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3 years ago

Why do you think football players are usually large in size?

2 answers:

8 0

Football players in the NFL are large in size due to their eating habits since as players they have to work out constantly to stay and shape and eat to gain weight and be strong, so its important for them to be strong against their opponents in gameplay. Football players sometimes take steroids which make their muscles stronger and harder, and also gives them a growth spurt but the effects are worse and makes them sicker, high blood pressure, and even more.

MA_775_DIABLO [31]3 years ago

5 0

I think practically all the football players you ever see are huge, because by the time you see them, they have already destroyed all the small ones.

You might be interested in

Suppose you exert a 25-N force to lift a ball 0.4 m in 2 s. How much work is done?

Sergio [31]

work is force x distance = 25 x 0.4

= 2.5x4 = 10joules

pwer would be 10j/2s watts .... 5 watts

3 0

3 years ago

You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac

zaharov [31]

Answer:

Explanation:

Given

Original Frequency $f=723\ Hz$

apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v_s^2-0=2\times 9.8\times s$

$s=\frac{12.8^2}{2\times 9.8}$

$s=8.35\ m$

5 0

3 years ago

The engine in the car from question 1 uses a force of 1200 N to cause the car to accelerate at 3.5 m/s2. What is the car's mass?

zalisa [80]

Answer:

Mass of car's engine = 342.85 kg (Approx.)

Explanation:

Given values:

Force applied by car = 1,200 N

Acceleration of car' engine = 3.5 m/s²

Find:

Mass of car's engine

Computation:

⇒ Mass = Force / Acceleration

⇒ Mass of car's engine = Force applied by car / Acceleration of car

⇒ Mass of car's engine = 1,200 / 3.5

⇒ Mass of car's engine = 342.85 kg (Approx.)

7 0

2 years ago

7) A long straight wire of radius 1.00x10-3 m (1 mm) carries uniform current density J such that the magnetic field at the edge

Vedmedyk [2.9K]

Answer:A)0.5 and 2 mm

Explanation:

Given

radius R of wire is 1 mm

magnetic Field at edge(surface) $=10^{-5} T$

Magnetic Field at a distance r' from Center is $B'=5\times 10^{-6} T$

and we know

$B=\frac{\mu _0I}{2\pi r}$

where $I= current$

$\mu _o=$ Permeability of free space

$r=$ distance from center

For $r=R$

$10^{-5}=\frac{\mu _0I}{2\pi R}$ ---1

For $r=r'$

$5\times 10^{-6}=\frac{\mu _0I}{2\pi r'}$ ---2

Divide 1 and 2

$\frac{10^{-5}}{0.5\times 10^{-5}}=\frac{r'}{R}$

$r'=2 R=2 mm$

If r is inside the wire then

$B=\frac{\mu _0rI}{2\pi R^2}$

for $r=R$

$10^{-5}=\frac{\mu _0R\cdot I}{2\pi R^2}$ ---3

for $r=r"$

$5\times 10^{-6}=\frac{\mu _0r"I}{2\pi R^2}$ ----4

Divide 3 and 4

$2=\frac{R}{r"}$

$r"=0.5 mm$

3 0

2 years ago

A spring stretches from an initial height of 5 cm to a final stretch of 10 cm. the spring constant is 800 n/m.How much work was

Aleks04 [339]

Answer:

Explanation:

F = kx so k = 800/((10-5)/100) = 16000 N/m

W = 1/2 kx^2 = 1/2 * 16000 * .05^2 = 20 J.

(sorry if it's wrong)

3 0

2 years ago