Answer:
1)true
2)false
3)true
4)true
5)false,cz amatter is made of atoms while light is an electromagnetic radiation.
Um tubo de raios-X é um tubo de vácuo que converte a energia elétrica em raios-X. A disponibilidade dessa fonte controlável de raios-X criou o campo da radiografia, a imagem de objetos parcialmente opacos com radiação penetrante. Em contraste com outras fontes de radiação ionizante, os raios X são produzidos apenas enquanto o tubo de raios X estiver energizado. Os tubos de raios-X também são utilizados em scanners de tomografia computadorizada, scanners de bagagem de aeroportos, cristalografia de raios-X, análise de materiais e estrutura e para inspeção industrial.
Answer:
i dont know
Explanation:
i dont know since you didn't provide something to base off of
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
