Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
Answer:
1.3823 rad/s
20.7345 m/s
28.66129935 m/s²

2006.29095 N radially outward
Explanation:
r = Radius = 15 m
m = Mass of person = 70 kg
g = Acceleration due to gravity = 9.81 m/s²
Angular velocity is given by

Angular velocity is 1.3823 rad/s
Linear velocity is given by

The linear velocity is 20.7345 m/s
Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²
Acceleration in terms of g


Centripetal force is given by

The centripetal force is 2006.29095 N radially outward
The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.
Answer:
t = 5 s
Explanation:
Data:
- Initial Velocity (Vo) = 7 m/s
- Acceleration (a) = 3 m/s²
- Final Velocity (Vf) = 22 m/s
- Time (t) = ?
Use formula:
Replace:
Solve the subtraction of the numerator:
It divides:
How much time did it take the car to reach this final velocity?
It took a time of <u>5 seconds.</u>
Answer:
B) 4500 Pa
Explanation:
As pressure is force per unit area,
P = F/A
It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.
The possible areas are
0.30(0.40) = 0.12 m²
0.30(0.50) = 0.15 m²
0.40(0.50) = 0.20 m²
The pressure when the face with the largest area (0.20 m²) is down is
P = 900 / 0.20 = 4500 N/m² or 4500 Pa
the other possible pressures would be
900/0.15 = 6000 Pa
900/0.12 = 7500 Pa
which are both larger than our solution.
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s