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Sloan [31]
2 years ago
15

A 50 kg athlete running at speed v grabs a light rope that hangs from a 10-meter-high platform and swings to a maximum of 1.8 m

above the ground. later, a 100 kg athlete, running at the same speed, grabs a similar rope hanging from a 5-meter-high platform. what is the maximum height to which the 100 kg athlete swings?
Physics
1 answer:
balu736 [363]2 years ago
4 0
The answer is: 1.8 meters.

Explanation:
An athlete swinging can be considered a pendulum.

The pendulum's maximum height is the point at which it changes direction, which means that its velocity is equal to zero. In this point, for the mechanical energy conservation, all its kinetic energy is transformed into potential energy. Similarly, when the pendulum is at its resting position (when the athlete grabs the rope), its energy is totally kinetic.

Therefore we can say that:
\frac{1}{2}m v_{1} ^{2}  = mgh_{max}

Solving for h:
h_{max} =  \frac{v^{2}}{g}

As we can see, the maximum height is independent on the mass and on the length of the rope, therefore it will be the same for the 100kg-athlete as it is for the 50kg-athlete, since their initial speeds are the same.

We know that the <span>50kg-athlete reached a height of 1.8 m, h</span>ence, the maximum height reached by the 100kg-athlete will be 1.8 m.
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A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
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Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

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A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta
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Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s

So, the final velocity of the bullet is 9 m/s.

5 0
2 years ago
A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the
antoniya [11.8K]
Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

6 0
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