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WARRIOR [948]
2 years ago
14

Work/Power problem with friction

Physics
2 answers:
Kisachek [45]2 years ago
3 0

Answer:

This is the formula

Explanation:

−(0.54 m/s)2

2(−1.67 m/s2)

so this is the step by taking the mean of the maximum value of friction with the minimum value you can get the average force of friction

cestrela7 [59]2 years ago
3 0

Answer:

Explanation & Steps:

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A student starts a food fight by throwing a 0.5 kg burrito at some girl he likes. He throws it kind-of hard so he accelerates it
Elenna [48]

Explanation:

m = mass of burrito thrown by the student = 0.5 kg

a = acceleration of the burrito thrown by the student = 3 m/s²

F = force applied by the student on the burrito = ?

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inserting the values

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F = 1.5 N

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2 years ago
Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
slava [35]

Answer:

69.68 N

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Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

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we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

7 0
3 years ago
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Answer:

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Neko [114]

Answer:

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Mechanical advantage = de/dr

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Mechanical advantage = 8/2

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5 0
2 years ago
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