All categories

3 years ago

Describe some precautions needed to increase the safety of participants in the sports of football.

1 answer:

8 0

Football is a popular ball game and so many people play football these days. Been a contact game, there is no way players will not sustain injuries, but this can be reduced to the lowest point by taking some precautions.
The precautions that could be taken include the following:
Wear the right shoes.
Wear shin guard and soccer socks.
Use the right and properly fitted equipment [helmets, shoulder pads, googles, mouth-guards, etc]. All these equipment should be made for the soccer sport.

You might be interested in

A 19.0-g bullet embeds itself in a 8.0-kg wooden block, which rests on a horizontal frictionless surface and is attached to a ho

Anton [14]

Answer:

d) 700 m/s

Explanation:

if k is the force constant and x is the maximum compression distance, then:

the potential energy the spring can acquire is given by:

U = 1/2×k×(x^2)

and, the kinetic energy system is given by:

K = 1/2×m×(v^2)

if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:

initial energy = final energy

Ki +Ui = Kf +Uf

Ui = 0 J and Kf = 0J

Ki = Uf

1/2×m×(v^2) = 1/2×k×(x^2)

m×(v^2) = k×(x^2)

v^2 = k×(x^2)/m

= (500)×((21×10^-2)^2)/(19×10^-3 + 8)

= 2.75

v = 1.66 m/s

the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:

m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.

m1×v1 = v×(m1 + M)

v1 = v×(m + M)/(m1)

= (1.66)×(19×10^-3 + 8)/(19×10^-3)

= 699.86 m/s

≈ 700 m/s

Therefore, the velocity of the bullet just before it hits the block is 700 m/s.

5 0

3 years ago

Un gallo se pavonea en línea recta en el techo. Su movimiento se muestra en la siguiente gráfica de la posición horizontal, x, c

Svetllana [295]

Answer:

Entonces, velocidad instantánea = | v | = 0,25 m / s

Explanation:

pregunta completa

Por favor, encontrar el archivo adjunto

Solución

La velocidad instantánea está dada por,

v = dx / dt = pendiente de la curva x-t

aquí, del gráfico:

en x = 1 m ---> t = 8 s

en x = 3 m ---> t = 0 s

entonces, dx = 1-3 = -2 m

dt = 8-0 = 8 s

Entonces,

v = -2/8

v = -0,25 m / s

Dado que, pseed es una cantidad escalar.

Entonces, velocidad instantánea = | v | = 0,25 m / s

8 0

3 years ago

A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of

zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(d) a = - 80 m/s²

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

α = - 1000 rad/s²

Negative sign represents deceleration.

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

θ = 3581 rotations

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

L = 1800 m

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

a = - 80 m/s²

4 0

3 years ago

Spaceship 1 and spaceship 2 have equal masses of 300 kg. spaceship 1 has an initial momentum magnitude of 600 kg-m/s. what is it

pickupchik [31]

2 m/s (apex)

I hope this was right

5 0

3 years ago

Read 2 more answers

Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d

lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

$d = d_1 + d_2 + d_3$

$d = 7 km + 5 km + 1 km$

$d = 13 km$

Now in order to find the displacement we will show all with their directions

$\vec d_1 = 7 + 5 = 12 km$ towards North

$\vec d_2 = 1 km$ towards East

So total displacement is

$\vec d = \vec d_1 + \vec d_2$

$\vec d = 12 \hat j + 1 \hat i$

so net displacement will be

$d = \sqrt{12^2 + 1^2} = 12.04 km$

so displacement is 12.04 km

6 0

4 years ago