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3 years ago

Where are the physics in football

Physics

1 answer:

sasho [114]3 years ago

5 0

Answer:

The branch of physics that is most relevant to football is mechanics, the study of motion and its causes.

Explanation:

When the ball leaves the punter's foot, it is moving with a given velocity (speed plus angle of direction) depending upon the force with which he kicks the ball. The ball moves in two directions, horizontally and vertically. Because the ball was launched at an angle, the velocity is divided into two pieces: a horizontal component and a vertical component.

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If the range of a projectile's trajectory is ten times larger than the height of the trajectory, then what was the angle of laun

Lunna [17]

Answer:

Angle of projection from the horizontal will be $21.80^{\circ}$

Explanation:

We have given that range of the projectile is 10 times the height of the projectile

Let the projectile is projected with velocity u at an angle $\Theta$

Range of the projectile is equal to $R=\frac{u^2sin2\Theta }{g}$

And height of the projectile is equal to $h=\frac{u^2sin^2\Theta }{2g}$

Now according to question range is 10 times of height

So $\frac{u^2sin2\Theta }{g}=10\times \frac{u^2sin^2\Theta }{2g}$

$sin2\Theta =5sin^2\Theta$

$2sin\Theta cos\Theta =5sin^2\Theta$

$tan\Theta =\frac{2}{5}=0.4$

$\Theta =tan^{-1}0.4=21.80^{\circ}$

So angle of projection from the horizontal will be $21.80^{\circ}$

3 0

3 years ago

As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, b

katovenus [111]

The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

<h3>What do you mean by sound radiates?</h3>

Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.

(Lw) = 10·log (W/Wo) dB

Given:

sound level, $\beta= 101 dB$

Area, A = $22\;m^{2}$

Time, $\triangle t = 20\;min=1200\;s$

Intensity, $I=1\times 10^{-12}\;W/m^{2}$

$r=1\;km=1000\;m$

(a)

We know that, Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Solving the above equation for sound intensity,

$I=I_{o} \times 10^{\frac{\beta}{10} }$

$I=1 \times 10^{-12} \times 10^{\frac{101}{10} }$

$I=0.0126\;W/m^{2}$

Therefore, The sound energy is,

$E=P\times \triangle t$

Substitute $P=I \times A$ in the above equation,

$E=I \times A \times \triangle t$

$E=0.0126 \times 22 \times 1200$

$E=332.6\;J$

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

$A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi$

Substitute the known value in the above equation ,

$A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi$

$A_{hemisphere} = 6283185\;m^{2}$

Sound Intensity is,

$I = \frac{P}{A_{hemisphere}}$

Substitute $P=I \times A$ in the above equation,

$I = \frac{I \times A}{A_{hemisphere}}$

Substitute the known value in the above equation,

$I = \frac{0.0126 \times 22}{6283185}$

$I = 4.4 \times 10^{-8}\;W/m^{2}$

Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Substitute the known value in the above equation,

$\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )$

$\beta=46.4\;dB$

To learn more about sound radiates, Visit:

brainly.com/question/20360072

#SPJ4

8 0

1 year ago

What can electricty from solar power be used for

Mandarinka [93]

Regular electrical needs such as heat, a/c, appliances, anything that would be used regularly with electricity. Most electrical companies use gas, coal, and other nonrenewable resources to power houses. Solar power is just a renewable way of gathering power from the sun and converting it to electricity.

7 0

3 years ago

A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of

Blizzard [7]

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

$F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T$

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

7 0

2 years ago

A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.

dlinn [17]

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

Force= 9.80665 x 0.5kg
Force= 4.90332 Newtons

Distance changed= 5-3
distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

5 0

2 years ago