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Flauer [41]
2 years ago
12

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. Th

e total mechanical energy is 25 J. The maximum speed of the block is:
Physics
1 answer:
wolverine [178]2 years ago
6 0

Answer:

Explanation:

easy way

when system is all kinetic energy, velocity is at a maximum

E = ½mv²

v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s

harder way

ω = √(k/m) = √(80/0.5) = √160 rad/s

When the system is entirely spring potential, the amplitude A is

E = ½kA²

A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m

maximum velocity is ωΑ = 0.79√160 = 10 m/s

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The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?
timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The <u>Drag Force</u> equation is:

F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2} (1)

Where:

F_{D} is the Drag Force

C_{D} is the Drag coefficient, which depends on the material

\rho is the density of the fluid where the bicycle is moving (<u>air in this case) </u>

A_{D} is the transversal area of the body or object

V the bicycle's velocity

Now, if we assume C_{D}, \rho and A_{D} do not change, we can rewrite (1) as:

F_{D}=C.V^{2} (2)

Where C groups all these coefficients.

So, if we have a new velocity V_{n} , which is the double of the former velocity:

V_{n}=2V (3)

Equation (2) is written as:

F_{D}=C.V_{n}^{2}=C.(2V)^{2}

F_{D}=4CV^{2} (4)

Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>

7 0
3 years ago
A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball
RideAnS [48]

Answer:

C=2.32\times 10^{-4}\ Ns^2/m^2

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, F_r=Cv^2

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

Cv^2=mg

C=\dfrac{mg}{v^2}

C=\dfrac{0.046\times 9.8}{(44)^2}

C=2.32\times 10^{-4}\ Ns^2/m^2

Hence, this is the required solution.

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