physics
the period of motion of an object-spring system is t = 0.474 s when an object of mass m = 296 g is attached to the spring.

(a) find the frequency of motion in hertz.
 \_\_\_\_\_\_\_hz

(b) find the force constant of the spring.
 \_\_\_\_\_\_\_n/m

(c) if the total energy of the oscillating motion is 0.150 j, find the amplitude of the oscillations.
 \_\_\_\_\_\_\_m
(a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters
u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2 ?solution 1cq


here, when one spring is compressed, the other is stretched and the potential energy will be stored in the compressed spring and so the oscillation continues and then the other spring compresses and the process goes on considering the surface to be a frictionless surface. the oscillation goes on forever if there are no dissipative forces. the relationship of wavelength and frequency will appear as:using the wavelength range given, 280-320 nm, frequency range will be calculated using the equation above and by converting the values to the proper units [readmore]each wavelength must be in meters:we need to calculate the frequency for both 280 and 320 nm:range of frequency will appear as 1.07x1015 to 9.38x1014 s-1 v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> hi trinity, let's do this one!newton's famous second law states, in math terms: f = ma applying to the first case, we have: 100 = m(4.00) => m = 100/4 = 25 kgnow apply the formula to the second case: f = 25 x 16 = 400 newtons physics 101 notes: 8/25 (week 2 continued)  force o defined as a push or a pull, and is a vector quantity, meaning that it has direction depending on which way it is being pushed or pulled o it has amount and direction (how much force is being applied and in which direction it is moving)  newton’s first law o first draft: “an object that is free of forces move in a straight line and covers equal distances in equal time” o final: “an object that is free of external forces moves at a constant velocity”  how does a skater start, stop, or turn? o acceleration: “the rate at which an object’s velocity is changing with time o acceleration, s the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. coefficient of friction is defined as the ratio of the force of friction acting between two objects and the force which is pushing them together. the value of the coefficient of friction varies from zero to one. there are two types of the coefficient of friction: static coefficient of friction kinetic (also known as sliding) coefficient of friction. therefore, the coefficient of friction depends on the nature of the surface. class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) answer: the molecular mass of an element is defined as the sum of the masses of the elements present in the molecule. molecular mass is obtained by multiplying the atomic mass of an element with the number of atoms in the molecule and then adding the masses of all the elements in the molecule. o2 is a diatomic molecule. the mass of an oxygen atom = 16 amu. ∴ mass of o2 molecule = 2 × 16 = 32 amu. mass of one molecule of oxygen → 32/ avogadro constant ( 6.023 × 1023) → 5.31 × 10-23g.