dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.
density = 1.20 kg/m3
(a)volume = lenght * breadth * height
= 30 * 15 * 5
= 2250 metre cube = 2.25 cubic meter
(b) mass of air = density * volume
mass of air = 1.2 * 2250
mass of air = 2700kg
weight = mass * 9.8
= 2700 * 9.8
= 26,460 N
- The definition of Density is the amount of matter in a given space, or volume
- Density = mass/volume
- units for density kg/m^3
- Density of water 1g/ml
- Salt water is denser that is why don't sink as easily.
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Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
Answer:
Maximum acceleration in the simple harmonic motion will be 
Explanation:
We have given amplitude of simple harmonic motion is A = 0.43 m
Time period of the oscillation is T = 3.9 sec
We have to find the maximum acceleration
For this we have to find the angular frequency
Angular frequency will be equal to 
Maximum acceleration is given by 
So maximum acceleration in the simple harmonic motion will be
Answer:
2.2 s
Explanation:
Hi!
Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by
We are looking for a time t for which x(t) = 0
Solving for t:
Rounding at the first decimal:
t = 2.2 s
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
