Answer: your correct answer is a i took the test
Please i need brainlist i need one more and i level up :)
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 .
The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.
<h3 /><h3>Velocity: </h3>
This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.
⇒ Formula:
- mv²/2 = ke²/2
- mv² = ke².................. Equation 1
⇒ Where:
- m = mass of the stuntman
- v = velocity of the stuntman
- k = force constant of the spring
- e = compression of the spring
⇒ Make v the subject of the equation
- v = √(ke²/m)................. Equation 2
From the question,
⇒ Given:
- m = 48 kg
- k = 75 N/m
- e = 4 m
⇒ Substitute these values into equation 2
- v = √[(75×4²)/48]
- v = √25
- v = 5 m/s.
Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.
The right option is O A. 5 m/s
Learn more about velocity here: brainly.com/question/10962624
The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as
<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>
Generally, the equation for Rate of flow of Liquid is mathematically given as
$$
Where dP is pressure difference r is the radius
is the viscosity of water
L is the length of the pipe
In $30s the quantity that flows out of the tube
In conclusion, the quantity that flows out of the tube
Read more about the flows rate
brainly.com/question/27880305
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