Indentify the 3 types of heat tranfer
2 answers:
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Answer:
Explanation:
The attached figure shows the whole description. Considering the applied force is 100 N.
The acceleration of both blocks A and B,
Firstly calculating the mass m using the second law of motion as :
F = ma
m is the mass
m = 125 kg
It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N
So, the new acceleration of the block is . Hence, this is the required solution.
Refer to the diagram shown below.
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)