How much gravitational potential energy does a 1.0 kg hammer have on a shelf 1.5 m above the ground?

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

At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai

zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

Motion of 0.50 kg ball:

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

Motion of 0.25 kg ball:

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

$\bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

m₁ = 0.50 kg

x₁ = 5.38 m

m₂ = 0.25 kg

x₂ = 10.38 m

Substituting

$\bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m$

The center of mass of the two-ball system is 7.05 m above ground.

8 0

3 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 54.0 N at the s

butalik [34]

Answer:

$m=5.51Kg$

Explanation:

The weight of an object on Earth is given by $W=mg$ , so we can calculate its mass by doing $m=W/g$ , which for our values is:

$m=(54N)/(9.8m/s^2)=5.51Kg$

Nothing is being asked about Io but if one wanted to know the weight W' of the watermelon there one just have to do:

$W'=ma=(5.51Kg)(1.81m/s^2)=9.97N$

4 0

3 years ago

A person on a merry go round is constantly changing direction

Salsk061 [2.6K]

Answer: yurp

Explanation:

because its spinning

7 0

3 years ago

Read 2 more answers

It would be really helpful if u help me solving this question. PLEASE!!!

sweet [91]

Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice

water

heatproof mat

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

Immerse the unmarked thermometer into the ice in the beaker.

When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.

You may add little water and continue to stir gently.

The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value

Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point

This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.

Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.

7 0

3 years ago