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prisoha [69]
3 years ago
10

Imagine a player throwing a football. Explain the action of what is happening to the hand and ball, as well as describe the acti

on force and the reaction force
Physics
1 answer:
garri49 [273]3 years ago
7 0

Answer:

Explanation:

When a player throws a football , he applies force on the ball , due to which ball accelerates in forward direction . This is the action force .

The ball also applies a force on the hand of the player . Due to it the player experiences a pushing force in backward direction . This is reaction force .

Due to this force , the player does not move in backward direction because of frictional force of the ground . The frictional force keeps him stationary .

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3 years ago
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A 9 cm tall object is placed 16 cm from a converging lens forming an image at 13 cm. What is the height of the image?
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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
4 years ago
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0
3 years ago
Dr. Suwary runs really fast. She is in a race and is running at 50 m/s. Dr. Suwary's enemy is
Ivenika [448]

Answer: 120 m/s?

Explanation:

3 0
3 years ago
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