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3 years ago

How does exercise help you body devolpe

1 answer:

8 0

An exercise strains the body and in response, your body heals/repairs itself and makes it better

example:
pushups strain the mucles in your arms and chest.
after the exercise, they might be sore. that's because they are damaged and need to heal. when they heal, they will be stronger and more durable

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You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-

Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

$E_P=\int dE \cos$

$E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}$

$\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ$

$\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}$

If we put an electron on point P, then force on point e is :

$\vec{F}=-|e|\vec{E_P}$

$F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}$

If r >> x , then $\frac{x^2}{r^2} \approx 0$

Then, $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0

2 years ago

What is the acceleration of an object with mass of 42.6 kg when an unbalanced force of 112 N is applied to it

lord [1]

The answer is 2.63m/s^2! You use the formula F=ma, 112 = 42.6(a), a= 2.63m/s^2.

7 0

3 years ago

Read 2 more answers

How much work is done when you push a crate horizontally with 150 N across a 13-m floor

Mila [183]

Work = (force) x (distance)

When a force of 150 N pushes through a distance of 13 meters,
it does

Work = (150 N) x (13 m) = 1,950 joules .

6 0

3 years ago

Buoyancy is a force that always acts in an

Margarita [4]

Buoyancy is a force that always acts in an upward direction exerted by a fluid on a body placed in the fluid

Hope this helps :)

7 0

2 years ago

Read 2 more answers

How much force is required to move a sled 5 meters if a person uses 60 j of work? 300N 12N 65N 30N

maksim [4K]

The answer is b 12N because
We know that
W = F × d × c o s(θ)
assuming theta=0 we then solve and have
F=W/d
substitute known values to get:
F=60J/5m=12N

4 0

2 years ago