First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed.
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use:
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
Answer:
The answer
Explanation:
Thinking together, Better friendship, Makes teacher happy.
Answer:Highest frequency =618.89Hz
Lowest frequency=582.22Hz
Explanation:
The linear velocity of a sound generator is related to angular velocity and is given as
Vs = rω where
r = the radius of circular path = 1.0 m
ω is the angular velocity of the sound generator. = 100 rpm
1 rev/min = 0.10472 rad/s
100rpm =10.472 rad/ s
Vs = rω
= 1m x 10.472rad/ s= 10.472m/s
A) Highest frequency heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,
f max = (v/ v-vs) fs
Where , v is the speed of the sound in air at 20 degrees celcius =
343 metres per second
vs is the linear velocity of the sound generator=10.472m/s
fs is the frequency of the sound generator= 600 Hz
f max = (343/ 343 - 10.472) x 600
=343/332.528) x600
=618.89Hz
B) Lowest frequency heard by a student in the classroom = Minimum frequency
f min = (v/ v+vs) fs
(343/ 343 + 10.472) x 600
=343/353.472) x 600
=582.22hz
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 .
<span>As long as both mirrors are set at 45% and the same size then you see the same as is reflected in the upper mirror </span>
<span>Put a lens in the middle of the tube </span>
<span>? </span>
<span>We use mirrors when we drive cars ect </span>
<span>Normally they are set across from a concealed entrance or one that is hard to see both ways like the inside of a hairpin bend. Sometimes only to help in one direction. </span>
<span>Sonar which is sound waves that are sent out at a set rate then reflected by objects. The longer the gap between the two the further away it is, They still use periscopes to target boats though. </span>
<span>The periscope can only reflect what is outside so if you could see it because there is enough light then Yes. If you could not see it because it is dark then No unless you get into Info-Red light or Image Intensifying systems as well </span>
