In a stationary situation, the weight of person is

This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).
When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is

This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:


where a is the acceleration of the elevator. If we solve for a, we find

The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
Answer:
<em>His angular velocity will increase.</em>
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system =
'ω'
where
' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = 
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to
.
From
'ω' =
ω
since
is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
Responda:
1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5
Explicação:
Dado o seguinte:
Carga (q) = 3uC = 3 × 10 ^ -6C
Força elétrica (Fe) = 18N
Intensidade do campo elétrico (E) =?
1)
Lembre-se:
Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)
Fe = qE; E = Fe / q
E = 18N / (3 × 10 ^ -6C)
E = 6N / 10 ^ -6C
E = 6 × 10 ^ 6NC ^ -1
2)
Lembre-se:
E = kQ / r ^ 2
E = intensidade do campo elétrico
Q = carga de origem
r = distância de espera = 30cm = 30/100 = 0,3m
K = 9,0 × 10 ^ 9
6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2
9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09
Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9
Q = 0,06 × 10 ^ (6-9)
Q = 0,06 × 10 ^ -3
Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC
Answer:
346.66 Hz
Explanation:
= Length of string which is unfingered = l
= Length of string which is vibrate when fingered = 
= Unfingered frequency = 260 Hz
= Fingered frequency
Frequency is inversely proportional to length

So,

The frequency of the fingered string is 346.66 Hz