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Jobisdone [24]
3 years ago
13

If this plastic cup is heated to its melting

Physics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

A. The particles will begin to move enough

that they slide past each other.

Explanation:

When the plastic cup is heated, the Kinetic energy of its particles starts increasing. As the temperature rises, the kinetic energy keeps increasing. With the increase of K.E, the particles start moving faster and faster. When the temperature finally reaches the melting point, the K.E of the molecules is enough to break the bonds and slide past each other.

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A car has an initial velocity of 10m/s. It accelerates at a rate of 2m/s2 until it reaches 30m/s. How far does it travel and for
Anna [14]

Answer:

acceleration=(v final-vinitial)/time

a=(vf-vi)/t

2=(30-10)/t

2=20/t

t=20/2=10 seconds

distance=v0t+at²/2

d=10 x 10+ 2x 10²/2=100+100= 200 m

Explanation:

7 0
2 years ago
A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m
ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

S_{el} = 414Mpa

Yield Strain of the Specimen

\epsilon_{el} = 0.002

Diameter of the test-specimen

d_0 = 12.8mm

Gage length of the Specimen

L_0 = 50mm

Modulus of elasticity

E = \frac{S_{el}}{\epsilon_{el}}

E = \frac{414Mpa}{0.002}

E = 207Gpa

Strain energy per unit volume at the elastic limit is

U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}

U'_{el} = \frac{1}{2} (414)(0.002)

U'_{el} = 414kN\cdot m/m^3

Considering that the net strain energy of the sample is

U_{el} = U_{el}' \cdot (\text{Volume of sample})

U_{el} =  U_{el}'(\frac{\pi d_0^2}{4})(L_0)

U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})

U_{el} = 2.663N\cdot m

Therefore the net strain energy of the sample is 2.663N\codt m

6 0
3 years ago
A ball at the top of a hill is an example of ______ and a ball rolling down the hill is an example of ______.
larisa [96]

Potential energy , Kinetic energy

<h3 /><h3>What is kinetic energy and potential energy?</h3>

Kinetic energy is a form of energy that posses by an object due to its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.

Potential energy is the energy held by an object due to its position relative to other objects, stresses within itself, its electric charge, or other factors.

According to given condition,

At first,

When a ball is at the top of a hill , it posses a potential energy because the ball is at some height from the surface  where  the energy held by ball is due to its position.

That is why,  a ball at the top of a hill is an example of Potential energy.

Secondly,

When a ball is rolling down the hill , it posses kinetic energy because the ball have some energy that posses by an object due to its motion.

That is why, a ball rolling down the hill is an example of  Kinetic energy.

Learn more about Potential energy and Kinetic energy here:

brainly.com/question/11749818

#SPJ1

6 0
2 years ago
How are electromagnetic waves produced? A) An accelerating electric point charge (electron) emits electromagnetic waves. B) A ba
shusha [124]
Hello,

Here is your answer:

The proper answer to this question will be option A "An accelerating electric point charge (electron) emits electromagnetic waves." That's because electromagnetic waves are formed by motion of electrically charged particles (electrons)!

Your answer is A!

If you need anymore help feel free to ask me!

Hope this helps!
8 0
3 years ago
Read 2 more answers
Satellite A orbits a planet at a distance d from the planet’s center with a centripetal acceleration a0. A second identical sate
leva [86]

To solve this problem it is necessary to use the concepts related to the Gravitational Force and Newton's Second Law, as far as we know:

F_g = \frac{GMm}{r^2}

Where,

G = Gravitational constant

M = Mass of earth (in this case)

m = mass of satellite

r = radius

In the other hand we have the second's newton law:

F = ma

Where,

m = mass

a = acceleration

Equation both equations we have,

ma = \frac{GMm}{r^2}

For the problem we have that,

<em>Satellite A:</em>

ma_A = \frac{GMm}{r^2}

<em>Satellite B:</em>

ma_B = \frac{GMm}{(2r)^2}

The ratio between the two satellites would be,

\frac{ma_A}{ma_B}= \frac{\frac{GMm}{r^2}}{\frac{GMm}{(2r)^2}}

Solving for a_B,

a_B = \frac{a_A}{4}

Therefore the centripetal acceleration of  A_B is a quarter of a_A

7 0
3 years ago
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