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4 years ago

Which object has elastic potential energy? a piece of bread a ball at rest a stretched rubber band a falling leaf

Physics

2 answers:

DaniilM [7]4 years ago

5 0

A stretched rubber band has elastic potential energy.

Savatey [412]4 years ago

3 0

A stretched rubber band, for example, when someone stretches it and they let go of one side, it goes flying across the room.

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Anders suffered a shock when his electric radio dropped into the bathtub--while anders was taking a bath. Anders argued that he

blsea [12.9K]

The claim Anders is most likely to make is the failure of the manufacturer to warn about such risk.

This is defined as the possibility of something bad happening and in this case it is electric shock when dropped into the bathtub.

Anders can decide to sue for not warning against risk of electric shock when in contact with water.

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2 years ago

HELP PLEASE

lisabon 2012 [21]

Sustainable buildings are generally less expensive over the lifetime of the building, due to lower energy costs in production of the materials and lower energy bills. Option A is correct.

<h3>What is sustainable buildings? </h3>

Sustainable buildings refer to both a structure and the implementation of environmentally responsible and resource-efficient methods.

Sustainable buildings are generally less expensive over the lifetime of the building, due to lower energy costs in production of the materials and lower energy bills,

Hence,option A is correct.

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5 0

2 years ago

Derive equation of motion s=ut+1/2at²

Pavel [41]

Recall the definitions of

• average velocity:

v[ave] = ∆x/∆t = (x[final] - x[initial])/t

Take the initial position to be the origin, so x[initial] = 0, and we simply write x[final] = s. So

v[ave] = s/t

• average acceleration:

a[ave] = ∆v/∆t = (v[final] - v[initial])/t

Assume acceleration is constant (a[ave] = a). Let v[initial] = u and v[final] = v, so that

a = (v - u)/t

Under constant acceleration, the average velocity is also given by

v[ave] = (v[final] + v[initial])/2 = (v + u)/2

Then

v[ave] = s/t = (v + u)/2 ⇒ s = (v + u) t/2

and

a = (v - u)/t ⇒ v = u + at

so that

s = ((u + at) + u) t/2

s = (2u + at) t/2

s = ut + 1/2 at²

4 0

2 years ago

55. (III) A uniform circular plate of radius 2R has a circular hole of radius R cut out of it. The center of the smaller circle

Aleksandr [31]

Answer:

P = 0.27R from the center

Explanation:

Given,

The radius of the uniform circular plate, R = 2R

The radius of the hole, r = R

The center of the smaller circle from the center is, d = 0.8R

The center of mass of a circular disc with a hole in it given by the formula

P = dr²/R² - r²

Where P is the distance from the center of mass located in the line joining the two centers opposite to the hole.

Substituting the given values in the above equation,

P = 0.8R x R² / 4R² - R²

= 0.27R³/R²

= 0.27R

Hence the center of mass of plate is at a distant P = 0.27R from the center

6 0

3 years ago

If it takes a planet 2.8 Ã— 108 s to orbit a star with a mass of 6.2 Ã— 1030 kg, what is the average distance between the planet

Shtirlitz [24]

The average distance between the planet and the star is:

R=9.36*10^11 m

Orbital velocity v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).

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The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.

4 0

1 year ago