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grandymaker [24]
2 years ago
13

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80 m

/s^2 . Ignore air resistance. A flea jumps straight up to a maximum height of 0.490 m. What is its initial velocity v0 as it leaves the ground?
Physics
1 answer:
leonid [27]2 years ago
4 0

The initial velocity used by the flea to jump straight upward is 3.1 m/s

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) = 0.49 m
  • Final velocity (v) = 0 m/s (at maximum height)
  • Initial velocity (u) =?

<h3>How to determine the initial velocity</h3>

The initial velocity used by the flea to jump upward can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = u² – (2 × 9.8 × 0.49)

0 = u² – 9.604

Collect like terms

u² = 0 + 9.604

u² = 9.604

Take the square root of both sides

u = √9.604

u = 3.1 m/s

Thus, the initial velocity is 3.1 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

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A bus initially moving at 20 m/s with an acceleration of -4m/s² for 5
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Answer:

50m; 0m/s.

Explanation:

Given the following data;

Initial velocity = 20m/s

Acceleration, a = - 4m/s²

Time, t = 5secs

To find the displacement, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Substituting into the equation, we have;

S =20*5 + \frac{1}{2}*(-4)*5^{2}

S =100 + (-2)*25

S =100 - 50

S = 50m

Next, to find the final velocity, we would use the third equation of motion;

V^{2} = U^{2} + 2aS

Where;

  • V represents the final velocity measured in meter per seconds.
  • U represents the initial velocity measured in meter per seconds.
  • a represents acceleration measured in meters per seconds square.

<em>Substituting into the equation, we have;</em>

V^{2} = 20^{2} + 2(-4)*50

V^{2} = 400 - 400

V^{2} = 0

V = 0m/s

<em>Therefore, the displacement of the bus is 50m and its final velocity is 0m/s.</em>

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