A spring scale hung from the ceiling stretches by 5.2 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl
2 answers:
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The time taken for him to move the bin 6.5 m is 2.30 s.
The given parameters;
- <em>weight of the load, w = 557 N</em>
- <em>force applied , F = 410 N</em>
- <em>angle of force, = 15°</em>
- <em>coefficient of kinetic friction = 0.46</em>
- <em>distance moved, d = 6.5 m</em>
The net horizontal force on the recycling bin is calculated as follows;
where;
- <em>m is the mass of the recycling bin</em>
- <em />
<em> is the frictional force </em>
W = mg
The net horizontal force on the recycling bin is calculated as;
The time taken for him to move the bin 6.5 m is calculated as follows;
Thus, the time taken for him to move the bin 6.5 m is 2.30 s.
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Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]