle $r=1\ m$

Force towards the center $F=1\ N$

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

$W=F\cdot s\cos 90$

$W=0$

4 0

3 years ago

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f

dem82 [27]

Answer:

Part a)

$Loss = 3.6 J$

Part b)

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

$U = mgH$

Energy Loss = 0.35 mgH[/tex]

$Loss = 0.35(0.70)(9.81)(1.5)$

$Loss = 3.6 J$

Part b)

Energy of the ball after first bounce

$U_1 = 0.65 mgH$

energy of ball after 2nd Bounce

$U_2 = 0.65(0.65 mgH)$

energy of the ball after 3rd bounce

$U_3 = (0.65)(0.65^2)mgH$

$U_3 = 0.65^3(0.70)(9.81)(1.5)$

$U_3 = 2.83 J$

Now we will have energy loss in fourth bounce given as

$Loss = 0.35 U_3$

$Loss = 0.35(2.83)$

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0

3 years ago

Four velcro-lined air-hockey disks collide with each other in a perfectly

Reil [10]

Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, $\Sigma P_{initial}$ is given as follows;

$\Sigma P_{initial}$ = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ $\Sigma P_{initial}$ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ $\Sigma P_{initial}$ = -20·i + 130·j

By the law of conservation of linear momentum, we have;

$\Sigma P_{initial} = \Sigma P _{final}$ = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

$\Sigma P _{final}$ = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, $\left | v \right |$ = √((-0.08)² + (0.52)²) ≈ 0.526

$\left | v \right |$ ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0

3 years ago