Answer:
Yes they can stop or bat the puck with their open hand
Explanation:
Answer:
Explanation:
Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .
Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²
Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²
Applying law of conservation of angular momentum
I₁ ω₁ = I₂ ω₂
Putting the values ,
3 x .75 = .27 x ω₂
ω₂ = 8.33 rad / s
New angular speed = 8.33 rad /s .
Answer:
Change in specific internal Energy
Explanation:
Given:
- Mass of the gas, m=0.4 lb
- Initial pressure and volume are

- Final pressure and temperature are

- Heat transfer from the gas is 2.1 Btu
Since the process is isotropic we have

So the final volume of the gas is calculated.
Work in any isotropic is given by w

According to the first law of thermodynamics we have

So the Specific Internal Change is given by

So the specific Change in Internal energy is calculated.
Answer:
0.0021576N
Explanation:
F=(k)(q1q2/r^2)
F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)
F=0.0021576N
Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π ×
................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second