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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Answer:
206.62313 m/s
Explanation:
u = Muzzle speed
g = Acceleration due to gravity = 9.8 m/s²
= Angle at which the bullet is fired = 30°
h = Maximum height = 544 m
Maximum height is given by
The muzzle speed is 206.62313 m/s
Answer:
The blood vessels dat carry blood away from the heart are non as arteries, while those dat carry blood back to the heart are veins.
Explanation:
The ARTERIES are major blood vessels connected to your heart.
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Take the object's starting direction of motion to be the positive direction, so that a stopping force acts in the opposite direction. By Newton's second law, the object undergoes an acceleration <em>a</em> such that
-15 N = (20 kg) <em>a</em>
Solve for <em>a</em> :
<em>a</em> = - (15 N) / (20 kg) = -0.75 m/s²
The object's velocity <em>v</em> at time <em>t</em> is then given by
<em>v</em> = 3 m/s + (-0.75 m/s²) <em>t</em>
so the time it takes for the object to slow to a rest is
0 = 3 m/s + (-0.75 m/s²) <em>t</em>
<em>t</em> = (3 m/s) / (0.75 m/s²) = 4.0 s
