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Anon25 [30]
2 years ago
9

By how much does the gravitational potential energy of a 54-kg pole vaulter change if her center of mass rises about 4.0 m durin

g the jump
Physics
2 answers:
kodGreya [7K]2 years ago
8 0
<span>the gravational potential energy of anything on the ground is zero. When calculating potential energy you take height in meters and multiply it by the mass of the object in kilograms and the acceleration of gravity to get a new unit called Joules. Any object at ground level has a potential energy of zero newtons becuase anything multiplied by zero is zero. An object with mass of 54 kg, 4 meters above the ground has a gravitatinal potential energy of 2116.8 Joules.</span>
dexar [7]2 years ago
7 0

Answer:

The gravitational potential energy of pole voulter is 2160\ J.

Explanation:

Given that,

Mass of the pole vaulter, m = 54 kg

Height it rises, h = 4 m

To find,

The gravitational potential energy.

Solution,

We know , gravitational potential energy of an object of mass m and height h is given by m\times g\times h.Here g is acceleration due to gravity=10 m/s^2.

E=m\times g\times h\\=54\times 10 \times 4\ J=2160\ J.

Therefore, the gravitational potential energy of pole voulter is 2160\ J.

Hence, this is the required solution.

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A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

\dfrac{dP}{dt}=794.11\ N

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
2 years ago
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
Serggg [28]

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

L=I\omega

where

I is the moment of inertia

\omega is the angular speed

Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

\omega_1 = 5.00 rad/s is the initial angular speed

I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

7 0
2 years ago
Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th
puteri [66]
Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.
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2 years ago
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1. Confident in Reasoning: Douglas is truthful of his reasoning skills to yield good judgments.

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A 21200 kg sailboat experiences an eastward force 42700 N due to the tide pushing its hull while the wind pushes the sails with
My name is Ann [436]

Answer:

2.95 m/s^{2}

Explanation:

The resultant force F

F=\sqrt {F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos135^{o}} where F_{1} is eastward force, F_{2} is force directed towards the North

F=\sqrt {42700^{2}+85000^{2}+(2*42700*85000)cos135^{o}}=62573.17217 N

F=62573.2 N

The magnitude of acceleration of sailboat is given by

a=\frac {F}{m}=\frac {62573.2}{21200}=2.95 m/s^{2}

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