I'd guess at valve B. more information about the interesting question would help.
The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.
<h3>
What is the tension in the cord?</h3>
The tension in the cord is calculated as follows;
T = ma + mg
where;
- a is the acceleration of the block
- g is acceleration due to gravity
- m is mass of the block
T = m(a + g)
T = 1.5(a + 9.8)
T = 1.5a + 14.7
Thus, the tension in the cord is (1.5a + 14.7) N.
If the block is at rest, the tension is 14.7 N.
<h3>Force of the force</h3>
The force with which the cord pulls is equal to the tension in the cord
F = T = m(a + g)
F = (1.5a + 14.7) N
If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.
Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.
Learn more about tension here: brainly.com/question/187404
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Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
in english please
i don't understand actually
