Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π 
f= 1/2π 
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
Answer : The correct option is (D).
Explanation :
Given that,
A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a race.
We have to find her position after she runs 45 meters.
From the attached figure,
Let A is the position of Juliet. O is the initial point such that OA = 10 m, AB = 45 m and OP = 100 m.
So, using simple mathematics, it is clear that the position of Juliet after running 45 meters will be 55 m. It is OB in the figure.
So, the correct option is (D) " 55 meters ".
Answer:
b) Nothing will happen, the sea saw will still be balanced.
Explanation:
b) Nothing will happen, the sea saw will still be balanced.
Reason:-
When two kids are balanced, the sum of torques on the seesaw will be zero.
if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.
Therefore the seesaw is balanced
Answer: 0.9264 kg
Explanation: [I'll use "cc" for cubic centimeter, instead of cm^3.
The volume is 6cm*4cm*2cm = 48 cm^3 (cc).
Density of Au is 19.3 g/cc
Mass of gold = (48 cc)*(9.3 g/cc) = 926.4 grams Au
1 kg = 1,000 g
(926.4 grams Au)*(1 kg/1,000 g) = 0.9264 kg, 0.93 kg to 2 sig figs
At gold's current price of $57,500/kg, this bar is worth $53,268. Keep it hidden from your lab partner (and instructor).
Earth's magnetic field, also known as the geomagnetic field, is the magnetic field that extends from the Earth's interior out into space, where it interacts with the solar wind, a stream of charged particles emanating from the Sun.
