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4 years ago

Which of the following is not a unit used to measure pressure?

Physics

2 answers:

Luden [163]4 years ago

7 0

Newtons are used to measure force

Ilya [14]4 years ago

4 0

Answer:

The answer is Newton.

Explanation:

In order to determine the unit that it is not used to measure pressure, we have to define the units for every option and then compare them.

Torr (symbol: Torr): is a unit of pressure based on an absolute scale, now defined as exactly 1/760 of a standard atmosphere (≈ 133.32 Pa).

Newton (symbol: N) is the International System of Units (SI) derived unit of force.

Atmosphere (symbol: atm): the standard atmosphere is a unit of pressure defined as 101325 Pa (1.013 bar)

Milimeter of mercury (symbol: mmHg): is a manometric unit of pressure, formerly defined as the extra pressure generated by a column of mercury one millimetre high, and currently defined as exactly 133.32 Pa.

Therefore, the physics unit that not represent a pressure measure is Newton.

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An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q

Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency $f = ( \dfrac{1}{2 \pi}) \omega$

where;

$\omega = \sqrt{\dfrac{k}{m} }$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )$

However;

$k = \dfrac{F}{x}$ and;

mass $m = m_{car } + m_{person}$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )$

where;

$F = m_{person}g$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )$

replacing the values;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )$

$\mathbf{f = 0.442 \ Hz}$

8 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago

Which order shows increasing energy of photons?

Alchen [17]

The correct answer is C. microwaves, infrared, visible, gamma rays

Electromagnetic waves arranged in order of frequencies and wavelengths form the electromagnetic spectrum. In the order of decreasing frequency we obtain the following sequence.

Gamma ray>x-ray>ultraviolet >visible light>infrared light>microwaves>radio waves.

According to the Einsteins equation, E=hf, Energy is directly proportional to frequency. the higher the frequency the higher the energy of the photon and vice-versa.

4 0

4 years ago

Read 2 more answers

15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

kolbaska11 [484]

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

5 0

3 years ago

What is the number one and most important rule to remember in soccer?

AlekseyPX

To score in the correct goal

8 0

3 years ago