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3 years ago

Instructions:Select all the correct answers.

2 answers:

3 0

Choice (3) would be one of the objects because when its stretching its waiting to release the energy. Then choice (1) would be the other object because when it rolls the energy stored in the ball would be released when stopped. Therefore there storing the energy until released.

V125BC [204]3 years ago

3 0

Answer:

2. A small rock sitting on the top of big rock.

3. a stretched rubber band.

Explanation:

The potential energy is defined as the energy by the virtue of the position of the object, its electric energy, stress which present in itself.

There are the two forms of potential energies which are more common:

Gravitational potential energy: It is a form of potential energy which depends on the mass of an object and the distance of that object from the center of the earth. In the given situation the rock which is sitting on the top of big rock posses gravitational potential energy.
Elastic potential energy: It is the form of energy which is produced by the virtue of extending spring. This energy in this is the store energy of the extending or compressed spring. In the given situation a stretched rubber band is the example of elastic potential energy.

Therefore, in the given problem a stretched rubber band and a small rock which is sitting on the top of big rock have stored energy.

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(no links or files) How can an electromagnetic wave transmit energy?

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Answer:B

Explanation: I think this is the answer

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3 years ago

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A sound wave can be considered as a displacement wave or a pressure wave? What phase difference exists between the displacement

netineya [11]

Where the displacement is a maximum the pressure is a minimum.
Where the displacement is zero, the pressure is a maximum.

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4 years ago

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same am

Tcecarenko [31]

Answer:

Incomplete question,

This is the complete question

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration ~a of the block after it begins to move. Express your answer in terms of some or all of the variables µs, µk, and m, as well as the acceleration due to gravity g.

Explanation:

Let the force that make the object to start moving be F,

Frictional force is opposing the motion, the body has to overcome two frictional forces acting in the opposite direction of the motion.

Also, weight and normal reaction are acting in vertical direction, the weight is acting downward while the reaction is acting upward.

Weight of the object is given as

W=mg

Analyzing the vertical motion i.e y-axis.

ΣF = ma

since the body is not moving upward, the a=0

N-W=0

Then, N=W

So, N=mg

So, from friction law

Fr=µN

For static

Fs=µsN

For kinetic or dynamic

Fk= µkN

Using newton law

Along x-axis

Before the body start moving we can get the Force and since the force is the same use to start the block in motion.

Then,

ΣF = ma

Since at static the body is not moving then, a=0

F-Fs=0

F=Fs

Since, Fs=µsN

F=Fs=µsN

Then, the force to keep the body in motion too is F=µsN

Now analyses when the body is in motion

ΣF = ma

F-Fk=ma

ma=F - Fk

Substituting F=µsN and Fk=µkN

ma=µsN - µkN

ma=N(µs - µk)

Since N=mg

Then, ma=mg(µs - µk)

m cancels out, then

a=g(µs - µk)

Then the acceleration of the body is given as "a=g(µs - µk)"

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3 years ago

Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is

lesya [120]

Answer:

Explanation:

Magnetic field at a a point R distance away

B = μ₀ / 4π X 2I / R where I is current

Magnetic field due to one current

= 10⁻⁷ x 2 x 24 / 1 x 10⁻³

48 x 10⁻⁴ T

Magnetic field due to other current

= 10⁻⁷ x 2 x 24 / 3x 10⁻³

16 x 10⁻⁴ T

Total magnetic field , as they act in opposite direction, is

= (48 - 16 ) x 10⁻⁴

32 x 10⁻⁴ T .

5 0

3 years ago

The electric field strength E 0 E 0 is measured at a perpendicular distance R R from an infinitely large, thin sheet that contai

IRISSAK [1]

Answer:

same value in R and 2R E = E₀ = σ / 2ε₀

Explanation:

For this exercise we use Gauss's law

Ф = E. dA = $q_{int}$ /ε₀

We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero

As the sheet has two faces

2E A = q_{int} /ε₀

The charge inside the cylinder is

σ = q_{int} / A

q_{int} = σ A

We substitute

E = σ / 2ε₀

We see that this expression is independent of the distance, so it has the same value in R and 2R

E = E₀ = σ / 2ε₀

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3 years ago