Find the acceleration if a 32.5 N force is used on an object that has a mass of 128.6 kg.

Aloiza [94]

Answer:

physical cosmology, the Big Rip is a hypothetical cosmological model concerning the ultimate fate of the universe, in which the matter of the universe, from stars and galaxies to atoms and subatomic particles, and even spacetime itself, is progressively torn apart by the expansion of the universe at a certain time

3 0

3 years ago

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A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

3 years ago

What causes the spiral structural in milky way?

Mrac [35]

The spiral structure of the milky way can be explained by long lived quasi-static density waves, according to the lin-shu hypothesis. Curiously, the waves of higher density gas and stars (seen as spiral arms) appear to remain static as stars move around the galaxy. This explained by assuming that the gravitational disturbances cause by the 'clumping' material in the arms does not affect the gravitational field of the galaxy as whole and is therefore negligible.

source: Astrophysicist

6 0

3 years ago

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x= cm ,T= 1

we know that time period of spring mas system given as

$T=2\pi \sqrt{\dfrac{m}{k}}$

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

5 0

3 years ago

A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ

emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

5 0

3 years ago