The answer is apparent weight is zero.
You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).
You still are being affected by the Earth's gravity.
Not all because of the previous two statements.
Not none because apparent weight is zero as you are falling.
Answer:
v_{ average} = 5.57
Explanation:
The most probable value of a measure is
v_average = 
∑ x_i
where N is the number of measurements
in tes case N = 3
v_{average} = ⅓ (5.63 +5.54 + 5.53)
V_{average} = 5,567
The number of significant figures must be equal to the number of figures that have the least in the readings.
v_{ average} = 5.57
Answer:
the no. of ejected electrons per second will increase.
Explanation:
In photoelectric effect, when a light is incident on a metal surface it ejects some electrons from the metal surface. The energy of photon of light must be equal to or greater than the work function of that metal. All the extra energy above the work potential appears as the kinetic energy of the ejected electrons. So, greater he energy of photon greater will be the kinetic energy of the ejected electrons.
A single photon interacts with a single electron and ejects it only if its energy is greater than work function. So, the increase in no. of photons per second means an increase in the intensity of laser beam. And greater no. of photons, will interact with greater no. of electrons. So, <u>the no. of ejected electrons per second will increase.</u>
Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
Hy tikki! I've asked some questions, so of you find the questions as easy, then answer it. I'll surely mark you as brainliest :)
