Explanation:
Joule (J) is the MKS unit of energy, equal to the force of one Newton acting through one meter.
Explanation:
solution: mass m = 5g = 0.005kg; extension e = 7cm = 0.07m; force f = 70 N; velocity = ?; using: work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone. 1/2 fe =...
Answer:sofa
Explanation:the weight of the mass of that object can be heavy (er) than the rest
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC.
Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules
Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used
q=mCΔT
11466.58 Joules = 65.1g x 4.18 J / g C x ΔT
11466.58/(65.1gx4.18)=ΔT
ΔT=42.14oC
So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>
Answer:
Frquency=3,994Hz
Explanation:
Tension =967N
Density of string (μ)=0.023g/cm
Length of the stretched spring=308cm
Fundamental frequency for nth harmonic :
Fn=n/2L(√T/μ)
Substituting the given values to find the frequency :
f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]
=6.16m[(√967N)/0.0023kg/m)]
=3,994.20Hz
Approximately,
The frequency will be =3,994Hz
