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2 years ago

If a football player collides with a goal post, what forces are at work?

1 answer:

7 0

When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.

You might be interested in

Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin

Vedmedyk [2.9K]

Answer:

The correct answer is $2.8*10^{-5}ms^{-1}$

Explanation:

The formula for the electron drift speed is given as follows,

$u=I/nAq$

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is $8.93/63.5 molcm^{-3}$ . Converting this number to m³ using very elementary unit conversion we get $140384molm^{-3}$ . If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

$n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}$

if we convert the area from mm³ to m³ we get $A=80*10^{-6}m^{2}$ .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

$u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}$

which is our final answer.

6 0

3 years ago

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Where,

$m_1$ = mass of ball

$m_2$ = mass of the person

$u_1$ = Velocity of ball before collision

$u_2$ = Velocity of the person before collision

$v_1$ = velocity of ball afer collision

$v_2$ = velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

$v_1 = v_2$

$m_1u_1 + m_2u_2 = (m_1+m_2)v_2$

Re-arrenge to find $v_2$ ,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

Our values are,

$m_1$ = 0.425kg

$u_1$ = 12m/s

$m_2$ = 68.5kg

$u_2$ = 0m/s

Substituting,

$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

$v_2 = 0.074m/s$

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0

3 years ago

If a force of 12 N is applied to an object

Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{12}{4} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

It's not D

DanielleElmas [232]

A. it provides support

5 0

2 years ago

A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.50 m long. The block is initia

jeka94

Answer:

34.83 m/s

Explanation:

From the law of conservation of momentum,

initial momentum of bullet = final momentum of block + bullet

mv₀ = (m + M)V

V = mv₀/(m + M)

where m = mass of bullet = 0.0120 kg, v₀ = initial momentum of bullet, M = mass of block = 0.775 kg, V = final velocity of block + bullet.

Now, since the block + bullet rise a height of 0.725 m, from the law of conservation of energy,

potential energy change of block + bullet = kinetic energy change of block + bullet.

So (m + M)gh - 0 = -1/2(m + M)(V₁² - V²) where h = vertical height moved = 0.725 m and V₁ = velocity at 0.725 m and it has zero potential energy initially.

gh = -1/2(V₁² - V²) (2)

Now, we obtain V₁ from

F = (m + M)V₁²/R since a centripetal force acts on the block + bullet at height 0.725 m. F = tension in chord = 4.88 N and R = length of cord = 1.50 m.

V₁ = √[FR/(m + M)]

Substituting V and V₁ into (2) above, we get

gh = -1/2(FR/(m + M) - [mv₀/(m + M)]²)

-2(m + M)²gh = FR(m + M) - (mv₀)²

v₀ = √([FR(m + M) + 2(m + M)²gh]/m)

substituting the values of the variables into v₀ we have

v₀ = √([4.88 N × 1.50 m × (0.0120 kg + 0.775 kg) + 2(0.0120 kg + 0.775 kg)² × 9.8 m/s² × 0.725 m]/0.0120 kg)

= √([7.32 × 0.787 + 2(0.787)² × 9.8 m/s² × 0.725 m]/0.0120 kg)

= √(5.76 + 8.80)/0.012 kg

= √14.56/0.012

= √1213.40

= 34.83 m/s

So the initial speed v₀ = 34.83 m/s

7 0

3 years ago