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3 years ago

Throwing a football is an example of force because

Physics

1 answer:

siniylev [52]3 years ago

4 0

Answer:

i found this provided by the San Francisco 49ers (found on Khan Academy)

Explanation:

I HOPE THIS HELPS!!!

So, what is a force? A force is a push or a pull exerted on one object from another. Forces make things move. You can make something start or stop when you push or pull an object.

There are many different types of forces in action in football. A player kicking a football is a force that makes the football fly through the air. A quarterback throwing a football is another example of a force that makes the football fly in a game.

When studying the concept of force, we can look to history to find mathematical principles that guide the laws of motion. Sir Isaac Newton was one of the most famous scientists of the 17th century to study the laws of forces and motion. Through careful study of how objects react to various forces, Newton developed the Three Laws of Motion. Below are explanations of each law and how these laws can be applied to football.

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At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 23 kn

Arturiano [62]

Answer:

30.66 knots

Explanation:

Distance of ship A from B at noon = 50 NM

$\frac{da}{dt}$ = Velocity of ship A = 22 knots = 22 NM/h

$\frac{db}{dt}$ = Velocity of ship B = 23 knots = 23 NM/h

Distance travelled by ship A from noon to 3 PM = 22×3 = 66 NM

a = Total distance travelled by ship A = 50+66 = 116 NM

b = Total distance travelled by ship B till 3 PM = 23×3 = 69 NM

c = Distance between Ship A and B at 3 PM = √(116²+69²) = 134.97 NM

a²+b² = c² (Pythagoras theorem)

Now differentiating with respect to time

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{116\times 22+69\times 23}{134.97}=\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=30.66\ NM/h$

∴ The velocity with which the distance is changing at 3 PM (3 hours later) is 30.66 knots

5 0

3 years ago

A trouble-making youth is standing on a bridge, and wants to drop a water balloon on an unsuspecting passerby. A man is jogging

DIA [1.3K]

Answer:

Explanation:

Time needed for a balloon to drop from vertical rest a distance of 11.6 m

t = √(2h/g) = √(2(11.6)/9.8) = 1.538618

d = vt = 4.2(1.538618) = 6.462197...

d = 6.5 m

8 0

3 years ago

If the pitch of the note becomes higher

VashaNatasha [74]

That tells us that the frequency of the sound wave increased, the period decreased, and the wavelength decreased. The guitar player may be twanging a higher string, OR he may be playing the same note but started walking toward us.

3 0

3 years ago

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

3 years ago

What is the work done by the friction when the body slides against a rough horizontal surfaces?

katovenus [111]

Negative work

Hope this helps :)

4 0

2 years ago