All categories

3 years ago

Give an example of the follow components of culture in your life and explain the role it plays.

Physics

2 answers:

kkurt [141]3 years ago

6 0

I believe in god and I go to church

lbvjy [14]3 years ago

3 0

I'm learning Arabic and so I'm able to communicate with others that also speak Arabic.

You might be interested in

O que levou a filosofia pensar em artes?

vredina [299]

Answer:

idk

Explanation:

4 0

3 years ago

A 0.35 m2 coil with 50 turns rotates at 5 radians per sec2 in a magnetic field of 0.6 Tesla. What is the value of the rms curren

soldier1979 [14.2K]

Answer:

11.25 amps

Explanation:

For transformers, the magnetic flux

$\Phi _{max} = \beta \times A$

Therefore;

$\Phi _{max} = 0.6 \times 0.35 = 0.21 \ Weber$

Ф = Фmax (cosωt) = 0.21·(cos(5·t))

From Faraday's law of induction, we have;

ε = -N × dΦ/dt

Which gives;

dΦ/dt = -1.05(sin (5t) )

ε = -N × dΦ/dt = -50× -1.05(sin (5t) )

ε = 52.5(sin (5t) )

I = ε/R = 52.5(sin (5t) )/3.3 = 15.9091(sin (5t) ) amps

The peak current is therefore = 15.9091 amps

The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.

5 0

3 years ago

How long does a lunar eclipse last? A)seconds B)Minutes C)Days D) Weeks

bonufazy [111]

Well, it happens a few weeks ahead, then for a total of 3 hours and 40 minutes.

3 0

3 years ago

Why can beta particles pass through materials more easily than alpha particles? (2 points)

never [62]

Beta particles are smaller

5 0

3 years ago

Read 2 more answers

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

2 years ago