The two possible distances that you might infer your friends swam while the lights were out are 25 m and 75 m.
Answer:
Explanation:
So you have to measure the distance covered by your friend in a time gap of 86 s. And the average velocity is given as 0.29 m/s.
Then as per the mathematical calculation of velocity, distance can be measured as the product of velocity with time interval.
Distance = Velocity × Time Interval
Distance = 0.29 m/s×86 s = 24.94 m.
So based on this calculation, one of the possible distance inferred by you will be 24.94 m.
Another possible distance can be guessed from the statements provided. So if the length of pool is 50 m, then covering halfway in opposite direction to his starting direction means completion of one full length i.e., 50 m and then halfway of that 50 m which is 25m, so totally 50 +25 = 75 m.
So in other way, we can assume that your friend has covered 75 m distance during the light out.
Thus, the two possible distances that you might infer your friends swam while the lights were out are 25 m and 75 m.
Answer:
2.10 N
Explanation:
Draw a free body diagram of the toy. There are four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force F pushing left,
Applied force 4.63 N pulling at an angle of 63.0°.
Sum of the forces in the x direction:
∑F = ma
4.63 N cos 63.0° − F = 0
F = 2.10 N
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is ![\tau = 34.3 \ N\cdot m](https://tex.z-dn.net/?f=%5Ctau%20%3D%2034.3%20%5C%20%20N%5Ccdot%20m)
Explanation:
From the question we are told that
The mass of the steel ball is ![m = 3.0 \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%203.0%20%5C%20%20kg)
The length of arm is ![l = 70 \ cm = 0.7 \ m](https://tex.z-dn.net/?f=l%20%3D%20%2070%20%5C%20cm%20%20%3D%200.7%20%5C%20%20m)
The mass of the arm is ![m_a = 4.0 \ kg](https://tex.z-dn.net/?f=m_a%20%20%3D%204.0%20%5C%20%20kg)
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as
![r = \frac{l}{2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Bl%7D%7B2%7D)
=>
=>
Generally the magnitude of torque about the athlete shoulder is mathematically represented as
![\tau = m_a * g * r + m * g * L](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%20m_a%20%2A%20g%20%2A%20r%20%20%2B%20m%20%2A%20g%20%2A%20%20L)
=> ![\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%204%20%2A%209.8%20%2A%200.35%20%2B%203%20%2A%209.8%20%2A%20%200.70)
=> ![\tau = 34.3 \ N\cdot m](https://tex.z-dn.net/?f=%5Ctau%20%3D%2034.3%20%5C%20%20N%5Ccdot%20m)
Answer:
Acute bronchitis can result from: a virus, for example, a cold or flu virus. a bacterial infection. exposure to substances that irritate the lungs, such as tobacco smoke, dust, fumes, vapors, and air pollution.
Explanation: