The average speed in m/s of a person that jogs eight complete laps around a 400m track in a total time of 15.1 min is 0.44m/s.
<h3>How to calculate average speed?</h3>
Average speed of a moving body can be calculated by dividing the distance moved by the time taken.
Average speed = Distance ÷ time
According to this question, a person jogs eight complete laps around a 400m track in a total time of 15.1 min. The average speed is calculated as follows:
15.1 minutes in seconds is as follows = 906 seconds
Average speed = 400m ÷ 906s
Average speed = 0.44m/s
Therefore, the average speed in m/s of a person that jogs eight complete laps around a 400m track in a total time of 15.1 min is 0.44m/s.
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Yes the plot dose make it with out baking out
Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/
. The correct answer is 4.90 m/
Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.
Let M = 1kg and m = 3 kg
Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.
When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,
M( g + a) = m(g - a)
1(9.8 + a) = 3( 9.8 - a)
9.8 +a = 29.4 - 3a
collect the like terms
4a = 19.6
a = 19.6/4
a = 4.9 m/
Therefore, the cart accelerates to the right with a magnitude of 4.9 m/. The correct answer is 4.90 m/
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Answer: a) 0.13*τ ; b) 2.08*τ
Explanation: In order to explain the discharg of a capacitor through a resistor, we have to consider the following:
Q(t)=Qo* exp(-t/τ) for a lose of 1/8-th of its charge
in this case, Q(t)=7/8*Qo=7/8*exp(-t/τ)
ln(7/8)*τ=-t
then, t= -ln(7/8)*τ =0.13
For a lose of 7/8 th of its charge , we have
Q(t)=1/7*Qo*exp(-t/τ)
t=-ln(1/8)*τ=2.08
Answer: sometimes when people start having an outbreak of it, i guess
Explanation:
