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katen-ka-za [31]

3 years ago

11

A typical radio frequency is 99.5 MHz, while a HD TV broadcast frequency is 600 MHz. Which one is your body position more likely

to interfere with and why

1 answer:

Leya [2.2K]3 years ago

8 0

The radio because u arent pushing as many watts out and it wont pass through as much

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A vertical piston-cylinder device initially contains 0.1 m^3 of air at 400 K and 100 kPa. At this initial condition, the piston

jenyasd209 [6]

Answer:

$Q=-38.15kJ$

Explanation:

From the question we are told that

Piston-cylinder initial Volume of air $v_1=0.1 m^3$

Piston-cylinder initial temperature $T_1=400k$

Piston-cylinder initial pressure $P_1= 100kpa$

Supply line temperature $T_s=400k$

Supply line pressure $P_s= 500kpa$

Valve final pressure $P_v=500kpa$

Piston movement pressure $P_m=200kpa$

Piston-cylinder final Volume of air $v_2=0.2 m^3$

Piston-cylinder final temperature $T_2=440k$

Piston-cylinder final pressure $P_2= 500kpa$

Generally the equation for conservation of mass is mathematically given by

$Q=m_2 \mu_2-m_1 \mu_1 +W-(m_2-m_1)h$

where

Initial moment

$m_1=\frac{p_1 V_1}{RT_1}$

$m_1=\frac{100*0.1}{0.287*400}$

$m_1=8.7*10^-^2kg$

Final moment

$m_2=\frac{p_2 V_2}{RT_2}$

$m_1=\frac{500*0.3}{0.287*440}$

$m_1=79*10^{-2}kg$

Work done by Piston movement pressure

$W=Pd$

$W=P(v_2-v_1)$

$W=200(0.2-0.1))$

$W=20000J$

Heat function

$h=cT_1$

$h=1.005(400)$

$h=402$

Therefore

$Q=(0.792*0.718(440)-0.0871*0.718(400)+20-(0.792-0.087)*402))$

$Q=-38.15kJ$

It is given mathematically that the system lost or dissipated Heat of

$Q=-38.15kJ$

7 0

3 years ago

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit

madam [21]

Answer:

The initial energy and final energy are the same

= 307.52m

≅ 308m

Explanation:

see the attached file

6 0

4 years ago

How much work must be done on a 5-kg sled to increase its speed from 10

kipiarov [429]

Explanation:

___________________

3 0

3 years ago

The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 {\rm rpm} .

damaskus [11]

Answer:

5833.33

Explanation:

$\alpha$ = Angular acceleration

$\theta$ = Number of revolutions

$\omega_i$ = Initial angular speed = 0

t = Time taken = 2 s

Final angular speed

$\omega_f=\dfrac{350000}{60}=5833.33\ rps$

From the equation of rotational motion we have

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{5833.33-0}{2}\\\Rightarrow \alpha=2916.665\ rev/s^2$

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2916.665\times 2^2\\\Rightarrow \theta=5833.33\ rev$

The number of revolutions is 5833.33

6 0

3 years ago

A 65 Kg roller-blade is accelerating at 5 m/s/s across the side walk. What force would be necessary for this acceleration to occ

Neporo4naja [7]

Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
$F=ma$
Using $m=65 kg$ and $a=5 m/s^2$ , we can find the value of the force applied to the roller-blade to obtain this acceleration:
$F=(65 kg)(5 m/s^2)=325 N$

3 0

3 years ago