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svp [43]
2 years ago
9

An athlete wants to refill an old punching bag used for training, as shown in the picture. When refilled, the bag will rip if it

is kicked with too much force

Physics
1 answer:
In-s [12.5K]2 years ago
5 0

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states action and reaction are equal and opposite. That is the force applied to an object is equal in magnitude to force experienced by the object but in opposite direction.

From the given question, when the bag exert a certain on the athlete, the athlete also exerts similar force to the bag but in opposite direction.

Thus, the complete sentence is as follows;

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

#SPJ1

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A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20
alexandr402 [8]

Answer:

the amplitude of  the subsequent oscillations is 0.11  m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, m_{1} = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, m_{2} = 200 g = 0.2 kg

the speed of glider,  v_{2} = 170 cm/s = 1.7 m/s

the amplitude of  the subsequent oscillations is A = 0.11  m

according to mechanical enery equation, we have

A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}

where

A is the amplitude and  v_{f} is the final speed.

to find v_{f}, we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

P_{f} = P_{i}

(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}

v_{1} = 0, thus

(0.75+0.2)v_{f} = (0.75)(0)+(0.2)(1.7)

0.95 v_{f} = 0.34

v_{f} = 0.36 m/s

Now we can calculate the amplitude

A = \sqrt{\frac{0.75 +0.2 }{10} }0.36

A = 0.11  m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π\sqrt{\frac{m_{1}+m_{2}  }{k} }

T = 2π\sqrt{\frac{0.75+0.2  }{10} }

T = 1.94 s

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