An object contains 4.9*10^7 excess electrons. What is its net charge?
1 answer:
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Answer:
<em>The shortest distance to an object that this radar can detect would be </em>
<em>111 m</em>
Explanation:
The shortest distance is the minimum distance that would be detected by the radar, The time of oscillation can be obtained thus;
T = 1/f ..........1
but f= v/λ substituting f into equation 1 we have;
T = λ/v...............2
Where λ is the wavelength = 2.2 m
v is the velocity of light since the radar is an electromagnetic wave
= 3 x m/s
T = 2.2 m / 3 x m/s
T = 7.33 x s
<u>Calculating the time interval required by the pulse</u>
Since the interval of the pulse (t) is 100 times greater than the period of oscillation, the time of the pulse is expressed as;
t = 100 x T
t = 100 x 7.33 x s
t = 7.33 x s
Therefor the time of the pulse is 7.33 x s
<u>Calculating for the shortest distance of the radar</u>
The shortest distance of the radar can be obtained using the equation below;
λ_s = (t/2) x v ............3
Substituting into equation 3 we have
λ_s = (7.33 x /2) x 3 x
m/s
λ_s = 111 m
Therefore the shortest distance to an object that this radar can detect would be 111 m
Answer:
266.7k
Explanation:
Given parameters:
Initial pressure = 36Pa
Initial temperature = 300k
New pressure = 32Pa
Unknown:
New temperature = ?
Solution:
To solve this problem, we have to apply the combined gas law when the volume is constant;
in this instant, when the volume is constant, the pressure of a given mass of gas varies directly with the absolute temperature.
=
P and T are pressure and temperature
1 and 2 are initial and final states
Insert the parameters and solve;
=
T₂ = 266.7k