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3 years ago

To squeeze a gas into a smaller place

Physics

1 answer:

IRINA_888 [86]3 years ago

7 0

Easy the answer is to Compress

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if an object traveling in one direction has s positive velocity , what kind of velocity would the same object traveling in the o

AveGali [126]

A negative velocity.

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3 years ago

Read 2 more answers

How much time passes when an object travels at a constant speed of 17 m/s over a distance of 323 m? Identify Variables, write fo

k0ka [10]

Answer: 19 s

Explanation:

The relationship between speed, distance and time is given by:

velocity= distance/time

time= distance/velocity= 323 m/17 m/s= 19 s

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4 years ago

In the Bohr model of the hydrogen atom, an electron (massm=9.1×10−31kg) orbits a proton at a distance of 5.3×10−11m. The proton

Marysya12 [62]

Answer:

$\omega = 6.557 \times 10^{16}\ rev/s$

Explanation:

GIVEN,

mass of electron = 9.1 x 10 kg

Radius = 5.3 x 10 m

pulling force = 8.2 x 10 N

Required centripetal for (Fe) for circular motion will be provided with electrical force (F)

$F = m_e\omega^2 r$

$\omega = \sqrt{\dfrac{F}{m_e\ r}}$

$\omega = \sqrt{\dfrac{8.2 \times 10^{-8}}{9.1 \times 10^{-31}\times 5.3 \times 10^{-11}}}$

$\omega = \sqrt{0.17 \times 10^{34}}$

ω = 4.12 x 10¹⁶ rad/s

$\omega = \dfrac{4.12 \times 10^{16}}{2\pi}$

$\omega = 6.557 \times 10^{16}\ rev/s$

6 0

3 years ago

The splitting of an atom is called Blank Space __________.

horsena [70]

Fission is the answer

8 0

3 years ago

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res

zhannawk [14.2K]

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

$w = \sqrt{\frac{k}{m}}$

$w = \sqrt{\frac{250}{0.5}}$

$w$ = 22.4 rad/s

$f$ = frequency

Angular frequency is also given as

$w$ = 2 π $f$

22.4 = 2 (3.14) f

$f$ = 3.6 Hz

$T$ = Time period

Time period is given as

$T = \frac{1}{f}$

$T = \frac{1}{3.6}$

$T$ = 0.28 sec

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

5 0

4 years ago