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Art [367]
3 years ago
15

A bicyclist of mass 90 kg drives around a circle with a centripetal acceleration

Physics
2 answers:
Serggg [28]3 years ago
5 0

given,

mass of bicyclist(m)=90Kg

centripetal acceleration(a)=1.5 m/s2

centripetal force(F)=ma= 90×1.5=145 N

Sati [7]3 years ago
5 0

Answer:

Force, F = 135 N

Explanation:

Given that,

Mass of the bicyclist, m = 90 kg

Centripetal acceleration of the bicyclist, a=1.5\ m/s^2

To find,

The centripetal force acting on the bicyclist.

Solution,

When an object moves in a circular track, it will under the action of centripetal acceleration and the exerted force is called centripetal force. It can be calculated as :

F=ma

F=90\ kg\times 1.5\ m/s^2

F = 135 N

So, the centripetal force acting on the bicyclist is 135 N.

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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

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= \frac{1}{2} * 4(3)^2

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Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

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2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

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8 0
3 years ago
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Vedmedyk [2.9K]

Answer:

Explanation:

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For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

v=v_0+at to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

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That's a and c. Now for b:

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V_x=25cos45 and

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e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):

Δx = vt so

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Δx = 65 meters.

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