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3 years ago

Which of the following is the most accurate statement concerning the properties of matter?

Physics

1 answer:

DIA [1.3K]3 years ago

8 0

Answer:

b. they can be observed and measured

Explanation:

Matter is anything that has weight and occupy space. There are three states of matter namely Solid, liquid and gas.

The properties of matter are both physical and chemical in nature. Both properties can be measured and observed. Phhysical properties are anything that can be measured without changing the state of the matter. Example of physical properties includes mass, volume, length, color etc.

Chemical properties is another properties of matter. This is the ability of the states of matters to combine with other substance to form a new product for example, rusting of iron, formation of salt etc.

All this as discussed are both measurable and can be observed.

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A 0.01-kg object is initially sliding at 9.0 m/s. It goes up a ramp (increasing its elevation by 1.5 m), and then moves horizont

barxatty [35]

Answer:

During this motion, 0.133 J of heat energy was created

Explanation:

Hi there!

Let´s calculate the energy of the object in each phase of the motion.

At first, the object has only kinetic energy (KE):

KE = 1/2 · m · v²

Where:

m = mass of the object.

v = velocity.

KE = 1/2 · 0.01 kg · (9 m/s)²

KE = 0.405 J

When the object goes up the ramp, it gains some gravitational potential energy (PE). Due to the conservation of energy, the object must convert some of its kinetic energy to obtain potential energy. By calculating the potential energy that the object acquires, we can know the loss of kinetic energy:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity (9.81 m/s²)

h = height.

PE = 0.01 kg · 9.81 m/s² · 1.5 m

PE = 0.147 J

The object "gives up" 0.147 J of kinetic energy to be converted into potential energy.

Then, after going up the ramp, the kinetic energy of the object will be:

0.405 J - 0.147 J = 0.258 J

When the object reaches the spring, kinetic energy is used to compress the spring and the object obtains elastic potential energy (EPE). Let´s calculate the EPE obtained by the object:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compression of the spring

EPE = 1/2 · 100 N/m · (0.05 m)² = 0.125 J

Then, only 0.125 J of kinetic energy was converted into elastic potential energy. The object is at rest at the end of the motion, i.e., the object does not have kinetic energy when it compresses the spring by 5.0 cm. Since energy can´t be lost, the rest of the kinetic energy, that was not used to compress the spring, had to be converted into heat energy:

Heat energy = initial kinetic energy - obtained elastic potential energy

Heat energy = 0.258 J - 0.125 J = 0.133 J

During this motion, 0.133 J of heat energy was created.

7 0

3 years ago

The vertical displacement of the wave is measured from the ?

sergey [27]

The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.

Equilibrium to the crest . . . that's the amplitude.

Crest to trough . . . that's double the amplitude.

Trough to trough . . . How did that get in here ? Yes, that's
   the wavelength, but it has nothing to do
   with vertical displacement.

Frequency . . . that's how many complete waves pass a mark
   on the ground every second. Doesn't belong here.

Notice that this has to be a transverse wave. If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.

It also has to be a vertically 'polarized' wave. If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement. It still has an amplitude, but the amplitude
is all horizontal.

6 0

3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have