J=joules, c=specific heat, q= energy, and the Tf and Ti are the final and initial temperatures cause I couldn't find a delta sign.

$Q=mc(T_{f}-T_{I}) \\ 33000j=2kg*c*80 \\\frac{33000j}{2kg*80} =c \\ c=206.25jkg^{-1}$

4 0

3 years ago

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

Marina86 [1]

Solution:

initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m

h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º

8 0

3 years ago

During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce

Anna [14]

Answer: a= 37m

Explanation: V= 15 m/s (Velocity) t= 0.41s (time) formula: a= v/t

15 m/s / 0.41 (15 divided by 0.41) = 36.583m

There are 2 significant digits, 36, you look at the third digit, either round up or down in this case up to 36. a= 37m

5 0

3 years ago

Which is a sub-atomic particle?

Natalka [10]

A particle that is smaller than an atom or a cluster of particles.

4 0

3 years ago