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mihalych1998 [28]

3 years ago

5

a 455n gymnast jumps upward a distance of 1.5 meters to reach the uneven parallel bars. how much work did she do before she even

began her routine

1 answer:

Taya2010 [7]3 years ago

3 0

Work = force x distance
Work = 455N (weight force) x 1.5m (distance)
Work done = 682.5J
Hope this helped!

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An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of

IrinaVladis [17]

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

$r_1$ = Distance between center of mass of person and pole = 0.45 m

$r_2$ = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

$\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s$

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

$L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s$

The new angular speed is 1.70344 rad/s

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3 years ago

Explain why the moon is always half illuminated and half dark no matter where it is in the lunar cycle.

BlackZzzverrR [31]

The easiest way I know to explain it is this:

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<em>Everything</em> in the solar system ... as long as it's shaped like a ball ... is
half illuminated by the sun and half dark.

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Answer:

Explanation:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

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3 years ago

As the object's particles move faster, the thermal energy increases, causing the _________ to also increase. *

ryzh [129]

Temperature that will be my answer number 1

7 0

3 years ago

Read 2 more answers

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(c) amax = 1.32m/s^2

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

3 years ago