Answer:
The voltage is 
Explanation:
From the question we are told that
The time that has passed is 
Here 
is know as the time constant
The voltage of the power source is 
Generally the voltage equation for charging a capacitor is mathematically represented as
![V = V_b [1 - e^{- \frac{t}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Cfrac%7B%5Ctau%7D%7B2%7D%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\tau}{2\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Ctau%7D%7B2%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{1}{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B1%7D%7B2%7D%20%7D%5D)
=>
Explanation:
The supermassive black holes that the Event Horizon Telescope is observing are far larger; Sagittarius A*, at the center of the Milky Way, is about 4.3 million times the mass of our sun and has a diameter of about 7.9 million miles (12.7 million km), while M87 at the heart of the Virgo A galaxy is about 6 billion solar ..
Answer:
Explanation:
As we know by radioactivity law
so here we will have
now we will have
now we also know that
A “real” image occurs when light rays actually intersect at the image, and become inverted, or turned upside down. ... In flat, or plane mirrors, the image is a virtual image, and is the same distance behind the mirror as the object is in front of the mirror. The image is also the same size as the object.
Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
