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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.

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Which of the following is true about the mass of an object?

Butoxors [25]

I believe it is the first one

5 0

3 years ago

A(n) _____ is a gap in the geologic record where some rock layers have been lost because of erosion.

Elenna [48]

I'm not sure, I think it's option A.

Let me know if I'm wrong!

4 0

4 years ago

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0

4 years ago

Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini

natka813 [3]

E=energy=5.09x10^5J = 509KJ
M=mass=2250g=2.25Kg 
C=specific heat capacity of water= 4.18KJ/Kg 
ΔT= change in temp= ? 
E=mcΔT 
509=(2.25)x(4.18)xΔT 
509=9.405ΔT 
ΔT=509/9.405=54.1degrees 
Initial temp = 100-54 = 46 degrees 
Hope this helps :)

5 0

3 years ago

What is the relationship between the buildup of electons and the sudden flow of electrons from one charged object to another?

melisa1 [442]

Static electricity

hope this helps :)

5 0

4 years ago

Read 2 more answers